initially top=0, if top reaches the size then stack is overflow.pdf
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initially top=0, if top reaches the size then stack is overflow, else push the element at top of the stack and increament the value top by 1 if (top == size) return false; // ** overflow error ** else { data[top] = newNode.deepCopy(); top = top + 1; return true; // push operation successful } Solution initially top=0, if top reaches the size then stack is overflow, else push the element at top of the stack and increament the value top by 1 if (top == size) return false; // ** overflow error ** else { data[top] = newNode.deepCopy(); top = top + 1; return true; // push operation successful }.
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![initially top=0,
if top reaches the size then stack is overflow,
else push the element at top of the stack and
increament the value top by 1
if (top == size)
return false; // ** overflow error **
else
{
data[top] = newNode.deepCopy();
top = top + 1;
return true; // push operation successful
}
Solution
initially top=0,
if top reaches the size then stack is overflow,
else push the element at top of the stack and
increament the value top by 1
if (top == size)
return false; // ** overflow error **
else
{
data[top] = newNode.deepCopy();
top = top + 1;
return true; // push operation successful
}](https://image.slidesharecdn.com/initiallytop0iftopreachesthesizethenstackisoverflow-230412092902-a2829191/85/initially-top-0-if-top-reaches-the-size-then-stack-is-overflow-pdf-1-320.jpg)
Recommended
1.Pros It is usually accepted today that almost all users wil.pdf
1.
Pros:
It is usually accepted today that almost all users will want web access and email ability, to this
end; it is of better convenience to the user if a web browser and email user are packaged with the
OS. Furthermore, coupling a web browser with the OS can give certain performance advantages.
For instance, because IE is coupled with Windows, it is cached while windows boots up - this
makes for quicker program loading, this is resisted to Mozilla which is not cached by Linux, and
so loads gradually every time it is invoked.
Cons:
The difficulties of Microsoft and its monopoly are obvious. It could also be disputed that the
function of an OS is to give a basis for requests and to act as mediator between a user and the
hardware. Thus addition of a web browser in the OS would be incorrect because it violates the
definition of an OS.
2.
Personal computers have all the need resources local to the machine and are able of handling all
requests locally. Network computers have very negligible resources locally and a minimalistic
operating system also. They depend on a network server for all their resource obligations. A
system that has a centralized storage that requires to be shared between several users is a good
scenario to utilize such a setup. For example, database class whole students require to work on a
central database server.
3.
Similarities
Differences
Solution
1.
Pros:
It is usually accepted today that almost all users will want web access and email ability, to this
end; it is of better convenience to the user if a web browser and email user are packaged with the
OS. Furthermore, coupling a web browser with the OS can give certain performance advantages.
For instance, because IE is coupled with Windows, it is cached while windows boots up - this
makes for quicker program loading, this is resisted to Mozilla which is not cached by Linux, and
so loads gradually every time it is invoked.
Cons:
The difficulties of Microsoft and its monopoly are obvious. It could also be disputed that the
function of an OS is to give a basis for requests and to act as mediator between a user and the
hardware. Thus addition of a web browser in the OS would be incorrect because it violates the
definition of an OS.
2.
Personal computers have all the need resources local to the machine and are able of handling all
requests locally. Network computers have very negligible resources locally and a minimalistic
operating system also. They depend on a network server for all their resource obligations. A
system that has a centralized storage that requires to be shared between several users is a good
scenario to utilize such a setup. For example, database class whole students require to work on a
central database server.
3.
Similarities
Differences.
The lewis structure for O2 is as follows .pdf
The lewis structure for O2 is as follows :
.. ..
: O = O :
Solution
The lewis structure for O2 is as follows :
.. ..
: O = O :.
The answer is e. carbon tetrachlorideIodine and carbon tetrachlori.pdf
The answer is e. carbon tetrachloride
Iodine and carbon tetrachloride are both nonpolar molecules with intermolecular dispersion (van
der Waal\'s) forces. Since their intermolecular forces are of the same type, their mutual solubility
will be greatest.
Vinegar, water and vodka contain polar molecules with hydrogen bonds. Their intermolecular
bonding is different from that of iodine, so their mutual solubility is lower.
Solution
The answer is e. carbon tetrachloride
Iodine and carbon tetrachloride are both nonpolar molecules with intermolecular dispersion (van
der Waal\'s) forces. Since their intermolecular forces are of the same type, their mutual solubility
will be greatest.
Vinegar, water and vodka contain polar molecules with hydrogen bonds. Their intermolecular
bonding is different from that of iodine, so their mutual solubility is lower..
range = 0mSolutionrange = 0m.pdf
The document provides a solution to set a range variable to 0 meters. It states the variable "range" and assigns it a value of 0 meters. No other information is given.
Moles = mass in grams over mass mole Lets have l0 grams total o.pdf
Moles = mass in grams over mass/ mole
Lets have l0 grams total of gas, 5 gms He( helium )and 5 gms Xe(xenon)
Moles He = 5g He( helium ) over 4 g/mole = 1.25 moles He
Moles Xe = 5 g Xe (xenon) over 131=.038 moles
total moles gases - .038 plus l.25 =1.288
pressure of each gas directly proportion to mole fraction.
Mole fraction He( helium ) = moles He ( helium ) over total moles
Mole fraction He( helium ) =1.25 over l.288 = .97
the mole fraction Xe (xenon) be 1-.97 or .03
X total pressure of 600 torr
Xe (xenon) its mole fraction x the total of 600 torr.
Solution
Moles = mass in grams over mass/ mole
Lets have l0 grams total of gas, 5 gms He( helium )and 5 gms Xe(xenon)
Moles He = 5g He( helium ) over 4 g/mole = 1.25 moles He
Moles Xe = 5 g Xe (xenon) over 131=.038 moles
total moles gases - .038 plus l.25 =1.288
pressure of each gas directly proportion to mole fraction.
Mole fraction He( helium ) = moles He ( helium ) over total moles
Mole fraction He( helium ) =1.25 over l.288 = .97
the mole fraction Xe (xenon) be 1-.97 or .03
X total pressure of 600 torr
Xe (xenon) its mole fraction x the total of 600 torr..
Manganese (VII)iodideCobalt (II) SulfiteSolutionManganese (V.pdf
This document lists two chemical compounds - Manganese (VII)iodide and Cobalt (II) Sulfite. It also mentions a solution but provides no other context or details about the compounds or solution.
Hello!MOLECULAR SOLIDS consist of atoms or molecules held together.pdf
Hello!
MOLECULAR SOLIDS consist of atoms or molecules held together by dipole-dipole forces,
London dispersion forces, and/or hydrogen bonds.
FALSE: Ionic solids have formula units in the point of the crystal lattice. Ionic solids have one
formula unit per unit cell.
What are the important intermolecular forces acting in CuSO4? London dispersion forces,
Dipole-Dipole Interactions and Ionic Bonding
I hope this helps!
Solution
Hello!
MOLECULAR SOLIDS consist of atoms or molecules held together by dipole-dipole forces,
London dispersion forces, and/or hydrogen bonds.
FALSE: Ionic solids have formula units in the point of the crystal lattice. Ionic solids have one
formula unit per unit cell.
What are the important intermolecular forces acting in CuSO4? London dispersion forces,
Dipole-Dipole Interactions and Ionic Bonding
I hope this helps!.
FunctionallyAs the name it self suggest LinkedList or single Linke.pdf
Functionally
As the name it self suggest LinkedList or single Linked list allows to traverse only in one
direction whereas in doubly linked list we can traverse both sides.
The information stored in single linked list would be info and pointer to to next node where as in
doubly linked list would be info, pointer to previous node , pointer to next node.
The element to be found if we know the possible nearest place where it could be found then
doubly linked list is better option for eg lets consider searching name which starts with a letter T
then we know that it will come in end in the directory hence we can loop from last directly but in
single list we have to loop from very first node sequentially as it moves only in one direction.
Performance
As signle linked list ony traverse in one direction hence it has only one pointer which takes less
memory than doubly linked list which as per design has two pointer to move in each direction.
Doubly linked list is faster in searching in comparision to single linked list.
Solution
Functionally
As the name it self suggest LinkedList or single Linked list allows to traverse only in one
direction whereas in doubly linked list we can traverse both sides.
The information stored in single linked list would be info and pointer to to next node where as in
doubly linked list would be info, pointer to previous node , pointer to next node.
The element to be found if we know the possible nearest place where it could be found then
doubly linked list is better option for eg lets consider searching name which starts with a letter T
then we know that it will come in end in the directory hence we can loop from last directly but in
single list we have to loop from very first node sequentially as it moves only in one direction.
Performance
As signle linked list ony traverse in one direction hence it has only one pointer which takes less
memory than doubly linked list which as per design has two pointer to move in each direction.
Doubly linked list is faster in searching in comparision to single linked list..
Recommended
1.Pros It is usually accepted today that almost all users wil.pdf
1.
Pros:
It is usually accepted today that almost all users will want web access and email ability, to this
end; it is of better convenience to the user if a web browser and email user are packaged with the
OS. Furthermore, coupling a web browser with the OS can give certain performance advantages.
For instance, because IE is coupled with Windows, it is cached while windows boots up - this
makes for quicker program loading, this is resisted to Mozilla which is not cached by Linux, and
so loads gradually every time it is invoked.
Cons:
The difficulties of Microsoft and its monopoly are obvious. It could also be disputed that the
function of an OS is to give a basis for requests and to act as mediator between a user and the
hardware. Thus addition of a web browser in the OS would be incorrect because it violates the
definition of an OS.
2.
Personal computers have all the need resources local to the machine and are able of handling all
requests locally. Network computers have very negligible resources locally and a minimalistic
operating system also. They depend on a network server for all their resource obligations. A
system that has a centralized storage that requires to be shared between several users is a good
scenario to utilize such a setup. For example, database class whole students require to work on a
central database server.
3.
Similarities
Differences
Solution
1.
Pros:
It is usually accepted today that almost all users will want web access and email ability, to this
end; it is of better convenience to the user if a web browser and email user are packaged with the
OS. Furthermore, coupling a web browser with the OS can give certain performance advantages.
For instance, because IE is coupled with Windows, it is cached while windows boots up - this
makes for quicker program loading, this is resisted to Mozilla which is not cached by Linux, and
so loads gradually every time it is invoked.
Cons:
The difficulties of Microsoft and its monopoly are obvious. It could also be disputed that the
function of an OS is to give a basis for requests and to act as mediator between a user and the
hardware. Thus addition of a web browser in the OS would be incorrect because it violates the
definition of an OS.
2.
Personal computers have all the need resources local to the machine and are able of handling all
requests locally. Network computers have very negligible resources locally and a minimalistic
operating system also. They depend on a network server for all their resource obligations. A
system that has a centralized storage that requires to be shared between several users is a good
scenario to utilize such a setup. For example, database class whole students require to work on a
central database server.
3.
Similarities
Differences.
The lewis structure for O2 is as follows .pdf
The lewis structure for O2 is as follows :
.. ..
: O = O :
Solution
The lewis structure for O2 is as follows :
.. ..
: O = O :.
The answer is e. carbon tetrachlorideIodine and carbon tetrachlori.pdf
The answer is e. carbon tetrachloride
Iodine and carbon tetrachloride are both nonpolar molecules with intermolecular dispersion (van
der Waal\'s) forces. Since their intermolecular forces are of the same type, their mutual solubility
will be greatest.
Vinegar, water and vodka contain polar molecules with hydrogen bonds. Their intermolecular
bonding is different from that of iodine, so their mutual solubility is lower.
Solution
The answer is e. carbon tetrachloride
Iodine and carbon tetrachloride are both nonpolar molecules with intermolecular dispersion (van
der Waal\'s) forces. Since their intermolecular forces are of the same type, their mutual solubility
will be greatest.
Vinegar, water and vodka contain polar molecules with hydrogen bonds. Their intermolecular
bonding is different from that of iodine, so their mutual solubility is lower..
range = 0mSolutionrange = 0m.pdf
The document provides a solution to set a range variable to 0 meters. It states the variable "range" and assigns it a value of 0 meters. No other information is given.
Moles = mass in grams over mass mole Lets have l0 grams total o.pdf
Moles = mass in grams over mass/ mole
Lets have l0 grams total of gas, 5 gms He( helium )and 5 gms Xe(xenon)
Moles He = 5g He( helium ) over 4 g/mole = 1.25 moles He
Moles Xe = 5 g Xe (xenon) over 131=.038 moles
total moles gases - .038 plus l.25 =1.288
pressure of each gas directly proportion to mole fraction.
Mole fraction He( helium ) = moles He ( helium ) over total moles
Mole fraction He( helium ) =1.25 over l.288 = .97
the mole fraction Xe (xenon) be 1-.97 or .03
X total pressure of 600 torr
Xe (xenon) its mole fraction x the total of 600 torr.
Solution
Moles = mass in grams over mass/ mole
Lets have l0 grams total of gas, 5 gms He( helium )and 5 gms Xe(xenon)
Moles He = 5g He( helium ) over 4 g/mole = 1.25 moles He
Moles Xe = 5 g Xe (xenon) over 131=.038 moles
total moles gases - .038 plus l.25 =1.288
pressure of each gas directly proportion to mole fraction.
Mole fraction He( helium ) = moles He ( helium ) over total moles
Mole fraction He( helium ) =1.25 over l.288 = .97
the mole fraction Xe (xenon) be 1-.97 or .03
X total pressure of 600 torr
Xe (xenon) its mole fraction x the total of 600 torr..
Manganese (VII)iodideCobalt (II) SulfiteSolutionManganese (V.pdf
This document lists two chemical compounds - Manganese (VII)iodide and Cobalt (II) Sulfite. It also mentions a solution but provides no other context or details about the compounds or solution.
Hello!MOLECULAR SOLIDS consist of atoms or molecules held together.pdf
Hello!
MOLECULAR SOLIDS consist of atoms or molecules held together by dipole-dipole forces,
London dispersion forces, and/or hydrogen bonds.
FALSE: Ionic solids have formula units in the point of the crystal lattice. Ionic solids have one
formula unit per unit cell.
What are the important intermolecular forces acting in CuSO4? London dispersion forces,
Dipole-Dipole Interactions and Ionic Bonding
I hope this helps!
Solution
Hello!
MOLECULAR SOLIDS consist of atoms or molecules held together by dipole-dipole forces,
London dispersion forces, and/or hydrogen bonds.
FALSE: Ionic solids have formula units in the point of the crystal lattice. Ionic solids have one
formula unit per unit cell.
What are the important intermolecular forces acting in CuSO4? London dispersion forces,
Dipole-Dipole Interactions and Ionic Bonding
I hope this helps!.
FunctionallyAs the name it self suggest LinkedList or single Linke.pdf
Functionally
As the name it self suggest LinkedList or single Linked list allows to traverse only in one
direction whereas in doubly linked list we can traverse both sides.
The information stored in single linked list would be info and pointer to to next node where as in
doubly linked list would be info, pointer to previous node , pointer to next node.
The element to be found if we know the possible nearest place where it could be found then
doubly linked list is better option for eg lets consider searching name which starts with a letter T
then we know that it will come in end in the directory hence we can loop from last directly but in
single list we have to loop from very first node sequentially as it moves only in one direction.
Performance
As signle linked list ony traverse in one direction hence it has only one pointer which takes less
memory than doubly linked list which as per design has two pointer to move in each direction.
Doubly linked list is faster in searching in comparision to single linked list.
Solution
Functionally
As the name it self suggest LinkedList or single Linked list allows to traverse only in one
direction whereas in doubly linked list we can traverse both sides.
The information stored in single linked list would be info and pointer to to next node where as in
doubly linked list would be info, pointer to previous node , pointer to next node.
The element to be found if we know the possible nearest place where it could be found then
doubly linked list is better option for eg lets consider searching name which starts with a letter T
then we know that it will come in end in the directory hence we can loop from last directly but in
single list we have to loop from very first node sequentially as it moves only in one direction.
Performance
As signle linked list ony traverse in one direction hence it has only one pointer which takes less
memory than doubly linked list which as per design has two pointer to move in each direction.
Doubly linked list is faster in searching in comparision to single linked list..
F2 and O2 are both diatomic molecules with no dip.pdf
F2 and O2 are both diatomic molecules with no dipole moment. The are not polar,
and are going to have weak intermolecular forces. CH2+ (radical) and CH2 (methylene group)
are different things, be certain which you are discussing. CH2 will likely have the strongest
intermolecular forces, as it will have a dipole moment within many other compounds.
Solution
F2 and O2 are both diatomic molecules with no dipole moment. The are not polar,
and are going to have weak intermolecular forces. CH2+ (radical) and CH2 (methylene group)
are different things, be certain which you are discussing. CH2 will likely have the strongest
intermolecular forces, as it will have a dipole moment within many other compounds..
ideal gas law= PV=nRT Daltons law of partial pr.pdf
ideal gas law= PV=nRT Dalton\'s law of partial pressure= P1+P2.....Pn (The
pressure of a mixture of gases is equal to the sum of the pressures of all of the constituent gases
alone)
Solution
ideal gas law= PV=nRT Dalton\'s law of partial pressure= P1+P2.....Pn (The
pressure of a mixture of gases is equal to the sum of the pressures of all of the constituent gases
alone).
Hey I couldnt click on it so I will explain it .pdf
The document explains how to determine the reaction mechanism and major product of a reaction. It involves a tertiary alkyl halide in an aprotic solvent. This indicates the reaction will proceed by an E1 mechanism, not Sn2, forming a carbocation intermediate. The most stable carbocation and most favored major product will be the secondary carbocation according to Zaitsev's rule.
H attached to N can form H bond with oxygen of w.pdf
H attached to N can form H bond with oxygen of water
Solution
H attached to N can form H bond with oxygen of water.
First, assign oxidation numbers.+1 +6 -2 +2 +1 .pdf
First, assign oxidation numbers.
+1 +6 -2 +2 +1 +3 +3
Na2Cr2O7 + Fe^2+ ------- Na^+ + Fe(OH)3 + Cr(OH)3
Note: If you do not know how to assign oxidation numbers, visit this link:
http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm
Next, decide what is being oxidized and what is being reduced.
Cr is being reduced because its oxidation number is decreasing.
Fe is being oxidized because its oxidation number is increasing.
Next, start oxidation half reaction.
Fe^2+ ------> Fe(OH)3
Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH\'s everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!
Next, start the reduction half reaction.
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won\'t matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equatio.
CountStringCharacters.javaimport java.util.Scanner; public cla.pdf
CountStringCharacters.java
import java.util.Scanner;
public class CountStringCharacters {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter a sentence or phrase: \");
String s = scan.nextLine();
int numOfCharacters = getNumOfCharacters(s);
System.out.println(\"You entered: \"+s);
System.out.println(\"Number of characters: \"+numOfCharacters);
outputWithoutWhitespace(s);
}
public static int getNumOfCharacters(String s){
int numOfCharCount = 0;
for(int i=0; i
Solution
CountStringCharacters.java
import java.util.Scanner;
public class CountStringCharacters {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter a sentence or phrase: \");
String s = scan.nextLine();
int numOfCharacters = getNumOfCharacters(s);
System.out.println(\"You entered: \"+s);
System.out.println(\"Number of characters: \"+numOfCharacters);
outputWithoutWhitespace(s);
}
public static int getNumOfCharacters(String s){
int numOfCharCount = 0;
for(int i=0; i.
chemical equation 2HF(aq) + Ca(OH)2(aq) = CaF2(s) + 2H2O(l) ion.pdf
chemical equation:
2HF(aq) + Ca(OH)2(aq) = CaF2(s) + 2H2O(l)
ionic equation:
2H+(aq) + 2F-(aq) + Ca2+(aq) + 2OH-(aq) = CaF2(s) + 2H2O(l)
Solution
chemical equation:
2HF(aq) + Ca(OH)2(aq) = CaF2(s) + 2H2O(l)
ionic equation:
2H+(aq) + 2F-(aq) + Ca2+(aq) + 2OH-(aq) = CaF2(s) + 2H2O(l).
B. Router(config-if)#ip address dhcpThe ip address dhcp command mu.pdf
B. Router(config-if)#ip address dhcp
The ip address dhcp command must be configured on the interface.
Solution
B. Router(config-if)#ip address dhcp
The ip address dhcp command must be configured on the interface..
2nd option is correctas tha no of elements in LHS set equal to tha.pdf
2nd option is correct
as tha no of elements in LHS set equal to tha no of element in RHS set
Solution
2nd option is correct
as tha no of elements in LHS set equal to tha no of element in RHS set.
3. Compound XII only. other compounds dont hav.pdf
3. Compound XII only. other compounds don\'t have a chiral center and hence
optically inactive
Solution
3. Compound XII only. other compounds don\'t have a chiral center and hence
optically inactive.
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Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
Temple of Asclepius in Thrace. Excavation results
The temple and the sanctuary around were dedicated to Asklepios Zmidrenus. This name has been known since 1875 when an inscription dedicated to him was discovered in Rome. The inscription is dated in 227 AD and was left by soldiers originating from the city of Philippopolis (modern Plovdiv).
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F2 and O2 are both diatomic molecules with no dip.pdf
F2 and O2 are both diatomic molecules with no dipole moment. The are not polar,
and are going to have weak intermolecular forces. CH2+ (radical) and CH2 (methylene group)
are different things, be certain which you are discussing. CH2 will likely have the strongest
intermolecular forces, as it will have a dipole moment within many other compounds.
Solution
F2 and O2 are both diatomic molecules with no dipole moment. The are not polar,
and are going to have weak intermolecular forces. CH2+ (radical) and CH2 (methylene group)
are different things, be certain which you are discussing. CH2 will likely have the strongest
intermolecular forces, as it will have a dipole moment within many other compounds..
ideal gas law= PV=nRT Daltons law of partial pr.pdf
ideal gas law= PV=nRT Dalton\'s law of partial pressure= P1+P2.....Pn (The
pressure of a mixture of gases is equal to the sum of the pressures of all of the constituent gases
alone)
Solution
ideal gas law= PV=nRT Dalton\'s law of partial pressure= P1+P2.....Pn (The
pressure of a mixture of gases is equal to the sum of the pressures of all of the constituent gases
alone).
Hey I couldnt click on it so I will explain it .pdf
The document explains how to determine the reaction mechanism and major product of a reaction. It involves a tertiary alkyl halide in an aprotic solvent. This indicates the reaction will proceed by an E1 mechanism, not Sn2, forming a carbocation intermediate. The most stable carbocation and most favored major product will be the secondary carbocation according to Zaitsev's rule.
H attached to N can form H bond with oxygen of w.pdf
H attached to N can form H bond with oxygen of water
Solution
H attached to N can form H bond with oxygen of water.
First, assign oxidation numbers.+1 +6 -2 +2 +1 .pdf
First, assign oxidation numbers.
+1 +6 -2 +2 +1 +3 +3
Na2Cr2O7 + Fe^2+ ------- Na^+ + Fe(OH)3 + Cr(OH)3
Note: If you do not know how to assign oxidation numbers, visit this link:
http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm
Next, decide what is being oxidized and what is being reduced.
Cr is being reduced because its oxidation number is decreasing.
Fe is being oxidized because its oxidation number is increasing.
Next, start oxidation half reaction.
Fe^2+ ------> Fe(OH)3
Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH\'s everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!
Next, start the reduction half reaction.
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won\'t matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equatio.
CountStringCharacters.javaimport java.util.Scanner; public cla.pdf
CountStringCharacters.java
import java.util.Scanner;
public class CountStringCharacters {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter a sentence or phrase: \");
String s = scan.nextLine();
int numOfCharacters = getNumOfCharacters(s);
System.out.println(\"You entered: \"+s);
System.out.println(\"Number of characters: \"+numOfCharacters);
outputWithoutWhitespace(s);
}
public static int getNumOfCharacters(String s){
int numOfCharCount = 0;
for(int i=0; i
Solution
CountStringCharacters.java
import java.util.Scanner;
public class CountStringCharacters {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter a sentence or phrase: \");
String s = scan.nextLine();
int numOfCharacters = getNumOfCharacters(s);
System.out.println(\"You entered: \"+s);
System.out.println(\"Number of characters: \"+numOfCharacters);
outputWithoutWhitespace(s);
}
public static int getNumOfCharacters(String s){
int numOfCharCount = 0;
for(int i=0; i.
chemical equation 2HF(aq) + Ca(OH)2(aq) = CaF2(s) + 2H2O(l) ion.pdf
chemical equation:
2HF(aq) + Ca(OH)2(aq) = CaF2(s) + 2H2O(l)
ionic equation:
2H+(aq) + 2F-(aq) + Ca2+(aq) + 2OH-(aq) = CaF2(s) + 2H2O(l)
Solution
chemical equation:
2HF(aq) + Ca(OH)2(aq) = CaF2(s) + 2H2O(l)
ionic equation:
2H+(aq) + 2F-(aq) + Ca2+(aq) + 2OH-(aq) = CaF2(s) + 2H2O(l).
B. Router(config-if)#ip address dhcpThe ip address dhcp command mu.pdf
B. Router(config-if)#ip address dhcp
The ip address dhcp command must be configured on the interface.
Solution
B. Router(config-if)#ip address dhcp
The ip address dhcp command must be configured on the interface..
2nd option is correctas tha no of elements in LHS set equal to tha.pdf
2nd option is correct
as tha no of elements in LHS set equal to tha no of element in RHS set
Solution
2nd option is correct
as tha no of elements in LHS set equal to tha no of element in RHS set.
3. Compound XII only. other compounds dont hav.pdf
3. Compound XII only. other compounds don\'t have a chiral center and hence
optically inactive
Solution
3. Compound XII only. other compounds don\'t have a chiral center and hence
optically inactive.
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F2 and O2 are both diatomic molecules with no dip.pdf
F2 and O2 are both diatomic molecules with no dip.pdf
ideal gas law= PV=nRT Daltons law of partial pr.pdf
ideal gas law= PV=nRT Daltons law of partial pr.pdf
Hey I couldnt click on it so I will explain it .pdf
Hey I couldnt click on it so I will explain it .pdf
H attached to N can form H bond with oxygen of w.pdf
H attached to N can form H bond with oxygen of w.pdf
First, assign oxidation numbers.+1 +6 -2 +2 +1 .pdf
First, assign oxidation numbers.+1 +6 -2 +2 +1 .pdf
CountStringCharacters.javaimport java.util.Scanner; public cla.pdf
CountStringCharacters.javaimport java.util.Scanner; public cla.pdf
chemical equation 2HF(aq) + Ca(OH)2(aq) = CaF2(s) + 2H2O(l) ion.pdf
chemical equation 2HF(aq) + Ca(OH)2(aq) = CaF2(s) + 2H2O(l) ion.pdf
B. Router(config-if)#ip address dhcpThe ip address dhcp command mu.pdf
B. Router(config-if)#ip address dhcpThe ip address dhcp command mu.pdf
2nd option is correctas tha no of elements in LHS set equal to tha.pdf
2nd option is correctas tha no of elements in LHS set equal to tha.pdf
3. Compound XII only. other compounds dont hav.pdf
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Chapter 2
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إضغ بين إيديكم من أقوى الملازم التي صممتها
ملزمة تشريح الجهاز الهيكلي (نظري 3)
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- 1. initially top=0, if top reaches the size then stack is overflow, else push the element at top of the stack and increament the value top by 1 if (top == size) return false; // ** overflow error ** else { data[top] = newNode.deepCopy(); top = top + 1; return true; // push operation successful } Solution initially top=0, if top reaches the size then stack is overflow, else push the element at top of the stack and increament the value top by 1 if (top == size) return false; // ** overflow error ** else { data[top] = newNode.deepCopy(); top = top + 1; return true; // push operation successful }