Ionic compounds have electropositive and electronegative elements pr.pdf
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Ionic compounds have electropositive and electronegative elements present and high melting points due to strong ionic bonding. Polar molecules have weak intermolecular dipole-dipole interaction and dispersion forces, and low melting points. Nonpolar molecules have weak intermolecular forces, and low melting points. LiF - ionic compound ionic bonding BeF2 - ionic compound ionic bonding BF3 - nonpolar molecule dispersion force CF4 - nonpolar molecule dispersion force NF3 - polar molecule dispersion force dipole-dipole interaction OF2 - polar molecule dispersion force dipole-dipole interaction F2 - nonpolar molecule dispersion force Solution Ionic compounds have electropositive and electronegative elements present and high melting points due to strong ionic bonding. Polar molecules have weak intermolecular dipole-dipole interaction and dispersion forces, and low melting points. Nonpolar molecules have weak intermolecular forces, and low melting points. LiF - ionic compound ionic bonding BeF2 - ionic compound ionic bonding BF3 - nonpolar molecule dispersion force CF4 - nonpolar molecule dispersion force NF3 - polar molecule dispersion force dipole-dipole interaction OF2 - polar molecule dispersion force dipole-dipole interaction F2 - nonpolar molecule dispersion force.
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Chemical bonds- Properties of Ionic and Covalent compounds
This slide was used in the microteaching practice conducted by Dr. Denis Andrew D. Lajium for Teaching Method I (Chemistry) - TK30103.
all right reserve.
Ne and F2 are both uncharged molecules so there c.pdf
Ne and F2 are both uncharged molecules so there can be no ionic forces at work.
both are nonpolar, moleuclar compounds so there is no uneven charge distribution (dipoles). All
compounds have van der waals forces present so these are hte ones present between Ne and F2.
Xe is a molecular compound with no charge so no ionic or dipole. SF4 is polar and thus has van
der waals and dipole dipole, but since Xe has only van der waals, only van der waals interactions
can exist between these two. Both NO3- and Ca2+ are charged particles so theres electrostatic,
ion-ion attraction here
Solution
Ne and F2 are both uncharged molecules so there can be no ionic forces at work.
both are nonpolar, moleuclar compounds so there is no uneven charge distribution (dipoles). All
compounds have van der waals forces present so these are hte ones present between Ne and F2.
Xe is a molecular compound with no charge so no ionic or dipole. SF4 is polar and thus has van
der waals and dipole dipole, but since Xe has only van der waals, only van der waals interactions
can exist between these two. Both NO3- and Ca2+ are charged particles so theres electrostatic,
ion-ion attraction here.
Recommended
Chemical bonds- Properties of Ionic and Covalent compounds
This slide was used in the microteaching practice conducted by Dr. Denis Andrew D. Lajium for Teaching Method I (Chemistry) - TK30103.
all right reserve.
Ne and F2 are both uncharged molecules so there c.pdf
Ne and F2 are both uncharged molecules so there can be no ionic forces at work.
both are nonpolar, moleuclar compounds so there is no uneven charge distribution (dipoles). All
compounds have van der waals forces present so these are hte ones present between Ne and F2.
Xe is a molecular compound with no charge so no ionic or dipole. SF4 is polar and thus has van
der waals and dipole dipole, but since Xe has only van der waals, only van der waals interactions
can exist between these two. Both NO3- and Ca2+ are charged particles so theres electrostatic,
ion-ion attraction here
Solution
Ne and F2 are both uncharged molecules so there can be no ionic forces at work.
both are nonpolar, moleuclar compounds so there is no uneven charge distribution (dipoles). All
compounds have van der waals forces present so these are hte ones present between Ne and F2.
Xe is a molecular compound with no charge so no ionic or dipole. SF4 is polar and thus has van
der waals and dipole dipole, but since Xe has only van der waals, only van der waals interactions
can exist between these two. Both NO3- and Ca2+ are charged particles so theres electrostatic,
ion-ion attraction here.
Use of the plates within a specific time is necessary as time of inc.pdf
Use of the plates within a specific time is necessary as time of incubation is related to growth of
the bacterial cells which grows with time initially and after a particular period starts declining.
Moisture can increase the process of diffusion and also can spread bacterial cells into a zone
which is known as zone of inhibition.
Solution
Use of the plates within a specific time is necessary as time of incubation is related to growth of
the bacterial cells which grows with time initially and after a particular period starts declining.
Moisture can increase the process of diffusion and also can spread bacterial cells into a zone
which is known as zone of inhibition..
The five terrestrial adaptations of the seed plants include, reducti.pdf
The five terrestrial adaptations of the seed plants include, reduction of the gametophyte
generation, heterospory, ovules, and pollen.
Pollen grains were an important adaptation because the evolution of pollen allowed for
pollination and contributed to the diversity of seed plants.
Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole
structure is protected from desiccation and can reach the female organs without dependence on
water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an
extension of a cell within the pollen grain.
Seeds can survive harsh conditions through dormancy, are distributed far from their parent
sporophyte, and are multicellular.
Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain
growth and a protective coat give seeds their superior evolutionary advantage. Several layers of
hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of
water.
Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone
abscisic acid—until conditions for growth become favorable. Whether blown by the wind,
floating on water, or carried away by animals, seeds are scattered in an expanding geographic
range, thus avoiding competition with the parent plant.
All these various reasons make seeded plants to be domonant in plant communities today.
Solution
The five terrestrial adaptations of the seed plants include, reduction of the gametophyte
generation, heterospory, ovules, and pollen.
Pollen grains were an important adaptation because the evolution of pollen allowed for
pollination and contributed to the diversity of seed plants.
Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole
structure is protected from desiccation and can reach the female organs without dependence on
water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an
extension of a cell within the pollen grain.
Seeds can survive harsh conditions through dormancy, are distributed far from their parent
sporophyte, and are multicellular.
Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain
growth and a protective coat give seeds their superior evolutionary advantage. Several layers of
hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of
water.
Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone
abscisic acid—until conditions for growth become favorable. Whether blown by the wind,
floating on water, or carried away by animals, seeds are scattered in an expanding geographic
range, thus avoiding competition with the parent plant.
All these various reasons make seeded plants to be domonant in plant communities today..
SELF THEORY OF LATE ADULT HOOD it describes the changes of a adult.pdf
SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being
when they are getting aged.according to this theory
1.as the age of the particular person grows they will be more dependent on the others for their
needs.
2.some feelings like guilt,shame and depression develop as a result of their dependency.
3.many of the people will be feeling lonely when their loving partner passes away which will
have a negative impact on their health.
4.according to a theory stated by erikson people with a positive attitude will feel integrated
where as people with negative attitude or thoughts will be desperate.
STRATIFICATION THEORIES : These are related to the persons social categeory his social
status and the individual choices.this stratification is going to limit in many ways.many theories
have been stated they are as follows
Disengagement theory : the person disengages himself or herself away from the society and
become passive.
Activity theory : states that elderly people have to remain active in various social activities like
spending time with relatives,friends and various community groups and will be withdrawn only
as a result of age.
DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the
legacy of stratification.these dynamic theories states that each persons life will be chaning and is
purely self propelled within specific special contexts which will be ever changing.
Solution
SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being
when they are getting aged.according to this theory
1.as the age of the particular person grows they will be more dependent on the others for their
needs.
2.some feelings like guilt,shame and depression develop as a result of their dependency.
3.many of the people will be feeling lonely when their loving partner passes away which will
have a negative impact on their health.
4.according to a theory stated by erikson people with a positive attitude will feel integrated
where as people with negative attitude or thoughts will be desperate.
STRATIFICATION THEORIES : These are related to the persons social categeory his social
status and the individual choices.this stratification is going to limit in many ways.many theories
have been stated they are as follows
Disengagement theory : the person disengages himself or herself away from the society and
become passive.
Activity theory : states that elderly people have to remain active in various social activities like
spending time with relatives,friends and various community groups and will be withdrawn only
as a result of age.
DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the
legacy of stratification.these dynamic theories states that each persons life will be chaning and is
purely self propelled within specific special contexts which will be ever changing..
PV = PMTiThe goal is to solve the formula for i, doing so by isol.pdf
PV = PMT/i
The goal is to solve the formula for i, doing so by isolating it to one side of the equation.
The first step would be to cross multiply:
PV/1 = PMT/i
PV
Solution
PV = PMT/i
The goal is to solve the formula for i, doing so by isolating it to one side of the equation.
The first step would be to cross multiply:
PV/1 = PMT/i
PV.
Normal gene regulation in MAT locus YeastThere are 3 regulatory p.pdf
Normal gene regulation in MAT locus Yeast:
There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2.
These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable
opposite attracting pheromones.
a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha
factor and provokes the signalling process and there by repress the same..This happens the same
viceversa from alpha to a cells.
alpha 1 - protein activator.
alpha 2 - protein repressor.
a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets
of genes.
a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein
shows no growth, and also mating/sporulation defects.
It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate
or repress the MAT locus gene regulation.
So it\'s presence does not initiate any alteration in the gene regulation of MAT locus
Solution
Normal gene regulation in MAT locus Yeast:
There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2.
These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable
opposite attracting pheromones.
a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha
factor and provokes the signalling process and there by repress the same..This happens the same
viceversa from alpha to a cells.
alpha 1 - protein activator.
alpha 2 - protein repressor.
a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets
of genes.
a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein
shows no growth, and also mating/sporulation defects.
It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate
or repress the MAT locus gene regulation.
So it\'s presence does not initiate any alteration in the gene regulation of MAT locus.
Grey matter in the spinal cord is known as the grey column which tra.pdf
Grey matter in the spinal cord is known as the grey column which travels down the spinal cord
and is distributed in three grey columns that are presented in an \"H\" shape.
1. Anterior grey column.
It contains motor neurons. These synapse with interneurons and the axons of cells that have
travelled down the pyramidal tract. These cells are responsible for the movement of muscles.
2. Posterior grey column .
The posterior grey column contains the points where sensory neurons synapse. These receives
sensory information from the body, including fine touch, proprioception, and vibration. This
information is sent from receptors of the skin, bones, and joints through sensory neurons whose
cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the
spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the
spinothalamic tract.
3. Lateral grey column.
The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which
are part of the autonomic nervous system. It primarily involved with activity in the sympathetic
division of the autonomic motor system.
Solution
Grey matter in the spinal cord is known as the grey column which travels down the spinal cord
and is distributed in three grey columns that are presented in an \"H\" shape.
1. Anterior grey column.
It contains motor neurons. These synapse with interneurons and the axons of cells that have
travelled down the pyramidal tract. These cells are responsible for the movement of muscles.
2. Posterior grey column .
The posterior grey column contains the points where sensory neurons synapse. These receives
sensory information from the body, including fine touch, proprioception, and vibration. This
information is sent from receptors of the skin, bones, and joints through sensory neurons whose
cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the
spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the
spinothalamic tract.
3. Lateral grey column.
The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which
are part of the autonomic nervous system. It primarily involved with activity in the sympathetic
division of the autonomic motor system..
DateDescriptionDebitCredit2014Jan. 4Delivery Truck2800.pdf
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delivery truck
11680
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(28000 – 4640) / 4
= 11680
2015
Jan. 6
Delivery Truck
49400
Cash
49400
April 1
Depreciation Expense-Delivery Truck
1947
Delivery truck
1947
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12
= 1947
April 1
Cash
15180
Delivery truck
14373
Profit on sale of truck
807
Nov 2
Truck Repair Expense
470
Cash
470
Dec 31
Depreciation Expense-Delivery Truck
15972
Delivery truck
15972
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(49400-9470) / 5
= 15972
2016
July. 1
Delivery Truck
55400
Cash
55400
Oct 2
Depreciation Expense-Delivery Truck
8984
Delivery truck
8984
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(49400- 15972 )-9470 / 4 * 9/ 12
= 8984
Oct 2
Cash
17498
Loss on sale of truck
6946
Delivery truck
24444
Dec 31
Depreciation Expense-Delivery Truck
10611
Delivery truck
10611
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(55400-12955) / 8
= 10611
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delivery truck
11680
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(28000 – 4640) / 4
= 11680
2015
Jan. 6
Delivery Truck
49400
Cash
49400
April 1
Depreciation Expense-Delivery Truck
1947
Delivery truck
1947
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12
= 1947
April 1
Cash
15180
Delivery truck
14373
Profit on sale of truck
807
Nov 2
Truck Repair Expense
470
Cash
470
Dec 31
Depreciation Expense-Delivery Truck
15972
Delivery truck
15972
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(49400-9470) / 5
= 15972
2016
July. 1
Delivery Truck
55400
Cash
55400
Oct 2
Depreciation Expense-Delivery Truck
8984
Delivery truck
8984
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(49400- 15972 )-9470 / 4 * 9/ 12
= 8984
Oct 2
Cash
17498
Loss on sale of truck
6946
Delivery truck
24444
Dec 31
Depreciation Expense-Delivery Truck
10611
Delivery truck
10611
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(55400-12955) / 8
= 10611
Solution
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delive.
c. A destructor to free any allocated objects.d. A copy constructo.pdf
c. A destructor to free any allocated objects.
d. A copy constructor that allocates its own new member storage and copies the contents of
member variables.
e. An assignment operator that deallocates old storage before allocating new storage and copying
all the member variables.
Solution
c. A destructor to free any allocated objects.
d. A copy constructor that allocates its own new member storage and copies the contents of
member variables.
e. An assignment operator that deallocates old storage before allocating new storage and copying
all the member variables..
Barrier to entryThe government intervention to the barriers to en.pdf
Barrier to entry:
The government intervention to the barriers to entry to exist as a result of, while others occur
naturally within the business world. Often, existing firms within an industry lobby for the
government to erect new barriers to entry. In this is done to done by the protect the integrity of
the industry and prevent fly-by-night operations from setting up shop and hawking inferior
products and services. In reality, firms favor barriers to entry when already comfortably
ensconced in an industry to limit competition and claim a larger share of the industry\'s revenue.
Other barriers to entry occur naturally, often evolving over time as certain industry players
establish dominance.
A company can use this technology, for example, to build a barrier to entry, to build in switching
costs, and even, sometimes, to completely change the basis of competition. some companies
have seized the advantage, while others, more complacent, have ended up playing the difficult
and expensive game of catch-up ball. He also points out that it is important for executives to
make this competitive analysis in assessing where IS fits in their companies, since in some cases
it appropriately plays a support role and can add only modestly to the value of a company’s
products, while in other settings it is at the core of their competitive survival.
The computer’s main purpose is to cut order-entry costs and to provide more flexibility to
customers in the time and process of order submission. The system yields a larger competitive
advantage, adding value for customers and a substantial rise in their sales. The resulting sharp
increase in the company’s market share forces a primary competitor into a corporate
reorganization and a massive systems development effort to contain the damage.
Natural Barriers to Entry:
In industries where customers incur high costs switching from one brand to another, this
becomes a de facto barrier to entry for new firms, as they face difficulty enticing prospective
customers to pay the money required to make a chahange. Brand identity and customer loyalty
serve as barriers to entry for outsiders. Barriers to entry can also form naturally as the dynamics
of an industry take shape of the natural barrier.
Barrier option:
A barrier option is a type of option whose payoff depends on whether or not the underlying asset
has reached or exceeded a predetermined price. A barrier option can be a knock-out, meaning it
can expire worthless if the underlying exceeds a certain price, limiting profits for the holder but
limiting losses for the writer. It can also be a knock-in, meaning it has no value until the
underlying reaches a certain price.
BREAKING DOWN \'Barrier Option\':
Barrier options are considered a type of exotic option because they are more complex than basic
American or European options. Barrier options are also considered a type of path-dependent
option because their value fluctuates as the underlying\'s value changes during the opt.
Answer D all of the aboveLipids like triglycerides are semi-solid.pdf
Answer: D all of the above
Lipids like triglycerides are semi-solid at room temperature.
Lipids are not soluble in water
Steroids are a class of lipid molecules
Solution
Answer: D all of the above
Lipids like triglycerides are semi-solid at room temperature.
Lipids are not soluble in water
Steroids are a class of lipid molecules.
An example of when a one-way ANOVA could be used is if you want to d.pdf
An example of when a one-way ANOVA could be used is if you want to determine if there is a
difference in the mean height of stalks of three different types of seeds. Since there is more than
one mean, you can use a one-way ANOVA since there is only one factor that could be making
the heights different.
Now, if take these three different types of seeds, and then add the possibility that four different
types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of
the stalks could be different for a combination of several reasons:
The types of seed could cause the change,
the types of fertilizer could cause the change, and/or
there is an interaction between the type of seed and the type of fertilizer.
There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then
you can use a two-way ANOVA.
Solution
An example of when a one-way ANOVA could be used is if you want to determine if there is a
difference in the mean height of stalks of three different types of seeds. Since there is more than
one mean, you can use a one-way ANOVA since there is only one factor that could be making
the heights different.
Now, if take these three different types of seeds, and then add the possibility that four different
types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of
the stalks could be different for a combination of several reasons:
The types of seed could cause the change,
the types of fertilizer could cause the change, and/or
there is an interaction between the type of seed and the type of fertilizer.
There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then
you can use a two-way ANOVA..
10.dry and hot conditions11.C4 plants12.bundle sheet cells14.p.pdf
10.dry and hot conditions
11.C4 plants
12.bundle sheet cells
14.plasmodesmata
15)PEP carboxylase
16) bundle sheet cells
17) pineapple
18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if
takes place carboxylase activity will involve
19)corn
Solution
10.dry and hot conditions
11.C4 plants
12.bundle sheet cells
14.plasmodesmata
15)PEP carboxylase
16) bundle sheet cells
17) pineapple
18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if
takes place carboxylase activity will involve
19)corn.
1. Off Page refference arrow2. Connector Symbol3. Process4.Del.pdf
1. Off Page refference arrow
2. Connector Symbol
3. Process
4.Delay Symbol
5 .Merge Symbol
Solution
1. Off Page refference arrow
2. Connector Symbol
3. Process
4.Delay Symbol
5 .Merge Symbol.
1) the values of potassium, chlorine and HCO3 values are abnormal an.pdf
1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic
alkalosis which is causing high HCO3 values.
2) the pH is alkalotic and it is abnormal due to metabolic alkalosis
Solution
1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic
alkalosis which is causing high HCO3 values.
2) the pH is alkalotic and it is abnormal due to metabolic alkalosis.
The intermediate is the compound A. There are a c.pdf
The intermediate is the compound A. There are a couple of reasons for that: 1. The
oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic.
Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom
(the one that bears OH and then gets into the cycle) does not change in the reaction (retention of
the configuration). That supports the formation of the intermediate A. If the intermediate B
participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion
of the configuration).
Solution
The intermediate is the compound A. There are a couple of reasons for that: 1. The
oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic.
Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom
(the one that bears OH and then gets into the cycle) does not change in the reaction (retention of
the configuration). That supports the formation of the intermediate A. If the intermediate B
participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion
of the configuration)..
When ionic solids dissolve in water theions that are adjacent to each.pdf
When ionic solids dissolve in water theions that are adjacent to each other in the solid become
surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion
and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the
partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are
oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the
cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to
large ions. If this formation of the hydratedions were the only factor than we would expect all
ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we
need to condsider the other factors in thesolution process. the solution process is governed bythe
solute-solute, solvent-solvent and solute-solventintermolecular attractive forces. So far we have
only consideredhydration of the ions by the water molecules, that is thesoluteÐsolvent
interactions. When we consider the soluteinteractions we begin to see some of the problems that
can arise.The ions in a crystal are strongly attracted to each other and todissolve it is necessary to
overcome the electrostatic attractionbetween the oppositely charged ions. The lattice energy of a
solidis a measure of the strength of those electrostatic attractions.The lattice energy works to
keep the ions in the solid state andionic compounds with large lattice energies are insoluble in
waterwhile compounds with small lattice energies are soluble. Whenpotassium iodide dissolves
in water the ions is the potassiumiodide solid must be separated; KI(s) -ÐH2O -->K+(g) + I-(g)
ÆH1 = +632kJ/mol Now the gaseous ions are distributed inwater according to the equation;
H2O(l) + K+(g) +I-(g) -Ð-> K+(aq) + IÐ(aq)ÆH2 + ÆH3 = -617 kJ The sum of these two
equations yields theoverall solution process for KI dissolving in water. The value ofÆH2 + ÆH3
is call the hydrationenergy. In this example the hydration energy is not as large, inabsolute terms
as the energy required to separate the ions in thesolute and the heat of solution isendothermic.
Solution
When ionic solids dissolve in water theions that are adjacent to each other in the solid become
surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion
and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the
partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are
oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the
cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to
large ions. If this formation of the hydratedions were the only factor than we would expect all
ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we
need to condsider the other factors in.
#include iostream #include fstream #include map #include.pdf
#include
#include
#include
#include
#include
using namespace std;
//Used from
int index_of_largest(const int array[], int startIndex, const int size);
void selectSort(int a[], const int size);
void swap(int& i1, int& it2);
void fileOption();
void inputOption();
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size);
void printFileScreen(const int uniqueArray[], const int countArray[], const int size);
int main()
{
// int size;
char userInput[5];
//
cout << \"Read file from Input file? (Y/N) \";
cin >> userInput;
if(toupper(userInput[0]) == \'Y\')
{
fileOption();
}
else
{
inputOption();
}
return 0;
}
void selectSort(int array[], const int size)
{
int smallestIndex;
for(int i = 0;i < size - 1;i++)
{
smallestIndex = index_of_largest(array , i , size);
swap(array[i] , array[smallestIndex]);
}
}
void swap(int& i1, int& i2)
{
int temp;
temp = i1;
i1 = i2;
i2 = temp;
}
int index_of_largest(const int array[], int startIndex, const int size)
{
int min = array[startIndex] , index_of_min = startIndex;
for(int i = startIndex + 1;i < size;i++)
if(array[i] > min)
{
min = array[i];
index_of_min = i;
}
return index_of_min;
}
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size)
{
int number;
int j = 0;
int index;
for(int i = 0;i < size;i++)
{
int flag = 0;
number = inputArray[i];
//for each element in inputArray, check if it is there in uniqueArray
for (int k = 0; k < j; ++k)
{
if(number == uniqueArray[k])
{
flag = 1;
index = k;
}
}
if(flag == 0)
{
//if so, increment the countArray by 1
countArray[j]++;
// copy the element to uniqueArray
uniqueArray[j] = number;
j++;
}
else
countArray[index]++;
}
}
void fileOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50]= {0};
ifstream in_stream;
char fileName[16];
cout << \"What file would you like to obtain input from? \";
cin >> fileName;
in_stream.open(fileName);
if(in_stream.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
int size = 0;
while(in_stream >> inputArray[size])
{
size++;
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void inputOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50] = {0};
int size;
cout << \"How many numbers would you like to enter? \";
cin >> size;
cout << \"Enter numbers:\" << endl;
for(int i = 0;i < size;i++)
{
cin >> inputArray[i];
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void printFileScreen(const int uniqueArray[], const int countArray[], const int size)
{
//cout << \"jaja\";
ofstream output_file;
output_file.open(\"lab3Exercise2.txt\");
if(output_file.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
output_file << \"N\\t\" << \"count\ \";
for(int i = 0;i < size;i++)
{
if(uniqueArray[i] != 0)
{
output_file << uniqueArray[i] << \"\\t\";
outpu.
2024.06.01 Introducing a competency framework for languag learning materials ...
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
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Use of the plates within a specific time is necessary as time of inc.pdf
Use of the plates within a specific time is necessary as time of incubation is related to growth of
the bacterial cells which grows with time initially and after a particular period starts declining.
Moisture can increase the process of diffusion and also can spread bacterial cells into a zone
which is known as zone of inhibition.
Solution
Use of the plates within a specific time is necessary as time of incubation is related to growth of
the bacterial cells which grows with time initially and after a particular period starts declining.
Moisture can increase the process of diffusion and also can spread bacterial cells into a zone
which is known as zone of inhibition..
The five terrestrial adaptations of the seed plants include, reducti.pdf
The five terrestrial adaptations of the seed plants include, reduction of the gametophyte
generation, heterospory, ovules, and pollen.
Pollen grains were an important adaptation because the evolution of pollen allowed for
pollination and contributed to the diversity of seed plants.
Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole
structure is protected from desiccation and can reach the female organs without dependence on
water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an
extension of a cell within the pollen grain.
Seeds can survive harsh conditions through dormancy, are distributed far from their parent
sporophyte, and are multicellular.
Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain
growth and a protective coat give seeds their superior evolutionary advantage. Several layers of
hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of
water.
Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone
abscisic acid—until conditions for growth become favorable. Whether blown by the wind,
floating on water, or carried away by animals, seeds are scattered in an expanding geographic
range, thus avoiding competition with the parent plant.
All these various reasons make seeded plants to be domonant in plant communities today.
Solution
The five terrestrial adaptations of the seed plants include, reduction of the gametophyte
generation, heterospory, ovules, and pollen.
Pollen grains were an important adaptation because the evolution of pollen allowed for
pollination and contributed to the diversity of seed plants.
Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole
structure is protected from desiccation and can reach the female organs without dependence on
water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an
extension of a cell within the pollen grain.
Seeds can survive harsh conditions through dormancy, are distributed far from their parent
sporophyte, and are multicellular.
Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain
growth and a protective coat give seeds their superior evolutionary advantage. Several layers of
hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of
water.
Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone
abscisic acid—until conditions for growth become favorable. Whether blown by the wind,
floating on water, or carried away by animals, seeds are scattered in an expanding geographic
range, thus avoiding competition with the parent plant.
All these various reasons make seeded plants to be domonant in plant communities today..
SELF THEORY OF LATE ADULT HOOD it describes the changes of a adult.pdf
SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being
when they are getting aged.according to this theory
1.as the age of the particular person grows they will be more dependent on the others for their
needs.
2.some feelings like guilt,shame and depression develop as a result of their dependency.
3.many of the people will be feeling lonely when their loving partner passes away which will
have a negative impact on their health.
4.according to a theory stated by erikson people with a positive attitude will feel integrated
where as people with negative attitude or thoughts will be desperate.
STRATIFICATION THEORIES : These are related to the persons social categeory his social
status and the individual choices.this stratification is going to limit in many ways.many theories
have been stated they are as follows
Disengagement theory : the person disengages himself or herself away from the society and
become passive.
Activity theory : states that elderly people have to remain active in various social activities like
spending time with relatives,friends and various community groups and will be withdrawn only
as a result of age.
DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the
legacy of stratification.these dynamic theories states that each persons life will be chaning and is
purely self propelled within specific special contexts which will be ever changing.
Solution
SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being
when they are getting aged.according to this theory
1.as the age of the particular person grows they will be more dependent on the others for their
needs.
2.some feelings like guilt,shame and depression develop as a result of their dependency.
3.many of the people will be feeling lonely when their loving partner passes away which will
have a negative impact on their health.
4.according to a theory stated by erikson people with a positive attitude will feel integrated
where as people with negative attitude or thoughts will be desperate.
STRATIFICATION THEORIES : These are related to the persons social categeory his social
status and the individual choices.this stratification is going to limit in many ways.many theories
have been stated they are as follows
Disengagement theory : the person disengages himself or herself away from the society and
become passive.
Activity theory : states that elderly people have to remain active in various social activities like
spending time with relatives,friends and various community groups and will be withdrawn only
as a result of age.
DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the
legacy of stratification.these dynamic theories states that each persons life will be chaning and is
purely self propelled within specific special contexts which will be ever changing..
PV = PMTiThe goal is to solve the formula for i, doing so by isol.pdf
PV = PMT/i
The goal is to solve the formula for i, doing so by isolating it to one side of the equation.
The first step would be to cross multiply:
PV/1 = PMT/i
PV
Solution
PV = PMT/i
The goal is to solve the formula for i, doing so by isolating it to one side of the equation.
The first step would be to cross multiply:
PV/1 = PMT/i
PV.
Normal gene regulation in MAT locus YeastThere are 3 regulatory p.pdf
Normal gene regulation in MAT locus Yeast:
There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2.
These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable
opposite attracting pheromones.
a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha
factor and provokes the signalling process and there by repress the same..This happens the same
viceversa from alpha to a cells.
alpha 1 - protein activator.
alpha 2 - protein repressor.
a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets
of genes.
a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein
shows no growth, and also mating/sporulation defects.
It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate
or repress the MAT locus gene regulation.
So it\'s presence does not initiate any alteration in the gene regulation of MAT locus
Solution
Normal gene regulation in MAT locus Yeast:
There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2.
These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable
opposite attracting pheromones.
a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha
factor and provokes the signalling process and there by repress the same..This happens the same
viceversa from alpha to a cells.
alpha 1 - protein activator.
alpha 2 - protein repressor.
a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets
of genes.
a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein
shows no growth, and also mating/sporulation defects.
It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate
or repress the MAT locus gene regulation.
So it\'s presence does not initiate any alteration in the gene regulation of MAT locus.
Grey matter in the spinal cord is known as the grey column which tra.pdf
Grey matter in the spinal cord is known as the grey column which travels down the spinal cord
and is distributed in three grey columns that are presented in an \"H\" shape.
1. Anterior grey column.
It contains motor neurons. These synapse with interneurons and the axons of cells that have
travelled down the pyramidal tract. These cells are responsible for the movement of muscles.
2. Posterior grey column .
The posterior grey column contains the points where sensory neurons synapse. These receives
sensory information from the body, including fine touch, proprioception, and vibration. This
information is sent from receptors of the skin, bones, and joints through sensory neurons whose
cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the
spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the
spinothalamic tract.
3. Lateral grey column.
The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which
are part of the autonomic nervous system. It primarily involved with activity in the sympathetic
division of the autonomic motor system.
Solution
Grey matter in the spinal cord is known as the grey column which travels down the spinal cord
and is distributed in three grey columns that are presented in an \"H\" shape.
1. Anterior grey column.
It contains motor neurons. These synapse with interneurons and the axons of cells that have
travelled down the pyramidal tract. These cells are responsible for the movement of muscles.
2. Posterior grey column .
The posterior grey column contains the points where sensory neurons synapse. These receives
sensory information from the body, including fine touch, proprioception, and vibration. This
information is sent from receptors of the skin, bones, and joints through sensory neurons whose
cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the
spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the
spinothalamic tract.
3. Lateral grey column.
The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which
are part of the autonomic nervous system. It primarily involved with activity in the sympathetic
division of the autonomic motor system..
DateDescriptionDebitCredit2014Jan. 4Delivery Truck2800.pdf
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delivery truck
11680
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(28000 – 4640) / 4
= 11680
2015
Jan. 6
Delivery Truck
49400
Cash
49400
April 1
Depreciation Expense-Delivery Truck
1947
Delivery truck
1947
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12
= 1947
April 1
Cash
15180
Delivery truck
14373
Profit on sale of truck
807
Nov 2
Truck Repair Expense
470
Cash
470
Dec 31
Depreciation Expense-Delivery Truck
15972
Delivery truck
15972
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(49400-9470) / 5
= 15972
2016
July. 1
Delivery Truck
55400
Cash
55400
Oct 2
Depreciation Expense-Delivery Truck
8984
Delivery truck
8984
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(49400- 15972 )-9470 / 4 * 9/ 12
= 8984
Oct 2
Cash
17498
Loss on sale of truck
6946
Delivery truck
24444
Dec 31
Depreciation Expense-Delivery Truck
10611
Delivery truck
10611
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(55400-12955) / 8
= 10611
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delivery truck
11680
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(28000 – 4640) / 4
= 11680
2015
Jan. 6
Delivery Truck
49400
Cash
49400
April 1
Depreciation Expense-Delivery Truck
1947
Delivery truck
1947
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12
= 1947
April 1
Cash
15180
Delivery truck
14373
Profit on sale of truck
807
Nov 2
Truck Repair Expense
470
Cash
470
Dec 31
Depreciation Expense-Delivery Truck
15972
Delivery truck
15972
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(49400-9470) / 5
= 15972
2016
July. 1
Delivery Truck
55400
Cash
55400
Oct 2
Depreciation Expense-Delivery Truck
8984
Delivery truck
8984
Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset
* ( period used/12 month)
= 2*[(49400- 15972 )-9470 / 4 * 9/ 12
= 8984
Oct 2
Cash
17498
Loss on sale of truck
6946
Delivery truck
24444
Dec 31
Depreciation Expense-Delivery Truck
10611
Delivery truck
10611
Depreciation as per double declining method = 2*(cost – residual value)/ life of asset
= 2*(55400-12955) / 8
= 10611
Solution
Date
Description
Debit
Credit
2014
Jan. 4
Delivery Truck
28000
Cash
28000
Nov 2
Truck Repair Expense
635
Cash
635
Dec 31
Depreciation Expense-Delivery Truck
11680
Delive.
c. A destructor to free any allocated objects.d. A copy constructo.pdf
c. A destructor to free any allocated objects.
d. A copy constructor that allocates its own new member storage and copies the contents of
member variables.
e. An assignment operator that deallocates old storage before allocating new storage and copying
all the member variables.
Solution
c. A destructor to free any allocated objects.
d. A copy constructor that allocates its own new member storage and copies the contents of
member variables.
e. An assignment operator that deallocates old storage before allocating new storage and copying
all the member variables..
Barrier to entryThe government intervention to the barriers to en.pdf
Barrier to entry:
The government intervention to the barriers to entry to exist as a result of, while others occur
naturally within the business world. Often, existing firms within an industry lobby for the
government to erect new barriers to entry. In this is done to done by the protect the integrity of
the industry and prevent fly-by-night operations from setting up shop and hawking inferior
products and services. In reality, firms favor barriers to entry when already comfortably
ensconced in an industry to limit competition and claim a larger share of the industry\'s revenue.
Other barriers to entry occur naturally, often evolving over time as certain industry players
establish dominance.
A company can use this technology, for example, to build a barrier to entry, to build in switching
costs, and even, sometimes, to completely change the basis of competition. some companies
have seized the advantage, while others, more complacent, have ended up playing the difficult
and expensive game of catch-up ball. He also points out that it is important for executives to
make this competitive analysis in assessing where IS fits in their companies, since in some cases
it appropriately plays a support role and can add only modestly to the value of a company’s
products, while in other settings it is at the core of their competitive survival.
The computer’s main purpose is to cut order-entry costs and to provide more flexibility to
customers in the time and process of order submission. The system yields a larger competitive
advantage, adding value for customers and a substantial rise in their sales. The resulting sharp
increase in the company’s market share forces a primary competitor into a corporate
reorganization and a massive systems development effort to contain the damage.
Natural Barriers to Entry:
In industries where customers incur high costs switching from one brand to another, this
becomes a de facto barrier to entry for new firms, as they face difficulty enticing prospective
customers to pay the money required to make a chahange. Brand identity and customer loyalty
serve as barriers to entry for outsiders. Barriers to entry can also form naturally as the dynamics
of an industry take shape of the natural barrier.
Barrier option:
A barrier option is a type of option whose payoff depends on whether or not the underlying asset
has reached or exceeded a predetermined price. A barrier option can be a knock-out, meaning it
can expire worthless if the underlying exceeds a certain price, limiting profits for the holder but
limiting losses for the writer. It can also be a knock-in, meaning it has no value until the
underlying reaches a certain price.
BREAKING DOWN \'Barrier Option\':
Barrier options are considered a type of exotic option because they are more complex than basic
American or European options. Barrier options are also considered a type of path-dependent
option because their value fluctuates as the underlying\'s value changes during the opt.
Answer D all of the aboveLipids like triglycerides are semi-solid.pdf
Answer: D all of the above
Lipids like triglycerides are semi-solid at room temperature.
Lipids are not soluble in water
Steroids are a class of lipid molecules
Solution
Answer: D all of the above
Lipids like triglycerides are semi-solid at room temperature.
Lipids are not soluble in water
Steroids are a class of lipid molecules.
An example of when a one-way ANOVA could be used is if you want to d.pdf
An example of when a one-way ANOVA could be used is if you want to determine if there is a
difference in the mean height of stalks of three different types of seeds. Since there is more than
one mean, you can use a one-way ANOVA since there is only one factor that could be making
the heights different.
Now, if take these three different types of seeds, and then add the possibility that four different
types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of
the stalks could be different for a combination of several reasons:
The types of seed could cause the change,
the types of fertilizer could cause the change, and/or
there is an interaction between the type of seed and the type of fertilizer.
There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then
you can use a two-way ANOVA.
Solution
An example of when a one-way ANOVA could be used is if you want to determine if there is a
difference in the mean height of stalks of three different types of seeds. Since there is more than
one mean, you can use a one-way ANOVA since there is only one factor that could be making
the heights different.
Now, if take these three different types of seeds, and then add the possibility that four different
types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of
the stalks could be different for a combination of several reasons:
The types of seed could cause the change,
the types of fertilizer could cause the change, and/or
there is an interaction between the type of seed and the type of fertilizer.
There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then
you can use a two-way ANOVA..
10.dry and hot conditions11.C4 plants12.bundle sheet cells14.p.pdf
10.dry and hot conditions
11.C4 plants
12.bundle sheet cells
14.plasmodesmata
15)PEP carboxylase
16) bundle sheet cells
17) pineapple
18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if
takes place carboxylase activity will involve
19)corn
Solution
10.dry and hot conditions
11.C4 plants
12.bundle sheet cells
14.plasmodesmata
15)PEP carboxylase
16) bundle sheet cells
17) pineapple
18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if
takes place carboxylase activity will involve
19)corn.
1. Off Page refference arrow2. Connector Symbol3. Process4.Del.pdf
1. Off Page refference arrow
2. Connector Symbol
3. Process
4.Delay Symbol
5 .Merge Symbol
Solution
1. Off Page refference arrow
2. Connector Symbol
3. Process
4.Delay Symbol
5 .Merge Symbol.
1) the values of potassium, chlorine and HCO3 values are abnormal an.pdf
1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic
alkalosis which is causing high HCO3 values.
2) the pH is alkalotic and it is abnormal due to metabolic alkalosis
Solution
1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic
alkalosis which is causing high HCO3 values.
2) the pH is alkalotic and it is abnormal due to metabolic alkalosis.
The intermediate is the compound A. There are a c.pdf
The intermediate is the compound A. There are a couple of reasons for that: 1. The
oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic.
Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom
(the one that bears OH and then gets into the cycle) does not change in the reaction (retention of
the configuration). That supports the formation of the intermediate A. If the intermediate B
participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion
of the configuration).
Solution
The intermediate is the compound A. There are a couple of reasons for that: 1. The
oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic.
Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom
(the one that bears OH and then gets into the cycle) does not change in the reaction (retention of
the configuration). That supports the formation of the intermediate A. If the intermediate B
participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion
of the configuration)..
When ionic solids dissolve in water theions that are adjacent to each.pdf
When ionic solids dissolve in water theions that are adjacent to each other in the solid become
surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion
and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the
partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are
oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the
cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to
large ions. If this formation of the hydratedions were the only factor than we would expect all
ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we
need to condsider the other factors in thesolution process. the solution process is governed bythe
solute-solute, solvent-solvent and solute-solventintermolecular attractive forces. So far we have
only consideredhydration of the ions by the water molecules, that is thesoluteÐsolvent
interactions. When we consider the soluteinteractions we begin to see some of the problems that
can arise.The ions in a crystal are strongly attracted to each other and todissolve it is necessary to
overcome the electrostatic attractionbetween the oppositely charged ions. The lattice energy of a
solidis a measure of the strength of those electrostatic attractions.The lattice energy works to
keep the ions in the solid state andionic compounds with large lattice energies are insoluble in
waterwhile compounds with small lattice energies are soluble. Whenpotassium iodide dissolves
in water the ions is the potassiumiodide solid must be separated; KI(s) -ÐH2O -->K+(g) + I-(g)
ÆH1 = +632kJ/mol Now the gaseous ions are distributed inwater according to the equation;
H2O(l) + K+(g) +I-(g) -Ð-> K+(aq) + IÐ(aq)ÆH2 + ÆH3 = -617 kJ The sum of these two
equations yields theoverall solution process for KI dissolving in water. The value ofÆH2 + ÆH3
is call the hydrationenergy. In this example the hydration energy is not as large, inabsolute terms
as the energy required to separate the ions in thesolute and the heat of solution isendothermic.
Solution
When ionic solids dissolve in water theions that are adjacent to each other in the solid become
surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion
and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the
partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are
oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the
cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to
large ions. If this formation of the hydratedions were the only factor than we would expect all
ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we
need to condsider the other factors in.
#include iostream #include fstream #include map #include.pdf
#include
#include
#include
#include
#include
using namespace std;
//Used from
int index_of_largest(const int array[], int startIndex, const int size);
void selectSort(int a[], const int size);
void swap(int& i1, int& it2);
void fileOption();
void inputOption();
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size);
void printFileScreen(const int uniqueArray[], const int countArray[], const int size);
int main()
{
// int size;
char userInput[5];
//
cout << \"Read file from Input file? (Y/N) \";
cin >> userInput;
if(toupper(userInput[0]) == \'Y\')
{
fileOption();
}
else
{
inputOption();
}
return 0;
}
void selectSort(int array[], const int size)
{
int smallestIndex;
for(int i = 0;i < size - 1;i++)
{
smallestIndex = index_of_largest(array , i , size);
swap(array[i] , array[smallestIndex]);
}
}
void swap(int& i1, int& i2)
{
int temp;
temp = i1;
i1 = i2;
i2 = temp;
}
int index_of_largest(const int array[], int startIndex, const int size)
{
int min = array[startIndex] , index_of_min = startIndex;
for(int i = startIndex + 1;i < size;i++)
if(array[i] > min)
{
min = array[i];
index_of_min = i;
}
return index_of_min;
}
void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size)
{
int number;
int j = 0;
int index;
for(int i = 0;i < size;i++)
{
int flag = 0;
number = inputArray[i];
//for each element in inputArray, check if it is there in uniqueArray
for (int k = 0; k < j; ++k)
{
if(number == uniqueArray[k])
{
flag = 1;
index = k;
}
}
if(flag == 0)
{
//if so, increment the countArray by 1
countArray[j]++;
// copy the element to uniqueArray
uniqueArray[j] = number;
j++;
}
else
countArray[index]++;
}
}
void fileOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50]= {0};
ifstream in_stream;
char fileName[16];
cout << \"What file would you like to obtain input from? \";
cin >> fileName;
in_stream.open(fileName);
if(in_stream.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
int size = 0;
while(in_stream >> inputArray[size])
{
size++;
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void inputOption()
{
int inputArray[50];
int countArray[50] = {0};
int uniqueArray[50] = {0};
int size;
cout << \"How many numbers would you like to enter? \";
cin >> size;
cout << \"Enter numbers:\" << endl;
for(int i = 0;i < size;i++)
{
cin >> inputArray[i];
}
selectSort(inputArray , size);
unique(inputArray , uniqueArray , countArray , size);
printFileScreen(uniqueArray , countArray , size);
}
void printFileScreen(const int uniqueArray[], const int countArray[], const int size)
{
//cout << \"jaja\";
ofstream output_file;
output_file.open(\"lab3Exercise2.txt\");
if(output_file.fail())
{
cout << \"Error: Could not open file!\" << endl;
exit(1);
}
output_file << \"N\\t\" << \"count\ \";
for(int i = 0;i < size;i++)
{
if(uniqueArray[i] != 0)
{
output_file << uniqueArray[i] << \"\\t\";
outpu.
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