Now a normal O2 molecule has 12 electrons in the valence shell so, O.pdf
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Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2 Solution Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2.
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Recall Each Oxygen has 4 Electrons in its 2p bonding orbital. But .pdf
Recall Each Oxygen has 4 Electrons in its 2p bonding orbital.
But Since it is 02- Total e- = 4(2)+1=9
Put 2 electrons in 2p, 4 in 2p with a remaining 3 electrons to be placed.
By Aufbau principal, the next 3 electrons must go in the lowest energy state which would be
2p*.
3/4 electrons fill the 2p* leaving 1 unpaired.
Bond order=(Bonding Electrons - Antibonding)/2
=(6-3)/2
B.O.=1.5
Therefore, we can conclude that O2- is relativly stable as it has 1 complete bond. It also has one
unpaired electron in 2p* antibonding orbital which gives it paramagnetic properties.
2s and 1 s orbitals both have 2 electrons in bonding and antibonding orbitals if you draw them
out, canceling out any bonds they might create. Only 9 electrons play a significant role in the
bonding of O2-
Solution
Recall Each Oxygen has 4 Electrons in its 2p bonding orbital.
But Since it is 02- Total e- = 4(2)+1=9
Put 2 electrons in 2p, 4 in 2p with a remaining 3 electrons to be placed.
By Aufbau principal, the next 3 electrons must go in the lowest energy state which would be
2p*.
3/4 electrons fill the 2p* leaving 1 unpaired.
Bond order=(Bonding Electrons - Antibonding)/2
=(6-3)/2
B.O.=1.5
Therefore, we can conclude that O2- is relativly stable as it has 1 complete bond. It also has one
unpaired electron in 2p* antibonding orbital which gives it paramagnetic properties.
2s and 1 s orbitals both have 2 electrons in bonding and antibonding orbitals if you draw them
out, canceling out any bonds they might create. Only 9 electrons play a significant role in the
bonding of O2-.
How To Write Electron Configurations
Step by step instructions for writing electron conficurations with examples
QUESTION 4 Manganese has an elec tron configuration 1s 2s 2p 3s 3p 4s.docx
QUESTION 4 Manganese has an elec tron configuration 1s 2s 2p 3s 3p 4s 3d3. Which electrons are considered valence electrons? 4s 3d O 3s23p 63d 3d 4s QUESTION 5 lling diagrams below correctly represents the electron configuration for the valence electrons of oxygen, O? 2s 2p 2s 2p 3s 3p 2s 2p 3s 3p
Solution
4)
valence shell is outermost shell
Here outermiost shell is n=4
SO, 4s2 is valance shell
Answer: 4s2
5)
Electronic configuration is 1s2 2s2 2p4
Noble gas having lesser electron than O and closest to it is Helium (He)
Number of electron in He = 2
electronic configuration of He is: 1s2
So, electronic configuration of O in noble gas form is: [He] 2s2 2p4
So, it should have 2 electrons in 2s and 4 electrons in 2p
Remember pairing of electrons do not start until all orbitals are singly occupied
So, option 4 is wrong
Answer: 1st option
.
Recommended
Recall Each Oxygen has 4 Electrons in its 2p bonding orbital. But .pdf
Recall Each Oxygen has 4 Electrons in its 2p bonding orbital.
But Since it is 02- Total e- = 4(2)+1=9
Put 2 electrons in 2p, 4 in 2p with a remaining 3 electrons to be placed.
By Aufbau principal, the next 3 electrons must go in the lowest energy state which would be
2p*.
3/4 electrons fill the 2p* leaving 1 unpaired.
Bond order=(Bonding Electrons - Antibonding)/2
=(6-3)/2
B.O.=1.5
Therefore, we can conclude that O2- is relativly stable as it has 1 complete bond. It also has one
unpaired electron in 2p* antibonding orbital which gives it paramagnetic properties.
2s and 1 s orbitals both have 2 electrons in bonding and antibonding orbitals if you draw them
out, canceling out any bonds they might create. Only 9 electrons play a significant role in the
bonding of O2-
Solution
Recall Each Oxygen has 4 Electrons in its 2p bonding orbital.
But Since it is 02- Total e- = 4(2)+1=9
Put 2 electrons in 2p, 4 in 2p with a remaining 3 electrons to be placed.
By Aufbau principal, the next 3 electrons must go in the lowest energy state which would be
2p*.
3/4 electrons fill the 2p* leaving 1 unpaired.
Bond order=(Bonding Electrons - Antibonding)/2
=(6-3)/2
B.O.=1.5
Therefore, we can conclude that O2- is relativly stable as it has 1 complete bond. It also has one
unpaired electron in 2p* antibonding orbital which gives it paramagnetic properties.
2s and 1 s orbitals both have 2 electrons in bonding and antibonding orbitals if you draw them
out, canceling out any bonds they might create. Only 9 electrons play a significant role in the
bonding of O2-.
How To Write Electron Configurations
Step by step instructions for writing electron conficurations with examples
QUESTION 4 Manganese has an elec tron configuration 1s 2s 2p 3s 3p 4s.docx
QUESTION 4 Manganese has an elec tron configuration 1s 2s 2p 3s 3p 4s 3d3. Which electrons are considered valence electrons? 4s 3d O 3s23p 63d 3d 4s QUESTION 5 lling diagrams below correctly represents the electron configuration for the valence electrons of oxygen, O? 2s 2p 2s 2p 3s 3p 2s 2p 3s 3p
Solution
4)
valence shell is outermost shell
Here outermiost shell is n=4
SO, 4s2 is valance shell
Answer: 4s2
5)
Electronic configuration is 1s2 2s2 2p4
Noble gas having lesser electron than O and closest to it is Helium (He)
Number of electron in He = 2
electronic configuration of He is: 1s2
So, electronic configuration of O in noble gas form is: [He] 2s2 2p4
So, it should have 2 electrons in 2s and 4 electrons in 2p
Remember pairing of electrons do not start until all orbitals are singly occupied
So, option 4 is wrong
Answer: 1st option
.
Lesson 8Chemical Bonding & NomenclatureYou should view all.docx
Lesson 8
Chemical Bonding & Nomenclature
You should view all lectures as a slide show. In the toolbar above, click “Slide Show” and “From Beginning” to start. Hit the space bar to forward to next slide or item.
1
*Read: Watch the Lecture as a slideshow, hit the space bar to move forward or just left click.
1
Topics:
Chemical Bonds
Lewis Dot Structures
Molecular Geometry
Polar Bonds
Naming Compounds
Writing Chemical Formulas
Naming Molecular Compounds
Lesson 8: Chemical Bonding and Nomenclature
2
2
Chemical Bonds
Sodium (Na) is a silver-colored metal that reacts so
violently with water that flames are produced when sodium
gets wet.
Chlorine (Cl) is a greenish-colored gas that is so poisonous that it was used as a weapon in World War I.
When the electrons of sodium metal and chlorine gas interact, the compound sodium chloride (NaCl) is formed, which is table salt.
Since the body is over 90% water, ingesting sodium metal would literally set someone on fire!
If we make a compound out of Na, we can make something totally different such as the table salt used to season our food.
3
Chemical Bonds
Atoms can interact with each other to form new substances called
compounds.
Compounds are formed when electrons in an atom’s last energy level interact and form chemical bonds. A chemical bond is an attractive force between atoms that holds them together.
An atom’s outermost energy level of electrons is called the valence shell (or valence level) and the electrons in the valence shell are called valence electrons.
4
Na Atom
Cl Atom
Sodium has 1 valence electron in the valence shell.
Chlorine has 7 valence electron in the valence shell.
Chemical Bonds
In ionic bonds, metals always lose electrons to nonmetals and become positive (or cations).
In ionic bonds, nonmetals always attract electrons from metals and become negative (anions).
Ionic compounds are neutral compounds made up of cations and anions.
Covalent bonds are formed between nonmetals and electrons are shared so no ions are formed.
Cl
nonmetal
Na
metal
O
nonmetal
O
nonmetal
Metal + Nonmetal = Ionic Compound
Nonmetal + Nonmetal = Covalent Compound
Chemical Bonds: Ionic
The octet rule states that atoms will gain, lose, or share
valence electrons in a way that will give each atom eight electrons in their valence shell.
Na has 1 electron in its valence shell and Cl has 7 electrons in its valence shell. Cl needs 1 valence electron to have 8 in its outer shell.
If Na transfers its 1 valence electron to Cl, Na’s second energy level becomes the valence shell, which already has 8 electrons.
Now, Na has a positive 1 charge and Cl has a negative 1 charge. But, the charge on NaCl is zero (the charges cancel each other out).
6
Na Atom
Cl Atom
Ionic Bond
Chemical Bonds: Ionic
An Ionic bond is formed when there is a transfer of electrons from a metal to a nonmetal. Compounds formed by ionic bonds are ...
Electrons have places around the atom that they are most likely to b.pdf
Electrons have places around the atom that they are most likely to be. An electron configuration
is a way of describing these places, like an address for electrons. Each part of what you just
wrote out describes a place where electrons can be found in oxygen.
Let\'s take a look at just the first part: 1s2.
The \'1\' is the shell. The shell number is used to indicate how close to the nucleus the electrons
are. The first shell is the *closest* possible spot for the electrons - the higher this number
becomes, the farther away from the nucleus they go. The \'1\' shell MUST be get its electrons
before the \'2\' shell or \'3\' shell, and so on - it gets priority. With me so far? :)
The \'s\' is the subshell, which refers to the SHAPE of the area where the electron might be. \'S\'
stands for \'spherical\', like a beach ball. Imagine a sphere like that around the nucleus - electrons
in the 1s area can be anywhere on that shell.
Finally, we\'ve got the \'2\' of 1s2. This indicates the number of electrons present. ALL \'S\'
SHELLS CAN HOLD NO MORE THAN 2 ELECTRONS. EVER. So when you get to 2s? It
should have 2 electrons as well, when it\'s full. 3s? Should also have two electrons. ANYTHING
with an \'s\' should have up to two.
So, the fact that 1s comes before 2s comes before 2p? That must be memorized, unfortunately.
There is a handy chart to help you remember the correct order; you can find it here:
http://education.jlab.org/qa/electron_config.html.
Now we come onto how you do an electron configuration for a specific atom. Let\'s say your
teacher tells you to draw the electron configuration for oxygen. You look on the Periodic Table
(go ahead) and see that the number of electrons in an oxygen atom is eight. Now your job is to
use this code we\'ve talked about to describe where each of those eight electrons is in an oxygen
atom.
You have to start putting electrons in the \'1\' shell because you always fill the \'lowest\' (closest
to the nucleus) level first. So you have 1s2, two electrons in the first shell, darting around over a
sphere. That means you have two electrons so far. But oxygen has eight all together - the PT says
so. So we keep going. Again with the memorization, we know that 2s is next, and all \'s\'
subshells have up to two electrons. So we now have 1s2, 2s2. That\'s still only four electrons all
together. We have to keep going.
Next is 2p. \'p\' subshells can hold up to six electrons. BUT, if we do six... 1s2, 2s2, 2p6...
adding up our electrons gives us 10. Oxygen only needs eight electrons! So we need: 1s2, 2s2,
2p4. Now we have 8 electrons all together.
ALL atoms are done in exactly the same way. You figure out how many electrons they have,
then keep going with the electron configuration \'addresses\' until you have reached the right
number of electrons (and remember - the number of electrons is represented by the number
AFTER the letter; not before).
Try these: Do the electron configuration for helium and neon and chlorine using the.
Okay, so you know that the number of protons and electronsboth are e.pdf
Okay, so you know that the number of protons and electronsboth are equal to the atomic number
(on the periodic table, it\'sthe smaller number of the two seen when you look at elements...itis
also the number that holds the order to them in the periodictable of elements).
a. Fe2+Now, you look up theelement Fe and see that it normally has 26 protons and 26
electronsby looking it up on the table of elements and seeing its atomicnumber is \"26\" But we
also see that the atom has a charge of +2meaning two electrons (1election = 1 negative charge)
have beenremoved, SO you get 26 protons and 24 electrons.Since it is postively charged, the
questionasks us to find the most common oxide product or ratio of the givenelement and oxygen
(which has a -2 charge)...Fe is +2 and O is -2,Therefore you get FeO.
b. Fe3+26 protons, 23electrons Fe is +3 and O is -2,Therefore you get F2O3 to get a +6 and -6
tocancel out
c. Ba2+56 protons and 54electrons BaO
d. Cs+ 55 protons and54 electrons Cs2O
e. S2-For negative atoms,it\'s the same process but you have a negative net charge so youmust
have more electrons. 16 protons and 18 electrons.Al ion has a +3 charge...soAl2S3 +6 and -6
cancel each otherout
f. P3-15 protons and 18electrons AlP
g. Br-35 protons and 36electrons AlBr3
h. N3-7 protons and 10electrons AlN
When doing ion charges, the only number that changes is theelectrons because of the electron
clouds. The electron clouds arethe reason ions form ionic bonds - they all want the coveted
\"8electron outer shell\"...just like those noble gases.
When doing ion charges, the only number that changes is theelectrons because of the electron
clouds. The electron clouds arethe reason ions form ionic bonds - they all want the coveted
\"8electron outer shell\"...just like those noble gases.
Solution
Okay, so you know that the number of protons and electronsboth are equal to the atomic number
(on the periodic table, it\'sthe smaller number of the two seen when you look at elements...itis
also the number that holds the order to them in the periodictable of elements).
a. Fe2+Now, you look up theelement Fe and see that it normally has 26 protons and 26
electronsby looking it up on the table of elements and seeing its atomicnumber is \"26\" But we
also see that the atom has a charge of +2meaning two electrons (1election = 1 negative charge)
have beenremoved, SO you get 26 protons and 24 electrons.Since it is postively charged, the
questionasks us to find the most common oxide product or ratio of the givenelement and oxygen
(which has a -2 charge)...Fe is +2 and O is -2,Therefore you get FeO.
b. Fe3+26 protons, 23electrons Fe is +3 and O is -2,Therefore you get F2O3 to get a +6 and -6
tocancel out
c. Ba2+56 protons and 54electrons BaO
d. Cs+ 55 protons and54 electrons Cs2O
e. S2-For negative atoms,it\'s the same process but you have a negative net charge so youmust
have more electrons. 16 protons and 18 electrons.Al ion has a +3 charge...soAl2S3 +6 and -6
cancel each otherout
f. .
A Rapid and Innovative Method for Prediction of Spin Multiplicity and Spin St...
Determination of spin multiplicity is very important for identification of spin state of different
molecules or ions. But, the conventional method of prediction of spin multiplicity by using formula (2S+1) is
time consuming. So, in this manuscript, I tried to show a rapid and simple method for prediction of spin
multiplicity value without calculating the total spin quantum number(S=∑s). This method is only applicable for
homo and hetero nuclear diatomic molecules and ions having total electron (01-20).
D081112730
This is my published paper in IOSR JAC novemboe 2015 edition in the field of inorganic chemictry.By this method we can easily find the spin multiplicity value of any diatomic molecules having total electron no 1-20 ,without calculation total spin quantum no.This is most easy way to calculate spin multiplicity value and also time saving method with simple concept.
Simple Thinking Makes Chemistry Metabolic and Interesting - A Review Article
First of all I give my love to all the students who reside in ‘Chemistry Phobia zone’. In this short article I shall try to share my simple thinking that I achieved during my last 16 years of journey in the route of Chemical Education. Before starting my journey it was a big encounter for me to change the conventional methods and makes chemistry easier and interesting to such student who belongs to Chemistry Paranoia Zone. I believe that students are just like flowers and it is our duty to nourish them properly as a gardener. In my infinitely small area of knowledge, I just try to innovate fourteen (14) teaching methodologies by including thirty four (34) completely new formulae in the chemistry world which were followed by different eminent writers in their books since last 90-95 years namely Sir G.Wilkinson, Prof. F. Albert Cotton, Prof. I.L.Finar, Prof. R.T. Morrison, Prof. R.N. Boyd, Prof. Solomons and Prof. Fryhle, Prof. J.G. Smith etc..
1) MODERN TIMES is a very well prepared movie that describes the mis.pdf
1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of
humar resource during the period of industrialization era or machine age worlwide. It is a mix of
social drama and comedy. The content is presented in a satire. How did people survived the great
depression of the world is being documented here. The conditions werre miserable like acute
shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic
representation of what happens to a human being caught up in the mechanical bureaucracy of the
intellectual state.
2) Economically it could be understood that what miser situations could depression bring up.
Food and jobs comes hard in hand. To which a labor is convict to do under skill jobs as well for
survival. At the time of the great depression Americans had to fight for what little they had, and
as the movie showed many people went hungry. Modern Times looks toward the future, but not
with enthusiasm. Often described as a satire of the machine age, It has in fact a broader theme:
the dehumanizing effects of many aspects of modernity, including industrialization, bureaucracy,
urbanization, and law enforcement.
3) From the movie a prospect employee should not forget about the social ills and failings of
own day even if present age is seems far better than that was shown in the film. A dilbert-like
early scene shows the Tramp trying to take a break during work hours in the washroom — only
to have a big-screen image of the boss’s head suddenly appear on the washroom wall and order
him to quit stalling and get back to work. Modern audiences watching this scene may reflect
ruefully on the everyday reality of electronic surveillance in the workplace. Now, obviously
there’s nothing new about widespread poverty and want, but unemployment as we know it today
is a comparatively recent development — a byproduct of the modern workforce economy that
requires everyone to \"get a job.\"
4) The major ethical dilemma consists whether out of povertty stealing is right or wrong.
However this movie glorifies the struggle of the poor rather than vilifying them for stealing food
to survive. By glorifying the struggle the movie takes on yet another social issue: when the poor
steal to eat it is not criminal, just survival. Minor ones include
Solution
1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of
humar resource during the period of industrialization era or machine age worlwide. It is a mix of
social drama and comedy. The content is presented in a satire. How did people survived the great
depression of the world is being documented here. The conditions werre miserable like acute
shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic
representation of what happens to a human being caught up in the mechanical bureaucracy of the
intellectual state.
2) Economically it could be understood that what miser situations could depress.
1. Identify five differences between DNA replication and gene transc.pdf
1. Identify five differences between DNA replication and gene transcription.
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
2. Identify four differences between DNA and RNA.
DNA
RNA
Deoxyribonucleic acid is the hereditary material in most of the organisms
Made of 2-deoxyribose sugar
Contains thymine as one of the 4 nitrogen base
It is double stranded
Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to
the ribosome.
Made of ribose sugar
Has Uracil as the nitrogen base instead of thymine
Single stranded
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
Solution
1. Identify five differences between DNA replication and gene transcription.
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
2. Identify four differences between DNA and RNA.
DNA
RNA
Deoxyribonucleic acid is the hereditary material in most of the organisms
Made of 2-deoxyribose sugar
Contains thymine as one of the 4 nitrogen base
It is double stranded
Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to
the ribosome.
Made of ribose sugar
Has Uracil as the nitrogen base instead of thymine
Single stranded
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation.
1.1Yearcash flowpresent value of cash inflow = cash flow(1+r).pdf
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
Macculay duration in years
total/face value of bond
3939.467/1000
3.939467
1.4
modified duration
macculay duration/(1+YTM)
3.94/1.05
3.752381
1.5
change in price if YTM increases by .2%
modified duration*change in YTM
3.75*.2%
0.75%
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
Macculay duration in years
total/face value of bond
3939.467/1000
3.939467
1.4
modified duration
macculay duration/(1+YTM)
3.94/1.05
3.752381
1.5
change in price if YTM increases by .2%
modified duration*change in YTM
3.75*.2%
0.75%
Solution
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
M.
More Related Content
Similar to Now a normal O2 molecule has 12 electrons in the valence shell so, O.pdf
Lesson 8Chemical Bonding & NomenclatureYou should view all.docx
Lesson 8
Chemical Bonding & Nomenclature
You should view all lectures as a slide show. In the toolbar above, click “Slide Show” and “From Beginning” to start. Hit the space bar to forward to next slide or item.
1
*Read: Watch the Lecture as a slideshow, hit the space bar to move forward or just left click.
1
Topics:
Chemical Bonds
Lewis Dot Structures
Molecular Geometry
Polar Bonds
Naming Compounds
Writing Chemical Formulas
Naming Molecular Compounds
Lesson 8: Chemical Bonding and Nomenclature
2
2
Chemical Bonds
Sodium (Na) is a silver-colored metal that reacts so
violently with water that flames are produced when sodium
gets wet.
Chlorine (Cl) is a greenish-colored gas that is so poisonous that it was used as a weapon in World War I.
When the electrons of sodium metal and chlorine gas interact, the compound sodium chloride (NaCl) is formed, which is table salt.
Since the body is over 90% water, ingesting sodium metal would literally set someone on fire!
If we make a compound out of Na, we can make something totally different such as the table salt used to season our food.
3
Chemical Bonds
Atoms can interact with each other to form new substances called
compounds.
Compounds are formed when electrons in an atom’s last energy level interact and form chemical bonds. A chemical bond is an attractive force between atoms that holds them together.
An atom’s outermost energy level of electrons is called the valence shell (or valence level) and the electrons in the valence shell are called valence electrons.
4
Na Atom
Cl Atom
Sodium has 1 valence electron in the valence shell.
Chlorine has 7 valence electron in the valence shell.
Chemical Bonds
In ionic bonds, metals always lose electrons to nonmetals and become positive (or cations).
In ionic bonds, nonmetals always attract electrons from metals and become negative (anions).
Ionic compounds are neutral compounds made up of cations and anions.
Covalent bonds are formed between nonmetals and electrons are shared so no ions are formed.
Cl
nonmetal
Na
metal
O
nonmetal
O
nonmetal
Metal + Nonmetal = Ionic Compound
Nonmetal + Nonmetal = Covalent Compound
Chemical Bonds: Ionic
The octet rule states that atoms will gain, lose, or share
valence electrons in a way that will give each atom eight electrons in their valence shell.
Na has 1 electron in its valence shell and Cl has 7 electrons in its valence shell. Cl needs 1 valence electron to have 8 in its outer shell.
If Na transfers its 1 valence electron to Cl, Na’s second energy level becomes the valence shell, which already has 8 electrons.
Now, Na has a positive 1 charge and Cl has a negative 1 charge. But, the charge on NaCl is zero (the charges cancel each other out).
6
Na Atom
Cl Atom
Ionic Bond
Chemical Bonds: Ionic
An Ionic bond is formed when there is a transfer of electrons from a metal to a nonmetal. Compounds formed by ionic bonds are ...
Electrons have places around the atom that they are most likely to b.pdf
Electrons have places around the atom that they are most likely to be. An electron configuration
is a way of describing these places, like an address for electrons. Each part of what you just
wrote out describes a place where electrons can be found in oxygen.
Let\'s take a look at just the first part: 1s2.
The \'1\' is the shell. The shell number is used to indicate how close to the nucleus the electrons
are. The first shell is the *closest* possible spot for the electrons - the higher this number
becomes, the farther away from the nucleus they go. The \'1\' shell MUST be get its electrons
before the \'2\' shell or \'3\' shell, and so on - it gets priority. With me so far? :)
The \'s\' is the subshell, which refers to the SHAPE of the area where the electron might be. \'S\'
stands for \'spherical\', like a beach ball. Imagine a sphere like that around the nucleus - electrons
in the 1s area can be anywhere on that shell.
Finally, we\'ve got the \'2\' of 1s2. This indicates the number of electrons present. ALL \'S\'
SHELLS CAN HOLD NO MORE THAN 2 ELECTRONS. EVER. So when you get to 2s? It
should have 2 electrons as well, when it\'s full. 3s? Should also have two electrons. ANYTHING
with an \'s\' should have up to two.
So, the fact that 1s comes before 2s comes before 2p? That must be memorized, unfortunately.
There is a handy chart to help you remember the correct order; you can find it here:
http://education.jlab.org/qa/electron_config.html.
Now we come onto how you do an electron configuration for a specific atom. Let\'s say your
teacher tells you to draw the electron configuration for oxygen. You look on the Periodic Table
(go ahead) and see that the number of electrons in an oxygen atom is eight. Now your job is to
use this code we\'ve talked about to describe where each of those eight electrons is in an oxygen
atom.
You have to start putting electrons in the \'1\' shell because you always fill the \'lowest\' (closest
to the nucleus) level first. So you have 1s2, two electrons in the first shell, darting around over a
sphere. That means you have two electrons so far. But oxygen has eight all together - the PT says
so. So we keep going. Again with the memorization, we know that 2s is next, and all \'s\'
subshells have up to two electrons. So we now have 1s2, 2s2. That\'s still only four electrons all
together. We have to keep going.
Next is 2p. \'p\' subshells can hold up to six electrons. BUT, if we do six... 1s2, 2s2, 2p6...
adding up our electrons gives us 10. Oxygen only needs eight electrons! So we need: 1s2, 2s2,
2p4. Now we have 8 electrons all together.
ALL atoms are done in exactly the same way. You figure out how many electrons they have,
then keep going with the electron configuration \'addresses\' until you have reached the right
number of electrons (and remember - the number of electrons is represented by the number
AFTER the letter; not before).
Try these: Do the electron configuration for helium and neon and chlorine using the.
Okay, so you know that the number of protons and electronsboth are e.pdf
Okay, so you know that the number of protons and electronsboth are equal to the atomic number
(on the periodic table, it\'sthe smaller number of the two seen when you look at elements...itis
also the number that holds the order to them in the periodictable of elements).
a. Fe2+Now, you look up theelement Fe and see that it normally has 26 protons and 26
electronsby looking it up on the table of elements and seeing its atomicnumber is \"26\" But we
also see that the atom has a charge of +2meaning two electrons (1election = 1 negative charge)
have beenremoved, SO you get 26 protons and 24 electrons.Since it is postively charged, the
questionasks us to find the most common oxide product or ratio of the givenelement and oxygen
(which has a -2 charge)...Fe is +2 and O is -2,Therefore you get FeO.
b. Fe3+26 protons, 23electrons Fe is +3 and O is -2,Therefore you get F2O3 to get a +6 and -6
tocancel out
c. Ba2+56 protons and 54electrons BaO
d. Cs+ 55 protons and54 electrons Cs2O
e. S2-For negative atoms,it\'s the same process but you have a negative net charge so youmust
have more electrons. 16 protons and 18 electrons.Al ion has a +3 charge...soAl2S3 +6 and -6
cancel each otherout
f. P3-15 protons and 18electrons AlP
g. Br-35 protons and 36electrons AlBr3
h. N3-7 protons and 10electrons AlN
When doing ion charges, the only number that changes is theelectrons because of the electron
clouds. The electron clouds arethe reason ions form ionic bonds - they all want the coveted
\"8electron outer shell\"...just like those noble gases.
When doing ion charges, the only number that changes is theelectrons because of the electron
clouds. The electron clouds arethe reason ions form ionic bonds - they all want the coveted
\"8electron outer shell\"...just like those noble gases.
Solution
Okay, so you know that the number of protons and electronsboth are equal to the atomic number
(on the periodic table, it\'sthe smaller number of the two seen when you look at elements...itis
also the number that holds the order to them in the periodictable of elements).
a. Fe2+Now, you look up theelement Fe and see that it normally has 26 protons and 26
electronsby looking it up on the table of elements and seeing its atomicnumber is \"26\" But we
also see that the atom has a charge of +2meaning two electrons (1election = 1 negative charge)
have beenremoved, SO you get 26 protons and 24 electrons.Since it is postively charged, the
questionasks us to find the most common oxide product or ratio of the givenelement and oxygen
(which has a -2 charge)...Fe is +2 and O is -2,Therefore you get FeO.
b. Fe3+26 protons, 23electrons Fe is +3 and O is -2,Therefore you get F2O3 to get a +6 and -6
tocancel out
c. Ba2+56 protons and 54electrons BaO
d. Cs+ 55 protons and54 electrons Cs2O
e. S2-For negative atoms,it\'s the same process but you have a negative net charge so youmust
have more electrons. 16 protons and 18 electrons.Al ion has a +3 charge...soAl2S3 +6 and -6
cancel each otherout
f. .
A Rapid and Innovative Method for Prediction of Spin Multiplicity and Spin St...
Determination of spin multiplicity is very important for identification of spin state of different
molecules or ions. But, the conventional method of prediction of spin multiplicity by using formula (2S+1) is
time consuming. So, in this manuscript, I tried to show a rapid and simple method for prediction of spin
multiplicity value without calculating the total spin quantum number(S=∑s). This method is only applicable for
homo and hetero nuclear diatomic molecules and ions having total electron (01-20).
D081112730
This is my published paper in IOSR JAC novemboe 2015 edition in the field of inorganic chemictry.By this method we can easily find the spin multiplicity value of any diatomic molecules having total electron no 1-20 ,without calculation total spin quantum no.This is most easy way to calculate spin multiplicity value and also time saving method with simple concept.
Simple Thinking Makes Chemistry Metabolic and Interesting - A Review Article
First of all I give my love to all the students who reside in ‘Chemistry Phobia zone’. In this short article I shall try to share my simple thinking that I achieved during my last 16 years of journey in the route of Chemical Education. Before starting my journey it was a big encounter for me to change the conventional methods and makes chemistry easier and interesting to such student who belongs to Chemistry Paranoia Zone. I believe that students are just like flowers and it is our duty to nourish them properly as a gardener. In my infinitely small area of knowledge, I just try to innovate fourteen (14) teaching methodologies by including thirty four (34) completely new formulae in the chemistry world which were followed by different eminent writers in their books since last 90-95 years namely Sir G.Wilkinson, Prof. F. Albert Cotton, Prof. I.L.Finar, Prof. R.T. Morrison, Prof. R.N. Boyd, Prof. Solomons and Prof. Fryhle, Prof. J.G. Smith etc..
Similar to Now a normal O2 molecule has 12 electrons in the valence shell so, O.pdf (19)
Unidad v 2da parte- hibridacion - geometria molecular
Unidad v 2da parte- hibridacion - geometria molecular
Lesson 8Chemical Bonding & NomenclatureYou should view all.docx
Lesson 8Chemical Bonding & NomenclatureYou should view all.docx
Electrons have places around the atom that they are most likely to b.pdf
Electrons have places around the atom that they are most likely to b.pdf
Okay, so you know that the number of protons and electronsboth are e.pdf
Okay, so you know that the number of protons and electronsboth are e.pdf
A Rapid and Innovative Method for Prediction of Spin Multiplicity and Spin St...
A Rapid and Innovative Method for Prediction of Spin Multiplicity and Spin St...
Simple Thinking Makes Chemistry Metabolic and Interesting - A Review Article
Simple Thinking Makes Chemistry Metabolic and Interesting - A Review Article
More from aquacareser
1) MODERN TIMES is a very well prepared movie that describes the mis.pdf
1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of
humar resource during the period of industrialization era or machine age worlwide. It is a mix of
social drama and comedy. The content is presented in a satire. How did people survived the great
depression of the world is being documented here. The conditions werre miserable like acute
shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic
representation of what happens to a human being caught up in the mechanical bureaucracy of the
intellectual state.
2) Economically it could be understood that what miser situations could depression bring up.
Food and jobs comes hard in hand. To which a labor is convict to do under skill jobs as well for
survival. At the time of the great depression Americans had to fight for what little they had, and
as the movie showed many people went hungry. Modern Times looks toward the future, but not
with enthusiasm. Often described as a satire of the machine age, It has in fact a broader theme:
the dehumanizing effects of many aspects of modernity, including industrialization, bureaucracy,
urbanization, and law enforcement.
3) From the movie a prospect employee should not forget about the social ills and failings of
own day even if present age is seems far better than that was shown in the film. A dilbert-like
early scene shows the Tramp trying to take a break during work hours in the washroom — only
to have a big-screen image of the boss’s head suddenly appear on the washroom wall and order
him to quit stalling and get back to work. Modern audiences watching this scene may reflect
ruefully on the everyday reality of electronic surveillance in the workplace. Now, obviously
there’s nothing new about widespread poverty and want, but unemployment as we know it today
is a comparatively recent development — a byproduct of the modern workforce economy that
requires everyone to \"get a job.\"
4) The major ethical dilemma consists whether out of povertty stealing is right or wrong.
However this movie glorifies the struggle of the poor rather than vilifying them for stealing food
to survive. By glorifying the struggle the movie takes on yet another social issue: when the poor
steal to eat it is not criminal, just survival. Minor ones include
Solution
1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of
humar resource during the period of industrialization era or machine age worlwide. It is a mix of
social drama and comedy. The content is presented in a satire. How did people survived the great
depression of the world is being documented here. The conditions werre miserable like acute
shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic
representation of what happens to a human being caught up in the mechanical bureaucracy of the
intellectual state.
2) Economically it could be understood that what miser situations could depress.
1. Identify five differences between DNA replication and gene transc.pdf
1. Identify five differences between DNA replication and gene transcription.
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
2. Identify four differences between DNA and RNA.
DNA
RNA
Deoxyribonucleic acid is the hereditary material in most of the organisms
Made of 2-deoxyribose sugar
Contains thymine as one of the 4 nitrogen base
It is double stranded
Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to
the ribosome.
Made of ribose sugar
Has Uracil as the nitrogen base instead of thymine
Single stranded
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
Solution
1. Identify five differences between DNA replication and gene transcription.
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
2. Identify four differences between DNA and RNA.
DNA
RNA
Deoxyribonucleic acid is the hereditary material in most of the organisms
Made of 2-deoxyribose sugar
Contains thymine as one of the 4 nitrogen base
It is double stranded
Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to
the ribosome.
Made of ribose sugar
Has Uracil as the nitrogen base instead of thymine
Single stranded
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation.
1.1Yearcash flowpresent value of cash inflow = cash flow(1+r).pdf
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
Macculay duration in years
total/face value of bond
3939.467/1000
3.939467
1.4
modified duration
macculay duration/(1+YTM)
3.94/1.05
3.752381
1.5
change in price if YTM increases by .2%
modified duration*change in YTM
3.75*.2%
0.75%
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
Macculay duration in years
total/face value of bond
3939.467/1000
3.939467
1.4
modified duration
macculay duration/(1+YTM)
3.94/1.05
3.752381
1.5
change in price if YTM increases by .2%
modified duration*change in YTM
3.75*.2%
0.75%
Solution
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
M.
this is my code to count the frequency of words in a text file#.pdf
this is my code to count the frequency of words in a text file
#define _GNU_SOURCE
#include
#include
#include
#include
/******************************
here is the Structure for storing words
received from each thread
******************************/
struct return_vall{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*****************************
here are Aggregates Frequency Sum of
different threads
*****************************/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/**************************
Function thta are executed by each
thread on seperate line.
**************************/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
//printf(\"Thread %d\ \",line_number);
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] = \"p\' \'file.txt\'\\0\";
char command[100];
command[0] = \'\\0\';
strcat(command,cmd_p1);
strcat(command,cmd_p2);
strcat(command,cmd_p3);
//usleep(line_number);
char cmd[100] = \" | tr [:space:] \'\\\ \' | grep -v \'^\\\\s*$\' | sort | uniq -c | sort\\0\";
strcat(command,cmd);
//printf(\"%s\ \",command);
//fflush(stdout);
FILE *in;
in= popen(command, \"r\");
rewind(in);
char buff[50];
int counter = 0;
while(fgets(buff,sizeof(buff),in))
{
char c=\' \';
int i = 0;
int cnt = atoi(buff);
arr[line_number-1].count[counter] = cnt;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+6];
i++;
}
int cnnt = 0;
while(c!=\' \')
{
c = buff[cnnt];
cnnt++;
}
i=0;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+cnnt];
i++;
}
sprintf(arr[line_number-1].wordlist[counter],\"%s\",buff);
//printf(\"%d %s\",arr[line_number-1].count[counter],arr[line_number-
1].wordlist[counter]);
counter++;
}
arr[line_number-1].count[counter] = -1;
fclose(in);
return NULL;
}
int main(void)
{
FILE *fp;
fp = fopen(\"file.txt\",\"r\");
char a[200];
lines = 0;
void *status = NULL;
while(fgets(a,sizeof(a),fp) !=NULL)
{
lines++;
}
fclose(fp);
arr = (struct return_vall*) calloc(lines, sizeof(struct return_vall) * lines);
threads = malloc(sizeof(pthread_t)*lines);
int i;
for(i=0;i= 0)
{
if(arr[i].count[ctr] > 0)
{
strcpy(final[final_ctr],arr[i].wordlist[ctr]);
final_cnt[final_ctr] = arr[i].count[ctr];
arr[i].count[ctr] = 0;
int sum = aggregate(arr[i].wordlist[ctr]);
final_cnt[final_ctr] += sum;
printf(\"%d %s\",final_cnt[final_ctr],final[final_ctr]);
}
ctr++; final_ctr++;
}
i++;
if(i==lines)
{
break;
}
}
free(arr);
return 0;
}
Solution
this is my code to count the frequency of words in a text file
#define _GNU_SOURCE
#include
#include
#include
#include
/******************************
here is the Structure for storing words
received from each thread
******************************/
struct return_vall{
char wordlist[100][100];
.
The ice allows the refluxed materials to crystall.pdf
The ice allows the refluxed materials to crystallize, allowing for formation of
crystals and solid product. Also, the ice allows purification via washing, since it gets rid of
contaminants.
Solution
The ice allows the refluxed materials to crystallize, allowing for formation of
crystals and solid product. Also, the ice allows purification via washing, since it gets rid of
contaminants..
1. b2. c (bed bug)3. d (caterpillars)4. c (lyme disease)5. e.pdf
1. b
2. c (bed bug)
3. d (caterpillars)
4. c (lyme disease)
5. e (chagas disease)
6. a (hives)
7. e (short lived)
8. d (plants)
9. a (1892)
10. e (parvoviridae)
11. mosquito
12. a (flies)
13. c (lyme disease)
14.
15. --- holometabolous is the type of metamorphosis in beetles
16. oligocene
17. e (6 or more)
18. e
Solution
1. b
2. c (bed bug)
3. d (caterpillars)
4. c (lyme disease)
5. e (chagas disease)
6. a (hives)
7. e (short lived)
8. d (plants)
9. a (1892)
10. e (parvoviridae)
11. mosquito
12. a (flies)
13. c (lyme disease)
14.
15. --- holometabolous is the type of metamorphosis in beetles
16. oligocene
17. e (6 or more)
18. e.
#In this project you will write a program play TicTacToe #using tw.pdf
#In this project you will write a program play TicTacToe
#using two players (labels 0,1) or one play (label 0) playing with the machine (label 1).
#The TicTacToe board has 9 integers board = [1,2,3,4,5,6,7,8,9]. The following
# are the modules for the program
#
#def reset() resets the board to the original values
# board = [1,2,3,4,5,6,7,8,9]
#
#def printBoard() print the current state of the board using the format
#
#The current TicTacToe Board
# | 1 | 2 | 3 |
# | 4 | 5 | 6 |
# | 7 | O | 9 |
#
#The current TicTacToe Board
# | X | 2 | 3 |
# | 4 | 5 | 6 |
# | 7 | O | 9 |
#Note from the above that player 0 and 1 have played numbers 8 and 1
#respectively and the board display O for player 0 and X from player 1
#
#def changeBoard(num1, player) using the chosen box number to change
#the value of the box to 0 or -1 depending on whether the player is 0 or 1,
#respectively.
#
#def play(player) prints the player number (0 or 1) and prompts the player
# to enter a box value that have not changed to \'O\' or \'X\'
#
#def checkRows(value) checks to see which of the rows of the board
# has the same value and returns True, otherwise, returns False
#
#def checkCols(value) checks to see which of the cols of the board
# has the same value and returns True, otherwise, returns False
#
#def checkDiagonal(value) checks to see which of the diagonals of the board
# has the same value and returns True, otherwise, returns False
#
#def win(player) checks if a player wins the game, returns True of the player wins
# and False otherwise
#
#def machinePlay(player) plays the role of the player using random number.
# this function generates numbers in the interval [1,9] and uses the first
# generated random number that has not been used to play the game. The
#
#def ticTacToe(numPlayers) accepts the number of players and simulates the
#ticTapToe, asking players to enter unused box numbers.
#
#def main() is the driver module that accepts the number of players from the user
# and calls the ticTacToe module
#Assignment: Complete the follwoing modules:
#checkCols
#checkRows
#checkDiagonal
#win
#Sample of the output is
#
#Project
#This game can be played by one or two players
#Enter the number of players, 1/2 for one/two players: 1
#
#The current TicTapToe Board
#| 1 | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | 8 | 9 |
#Player 0 Enter a box value: 9
#
#The current TicTapToe Board
#| 1 | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | 8 | O |
#Player 1 ***Computer*** playing
#The current TicTapToe Board
#| 1 | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | X | O |
#Player 0 Enter a box value: 1
#
#The current TicTapToe Board
#| O | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | X | O |
#Player 1 ***Computer*** playing
#
#The current TicTapToe Board
#| O | X | 3 |
#| 4 | 5 | 6 |
#| 7 | X | O |
#Player 0 Enter a box value: 5
#
#The current TicTapToe Board
#| O | X | 3 |
#| 4 | O | 6 |
#| 7 | X | O |
#Player 0 Wins
#Do you want to quit?
#Begin Program
#import random number generator
#from random library
#from random import randint, seed
#Global va.
Web sever environmentA Web server is a program that uses HTTP (Hy.pdf
Web sever environment:
A Web server is a program that uses HTTP (Hypertext Transfer Protocol) to serve the files that
form Web pages to users, in response to their requests, which are forwarded by their computers\'
HTTP clients. Dedicated computers and appliances may be referred to as Web servers as
well.The process is an example of the client/server model. All computers that host Web sites
must have Web server programs. Leading Web servers include Apache (the most widely-
installed Web server), Microsoft\'s Internet Information Server (IIS) and nginx
(pronouncedengine X) from NGNIX. Other Web servers include Novell\'s NetWare server,
Google Web Server (GWS) and IBM\'s family of Domino servers. Web servers often come as
part of a larger package of Internet- and intranet-related programs for serving email,
downloading requests for File Transfer Protocol (FTP) files, and building and publishing Web
pages. Considerations in choosing a Web server include how well it works with the operating
system and other servers, its ability to handle server-side programming, security characteristics,
and the particular publishing, search engine and site building tools that come with it.
Advantages of using a web server within your development environment:
Problems posed by web server environment and methods to solve:
Various high-profile hacking attacks have proven that web security remains the most critical
issue to any business that conducts its operations online. Web servers are one of the most
targeted public faces of an organization, because of the sensitive data they usually host. Securing
a web server is as important as securing the website or web application itself and the network
around it. If you have a secure web application and an insecure web server, or vice versa, it still
puts your business at a huge risk. Your company’s security is as strong as its weakest point.
Although securing a web server can be a daunting operation and requires specialist expertise, it is
not an impossible task. Long hours of research and an overdose of coffee and take away food,
can save you from long nights at the office, headaches and data breaches in the future. Irrelevant
of what web server software and operating system you are running, an out of the box
configuration is usually insecure. Therefore one must take some necessary steps in order to
increase web server security. Below is a list of tasks one should follow when securing a web
server.
1. Remove Unnecessary Services
Default operating system installations and configurations, are not secure. In a typical default
installation, many network services which won’t be used in a web server configuration are
installed, such as remote registry services, print server service, RAS etc. The more services
running on an operating system, the more ports will be left open, thus leaving more open doors
for malicious users to abuse. Switch off all unnecessary services and disable them, so next time
the server is rebooted, they .
We ca use HashSet data structure to remove duplicats from an array i.pdf
We ca use HashSet data structure to remove duplicats from an array in O(n) time.
HashSet does not allow duplicate elements. So add all elements from an array to HashSet. While
adding first check whether current element is already in Set or not. If it is already there then
remove from an array.
PSUDO CODE:
void removeDuplicateFromArray(int []array, int size):
1. create a HashSet: HashSet set = new HashSet();
2. Itereate give array and add element in Set one by one:
int k = 0;
for i=0 to size-1:
if(!set.contains(array[i]) // if current element is not in Set, then add it in set and include in array
also
array[k++] = array[i];
set.add(array[i])
Runnning TIme: O(n)
Solution
We ca use HashSet data structure to remove duplicats from an array in O(n) time.
HashSet does not allow duplicate elements. So add all elements from an array to HashSet. While
adding first check whether current element is already in Set or not. If it is already there then
remove from an array.
PSUDO CODE:
void removeDuplicateFromArray(int []array, int size):
1. create a HashSet: HashSet set = new HashSet();
2. Itereate give array and add element in Set one by one:
int k = 0;
for i=0 to size-1:
if(!set.contains(array[i]) // if current element is not in Set, then add it in set and include in array
also
array[k++] = array[i];
set.add(array[i])
Runnning TIme: O(n).
this eample is of t-test for comparing two meanshere mean1 = m.pdf
this eample is of t-test for comparing two means
here mean1 = mean of FIFO and mean 2 : Mean of LIFO
so here we use student t-test for comparision of 2 means to evaluate both methods in this
procedure
ans: Student t-test for comparision of means
Solution
this eample is of t-test for comparing two means
here mean1 = mean of FIFO and mean 2 : Mean of LIFO
so here we use student t-test for comparision of 2 means to evaluate both methods in this
procedure
ans: Student t-test for comparision of means.
The middle ear bones in mammals are derived from bones in the dentar.pdf
The middle ear bones in mammals are derived from bones in the dentary bone (mandible) of
Therapsids. In the Therapsid dentary, there were three bones used in the jaw hinge: the angular,
articular and quadrate. The tympanic, malleus, and incus are homologous to the angular,
articular, and quadrate
Tympanic = Angular
Malleus = Articular
Incus = Quadrate
In early Therapsids, the quadrate, angular and articular were directly connected to the dentary
bone in the jaw hinge. These bones were essential for jaw movement. These bones were also
utilized for crude hearing capabilities. The next stage of evolution came with the separation of
these bones and the dentary. After the bones separated from the jaw, they formed the postdentary
rod. It is believed that these the mammalian middle ear bones evolved at least twice. It is
believed that monotreme and therian ancestors evolved very similar middle ear bones, but in two
separate paths. In monotreme ancestors, there is a mandibular trough, that is believed to have
accommodated the postdentary rod. This rod is believed to have been connected to the trough by
ligaments. This rod was used for hearing and the quadrate, angular, and articular were no longer
needed in the jaw hinge. In therians, the mandibular trough no longer held the quadrate, angular,
and articular. These bones were held to the jaw by ossified Meckel’s cartilage. This bridge of
cartilage and bone extended away from the dentary and approached the area that would make up
the modern ear cavity.
The evolution of the mammalian middle ear and jaw joint were pivotal steps in the evolution of
mammals. It is also a great example of how classical comparative anatomy, paleontology and
developmental biology have come together to piece together how this remarkable transformation
of jaw joint to ear ossicles was able to come about. The homologies of the malleus, incus and
stapes to the articular, quadrate and columella, and tympanic ring and gonial to the angular and
prearticular suggested by comparative anatomy 175 years ago have been recently confirmed by
molecular and developmental biology. The recent discovery of new mammali form fossils has
allowed careful documentation of the shift from primary to secondary jaw articulation, creating
an opportunity to follow the transformation of the post-dentary skeletal elements. This fossil data
has been complemented by the study of marsupial development, providing insight into the
changing role of the malleus and incus, and the relationship of the primary and secondary jaw
joints. We now have a number of unanswered questions. What are the signalling molecules
involved in the interactions between the condylar process and the glenoid fossa that create the
novel mammalian jaw joint? What are the mechanisms that lead to transformation of Meckel’s
cartilage into a ligament allowing isolation of the ear from the jaw? What controls the
distribution of secondary cartilages? What controls the timing of differentiation an.
There is an old saying “Do not put all eggs in one basket”WhyBe.pdf
There is an old saying “Do not put all eggs in one basket”
Why?
Because risk is high.By keeping eggs in different baskets, we reduce risk
We try to do same thing through diversification.
By investing our capital in different assets , we try to reduce risk.
Diversified folios reduce the risk and also the ratio of Risk to reward.
If w1, w2 , w3 …wn are weight in the portfolio for assets 1, 2,3 ….n
Then,w1+w2+w3+……………………+wn=1
R1, R2,R3,…….Rn are the return of the assets 1, 2 , 3 ….n
S1, S2, S3……Sn are the standard deviation of the assets 1, 2, 3 …n
Portfolio Return=w1R1+w2R2+w3R3+…….+wnRn
Portfolio
Variance=(w1^2)*(S1^2)+(w2^2)(S2^2)+………….(wn^2)*(Sn^2)+2w1w2*Cov(1,2)+2w1w3*
Cov(1,3)+………+w(n-1)wn*Cov(n,(n-1)
Cov(1,2)=Covariance of returns of asset1 and asset2
Portfolio Standard Deviation =Square root of Portfolio variance
Risk of a stock is measured by standard deviation.
Hence reduction of standard deviation through diversification means reduction of risk.
We can take a simple example of two assets 1 and 2
Return of asset1=R1=15%
Return of asset2=R2=12%
Standard deviation of asset 1=S1=10%
Standard deviation of asset 2=S2=8%
Correlation of asset 1 and 2=Corr(1,2)=0.1
Covariance(1,2)=Corr(1,2)*S1*S2=0.1*10*8=8
Assume for simplicity, equal amount is invested in asset 1 and asset 2
Hence, w1=w2=0.5
Portfolio Return;
0.5*15+0.5*12=13.5%
Portfolio Variance=(0.5^2)*(10^2)+(0.5^2)*(8^2)+2*0.5*0.5*8=45
Portfolio Standard Deviation=Square root of Variance=(45^0.5)= 6.708204
We can see, the risk of portfolio as measured by Standard Deviation has reduced significantly to
6,7 whereas the assets in the portfolio had standard deviation of 10 and 8
Risk / Return ratio of the portfolio=6.7/13.5=0.496
Risk/Return ratio of asset1=10/15= 0.666667
Risk/Return ratio of asset2=8/12= 0.666667
Risk return ratio of the portfolio is lower
WHY INVEST IN DIFFERENT INDUSTRIES TO DIVERSIFY?
Return of same types of assets are highly correlated. Suppose, if you invest in 10 different
Technology Companies, you may not be able to reduce risk, because all the companies will have
similar return and highly correlated.
We have seen in the equation of Portfolio Variance, one factor=2 w1*w2*Cov(1,2)
If there is high correlation between return of asset1 and 2, the Covariance(1,2) will be high .
Hence this factor 2w1w2Cov(1,2) will be high. As a result portfolio standard deviation will be
high and diversification will not serve any purpose.
Hence, in order to get benefit of diversification, we need to invest in different type of industries
so that the correlation between different assets are low. Consequently, the portfolio risk
(measured through standard deviation) is low.
Solution
There is an old saying “Do not put all eggs in one basket”
Why?
Because risk is high.By keeping eggs in different baskets, we reduce risk
We try to do same thing through diversification.
By investing our capital in different assets , we try to reduce risk.
Diversified folios reduce the risk and also the rati.
The ER Model is focussed to be a description of real-world entities..pdf
The ER Model is focussed to be a description of real-world entities. Although its
development/construction is done in such a way as to allow fair and easy translation to the
relational schema model, this is not fully trivial process. The ER diagram represents the
conceptual level of database design while the relational schema represents the logical level for
the database design. Following simple rules are followed:
ADDRESS( cityAddress, city, state ) candidate key:cityAddress, city , state
PERSON( pid,name ) candidate key: pid
OBSERVATION (oid, date, temperature, humidity, precip. ,city), candidate key: oid foreign key:
city
OBSERVER ( pid ), candidate key: pid
METEREOLOGIST( pid ), candidate key: pid
CONSUMER( pid ), candidate key: pid
WARNINGS( wwid,description,date,city) candidate key: wwid
FORECAST( fid,city,date,low,high) candidate key: fid
WEATHERPROGRAM( wid,name ) candidatekey; wid
LIVESAT(city, streerAddress, state,pid)
RECORDS(oid,pid,method)
ISSUES(wwid.pid)
PRESENTS(fid,pid)
VIEWS(pid,fid)
MAKES(wid,fid,oid)
15 tables will be created in all along with relationship tables
and only 9 tables if relationships are shown as foreign keys
Solution
The ER Model is focussed to be a description of real-world entities. Although its
development/construction is done in such a way as to allow fair and easy translation to the
relational schema model, this is not fully trivial process. The ER diagram represents the
conceptual level of database design while the relational schema represents the logical level for
the database design. Following simple rules are followed:
ADDRESS( cityAddress, city, state ) candidate key:cityAddress, city , state
PERSON( pid,name ) candidate key: pid
OBSERVATION (oid, date, temperature, humidity, precip. ,city), candidate key: oid foreign key:
city
OBSERVER ( pid ), candidate key: pid
METEREOLOGIST( pid ), candidate key: pid
CONSUMER( pid ), candidate key: pid
WARNINGS( wwid,description,date,city) candidate key: wwid
FORECAST( fid,city,date,low,high) candidate key: fid
WEATHERPROGRAM( wid,name ) candidatekey; wid
LIVESAT(city, streerAddress, state,pid)
RECORDS(oid,pid,method)
ISSUES(wwid.pid)
PRESENTS(fid,pid)
VIEWS(pid,fid)
MAKES(wid,fid,oid)
15 tables will be created in all along with relationship tables
and only 9 tables if relationships are shown as foreign keys.
Please find the answers belowAnswer 16 Option A (IEP). (IEP or i.pdf
Please find the answers below:
Answer 16: Option A (IEP). (IEP or individual educational plan is an official document which
carries the information depicting the personal/specific educatinal goals of children with special
needs. It is not only a document, but refers to a protocol/programme which clearly demarcates
the specific needs of a student. It has some features such as specilization, personalized, validity,
usability as a shared document, information about requirement of a specific teacher/mentor or
subject expert etc.)
Answer 17: The statement is true. Technically, based upon the physical, physiological and
psychological development of children, the age group of 6-11 has been globally accepted as the
time of adolescence for growing children. This age group is associated with discrete phsycial
growth (puberty), psycholgoical growth (feelings, emotions etc) and physiological growth
(activation of reproductive organs).
Solution
Please find the answers below:
Answer 16: Option A (IEP). (IEP or individual educational plan is an official document which
carries the information depicting the personal/specific educatinal goals of children with special
needs. It is not only a document, but refers to a protocol/programme which clearly demarcates
the specific needs of a student. It has some features such as specilization, personalized, validity,
usability as a shared document, information about requirement of a specific teacher/mentor or
subject expert etc.)
Answer 17: The statement is true. Technically, based upon the physical, physiological and
psychological development of children, the age group of 6-11 has been globally accepted as the
time of adolescence for growing children. This age group is associated with discrete phsycial
growth (puberty), psycholgoical growth (feelings, emotions etc) and physiological growth
(activation of reproductive organs)..
Question 1 answer is A Global trade allows wealthy countries to use .pdf
Question 1 answer is A Global trade allows wealthy countries to use their resources - whether
labor, technology or capital - more efficiently. Because countries are endowed with different
assets and natural resources (land, labor, capital and technology), some countries may produce
the same good more efficiently and therefore sell it more cheaply than other countries. If a
country cannot efficiently produce an item, it can obtain the item by trading with another country
that can. This is known as specialization in international trade.
Question 2 answer is C All options are correct
Question 3 answer is D Revenue tariffs are levied to raise money.
Question 4 answer is D.(cannot promote third world development, unless governments offer
some incentive for importing from third world economies).
Question 5 answer is A
Question 6 answer is C
Solution
Question 1 answer is A Global trade allows wealthy countries to use their resources - whether
labor, technology or capital - more efficiently. Because countries are endowed with different
assets and natural resources (land, labor, capital and technology), some countries may produce
the same good more efficiently and therefore sell it more cheaply than other countries. If a
country cannot efficiently produce an item, it can obtain the item by trading with another country
that can. This is known as specialization in international trade.
Question 2 answer is C All options are correct
Question 3 answer is D Revenue tariffs are levied to raise money.
Question 4 answer is D.(cannot promote third world development, unless governments offer
some incentive for importing from third world economies).
Question 5 answer is A
Question 6 answer is C.
Precipitation hardening, or age hardening, provides one of the most .pdf
Precipitation hardening, or age hardening, provides one of the most widely used mechanisms for
the strengthening of metal alloys. The fundamental understanding and basis for this technique
was established in early work at the U. S. Bureau of Standards on Duralumin.
The importance of theoretical suggestion for the development of new alloys is clear from the
historical record. At the end of the 19th century, cast iron was the only important commercial
alloy not already known to western technology at the time of the Romans. When age hardening
of aluminum was discovered accidentally by Wilm, during the years 1903 -1911, it quickly
became an important commercial alloy under the trade name Duralumin.
The strength and hardness of some metal alloys may be enhanced by the formation of extremely
small uniformly dispersed second-phase particles within the original phase matrix in a process
known as precipitation or age hardening. The precipitate particles act as obstacles to dislocation
movement and thereby strengthen the heat-treated alloys. Many aluminum based alloys, copper-
tin, certain steels, nickel based super-alloys and titanium alloys can be strengthened by age
hardening processes.
In order for an alloy system to be able to be precipitation-strengthened, there must be a terminal
solid solution that has a decreasing solid solubility as the temperature decreases. The Al-Cu
(Duralumin is an aluminum alloy of 2XXX group) phase diagram shown in Figure 1 shows this
type of decrease along the solvus between the and + regions. Consider a 96wt%Al – 4wt%Cu
alloy which is chosen since there is a large degrease in the solid solubility of solid solution in
decreasing the temperature from 550°C to 75°C.
Solution
Precipitation hardening, or age hardening, provides one of the most widely used mechanisms for
the strengthening of metal alloys. The fundamental understanding and basis for this technique
was established in early work at the U. S. Bureau of Standards on Duralumin.
The importance of theoretical suggestion for the development of new alloys is clear from the
historical record. At the end of the 19th century, cast iron was the only important commercial
alloy not already known to western technology at the time of the Romans. When age hardening
of aluminum was discovered accidentally by Wilm, during the years 1903 -1911, it quickly
became an important commercial alloy under the trade name Duralumin.
The strength and hardness of some metal alloys may be enhanced by the formation of extremely
small uniformly dispersed second-phase particles within the original phase matrix in a process
known as precipitation or age hardening. The precipitate particles act as obstacles to dislocation
movement and thereby strengthen the heat-treated alloys. Many aluminum based alloys, copper-
tin, certain steels, nickel based super-alloys and titanium alloys can be strengthened by age
hardening processes.
In order for an alloy system to be able to be precipitation-strengthene.
Protists are beneficial to the ecosystem in generating oxygen as the.pdf
Protists are beneficial to the ecosystem in generating oxygen as they are photosynthetic & single
celled in aquatic system result in higher dissolved oxygen finally other trophic levels are in
equilibrium
Some protists are natural scavangers & profoundly promote \"decomposition of natural organic
matter & dead bodies\" therefore nitrogen balance in the ecosystem
Some protist are harmful to the ecosystem and they cause disease to various animal and humans
result in ecological imbalance in trophic levels of food web
Economic importance of protists:
Bio-medical importance of protists:
Ecological importance of protists:
Ques-4:
Green algae: These are the algae, which were originated predominantly from the endosymbiotic
cyanobacteria, and these algae possess both plastids and chloroplasts for photosynthesis to
perform their nutrients synthesis.
Volvox is multicellular green algae and possesses cell differentiation as an evolutionary
reproductive adaptation. The cells are associated with flagella outside for motility and vacuoles
& chloroplasts inside. They mediate cell –to- cell communication through sensory perception
obtained from “eyespot” finally neighboring cells join through “cytoplasmic strands” that
promote cell –to-cell communication. Recent research has illustrated that cellular adhesion
molecules (Algal-CAMs) are found in multicellular Volvox. However, the bulk transport has not
observed in the Volvox, a primitive feature still yet to develop in them.
Brown algae are the algae, which possess secondary chloroplasts as their functional units to
perform their photosynthesis and these algae are commonly originated from the endosymbiotic
red algae. Examples are diatoms. Phylogenetic relationship with land plants:
Red algae do not possess property of photosynthesis & as they do not contain green pigments
Ques-3:
Outer coverings:
Sphingolipids
Pellicle
Phospholipids & glycolipids
Solution
Protists are beneficial to the ecosystem in generating oxygen as they are photosynthetic & single
celled in aquatic system result in higher dissolved oxygen finally other trophic levels are in
equilibrium
Some protists are natural scavangers & profoundly promote \"decomposition of natural organic
matter & dead bodies\" therefore nitrogen balance in the ecosystem
Some protist are harmful to the ecosystem and they cause disease to various animal and humans
result in ecological imbalance in trophic levels of food web
Economic importance of protists:
Bio-medical importance of protists:
Ecological importance of protists:
Ques-4:
Green algae: These are the algae, which were originated predominantly from the endosymbiotic
cyanobacteria, and these algae possess both plastids and chloroplasts for photosynthesis to
perform their nutrients synthesis.
Volvox is multicellular green algae and possesses cell differentiation as an evolutionary
reproductive adaptation. The cells are associated with flagella outside for motility and vacuoles
& chloroplasts inside. They mediate.
P1.javaimport java.util.Scanner;keyboard inputting package pub.pdf
P1.java
import java.util.Scanner;//keyboard inputting package
public class P1 {//main class
public static void main(String[] args) {//main method
Scanner input = new Scanner (System.in);
//User will input values
System.out.print(\"Enter your weght in pounds:\");
int weight = input.nextInt();//keyboard inputting
//User will input values
System.out.print(\"Enter your feet and inches:\");
int feet = input.nextInt();//keyboard inputting
int inches=input.nextInt();
int totalheightininches=0;
totalheightininches=(feet*12)+inches;//converting feet and inches into total inches
float bmi;
float constant=703f;
bmi = (float) (weight /Math.pow(totalheightininches,2))*constant;//bmi calculation
//Print out results on display
System.out.println(\"BMI is\" + bmi);
if(bmi <18.5)
System.out.println(\"Underweight\");
else if (bmi <25)
System.out.println(\"Normal\");
else if (bmi < 30)
System.out.println(\"Overweight\");
else
System.out.println(\"Obese\");
}
}
output
Enter your weght in pounds:150
Enter your feet and inches:5 5
BMI is24.95858
Normal
Solution
P1.java
import java.util.Scanner;//keyboard inputting package
public class P1 {//main class
public static void main(String[] args) {//main method
Scanner input = new Scanner (System.in);
//User will input values
System.out.print(\"Enter your weght in pounds:\");
int weight = input.nextInt();//keyboard inputting
//User will input values
System.out.print(\"Enter your feet and inches:\");
int feet = input.nextInt();//keyboard inputting
int inches=input.nextInt();
int totalheightininches=0;
totalheightininches=(feet*12)+inches;//converting feet and inches into total inches
float bmi;
float constant=703f;
bmi = (float) (weight /Math.pow(totalheightininches,2))*constant;//bmi calculation
//Print out results on display
System.out.println(\"BMI is\" + bmi);
if(bmi <18.5)
System.out.println(\"Underweight\");
else if (bmi <25)
System.out.println(\"Normal\");
else if (bmi < 30)
System.out.println(\"Overweight\");
else
System.out.println(\"Obese\");
}
}
output
Enter your weght in pounds:150
Enter your feet and inches:5 5
BMI is24.95858
Normal.
operating system linux,ubuntu,Mac#include iostream #include .pdf
//operating system linux,ubuntu,Mac
#include
#include
#include
#include // for Time
using namespace std;
struct Node {
int number;
struct Node* nextNode;
};
//append Method Implementation
void append(struct Node** head, int num)
{
if (*head == NULL) { // checking if first node is NULL or not
(*head) = (struct Node*)malloc(sizeof(struct Node));
(*head)->number = num;
(*head)->nextNode = NULL;
}
else {
struct Node* temp = *head;
while (temp->nextNode != NULL) {
temp = temp->nextNode;
}
struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically
allocation of memory
newNode->number = num;
newNode->nextNode = NULL;
temp->nextNode = newNode;
}
}
/* Getting Node with value from List*/
Node* search(struct Node** head, int value)
{
clock_t begin = clock();
std::cout<number == value) {
clock_t end = clock();
std::cout<nextNode;
}
return NULL;
}
/*Display List Method Implementation*/
void printList(struct Node* head)
{
while (head != NULL) {
std::cout << head->number << \" \";
head = head->nextNode;
}
std::cout << endl;
}
int main(int argc, char *argv[])
{
struct Node* head = NULL,*temp;
srand(time(0));
if ( argc != 3 ) // argc should be 2 for correct execution
// We print error message
std::cout<<\"Please Enter 3 argument\"<
Solution
//operating system linux,ubuntu,Mac
#include
#include
#include
#include // for Time
using namespace std;
struct Node {
int number;
struct Node* nextNode;
};
//append Method Implementation
void append(struct Node** head, int num)
{
if (*head == NULL) { // checking if first node is NULL or not
(*head) = (struct Node*)malloc(sizeof(struct Node));
(*head)->number = num;
(*head)->nextNode = NULL;
}
else {
struct Node* temp = *head;
while (temp->nextNode != NULL) {
temp = temp->nextNode;
}
struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically
allocation of memory
newNode->number = num;
newNode->nextNode = NULL;
temp->nextNode = newNode;
}
}
/* Getting Node with value from List*/
Node* search(struct Node** head, int value)
{
clock_t begin = clock();
std::cout<number == value) {
clock_t end = clock();
std::cout<nextNode;
}
return NULL;
}
/*Display List Method Implementation*/
void printList(struct Node* head)
{
while (head != NULL) {
std::cout << head->number << \" \";
head = head->nextNode;
}
std::cout << endl;
}
int main(int argc, char *argv[])
{
struct Node* head = NULL,*temp;
srand(time(0));
if ( argc != 3 ) // argc should be 2 for correct execution
// We print error message
std::cout<<\"Please Enter 3 argument\"<.
Network Protocol1) A Protocol is used for the communication betwe.pdf
Network Protocol:
1) A Protocol is used for the communication between two or more machines.
2) We can even define the network protocol as, it is for the communication between two network
devices.
3) Network protocols include some mechanisms such as to identify the devices and make
connections with each other.
4) After the connection is obtained, data is packaged into messages are sent and received.
5) Some protocols even support message acknowledgement and data compression is used for
high-performance.
What is a network service:
1) A network service is an application.
2) This runs at application layer and above.
3) This is used in computer networking.
4) Network service provides data Storage, manipulation ,presentation , communication
5) This is implemented using client-server architecture or peer-to-peer architecture.
Service Interface:
1) This interface is used to implement a Service-Oriented Architecture(SOA).
2) This is implemented to achieve interoperability among the applications among various
technologies.
Solution
Network Protocol:
1) A Protocol is used for the communication between two or more machines.
2) We can even define the network protocol as, it is for the communication between two network
devices.
3) Network protocols include some mechanisms such as to identify the devices and make
connections with each other.
4) After the connection is obtained, data is packaged into messages are sent and received.
5) Some protocols even support message acknowledgement and data compression is used for
high-performance.
What is a network service:
1) A network service is an application.
2) This runs at application layer and above.
3) This is used in computer networking.
4) Network service provides data Storage, manipulation ,presentation , communication
5) This is implemented using client-server architecture or peer-to-peer architecture.
Service Interface:
1) This interface is used to implement a Service-Oriented Architecture(SOA).
2) This is implemented to achieve interoperability among the applications among various
technologies..
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- 1. Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2 Solution Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2