B. Router(config-if)#ip address dhcpThe ip address dhcp command mu.pdf
•
0 likes•3 views
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..
Report
Share
Report
Share
Download to read offline
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.
initially top=0, if top reaches the size then stack is overflow.pdf
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
}.
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!.
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.
initially top=0, if top reaches the size then stack is overflow.pdf
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
}.
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).
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.
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Reimagining Your Library Space: How to Increase the Vibes in Your Library No ...
Librarians are leading the way in creating future-ready citizens – now we need to update our spaces to match. In this session, attendees will get inspiration for transforming their library spaces. You’ll learn how to survey students and patrons, create a focus group, and use design thinking to brainstorm ideas for your space. We’ll discuss budget friendly ways to change your space as well as how to find funding. No matter where you’re at, you’ll find ideas for reimagining your space in this session.
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
The Diamonds of 2023-2024 in the IGRA collection
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
BBR 2024 Summer Sessions Interview Training
Qualitative research interview training by Professor Katrina Pritchard and Dr Helen Williams
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...Nguyen Thanh Tu Collection
https://app.box.com/s/tacvl9ekroe9hqupdnjruiypvm9rdaneMore Related Content
More from premsrivastva8
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).
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.
More from premsrivastva8 (11)
FunctionallyAs the name it self suggest LinkedList or single Linke.pdf
FunctionallyAs the name it self suggest LinkedList or single Linke.pdf
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
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
3. Compound XII only. other compounds dont hav.pdf
Recently uploaded
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Reimagining Your Library Space: How to Increase the Vibes in Your Library No ...
Librarians are leading the way in creating future-ready citizens – now we need to update our spaces to match. In this session, attendees will get inspiration for transforming their library spaces. You’ll learn how to survey students and patrons, create a focus group, and use design thinking to brainstorm ideas for your space. We’ll discuss budget friendly ways to change your space as well as how to find funding. No matter where you’re at, you’ll find ideas for reimagining your space in this session.
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
The Diamonds of 2023-2024 in the IGRA collection
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
BBR 2024 Summer Sessions Interview Training
Qualitative research interview training by Professor Katrina Pritchard and Dr Helen Williams
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...Nguyen Thanh Tu Collection
https://app.box.com/s/tacvl9ekroe9hqupdnjruiypvm9rdaneWound healing PPT
This document provides an overview of wound healing, its functions, stages, mechanisms, factors affecting it, and complications.
A wound is a break in the integrity of the skin or tissues, which may be associated with disruption of the structure and function.
Healing is the body’s response to injury in an attempt to restore normal structure and functions.
Healing can occur in two ways: Regeneration and Repair
There are 4 phases of wound healing: hemostasis, inflammation, proliferation, and remodeling. This document also describes the mechanism of wound healing. Factors that affect healing include infection, uncontrolled diabetes, poor nutrition, age, anemia, the presence of foreign bodies, etc.
Complications of wound healing like infection, hyperpigmentation of scar, contractures, and keloid formation.
South African Journal of Science: Writing with integrity workshop (2024)
South African Journal of Science: Writing with integrity workshop (2024)Academy of Science of South Africa
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
-------------------------------------------------------------------------------
Find out more about ISO training and certification services
Training: ISO/IEC 27001 Information Security Management System - EN | PECB
ISO/IEC 42001 Artificial Intelligence Management System - EN | PECB
General Data Protection Regulation (GDPR) - Training Courses - EN | PECB
Webinars: https://pecb.com/webinars
Article: https://pecb.com/article
-------------------------------------------------------------------------------
For more information about PECB:
Website: https://pecb.com/
LinkedIn: https://www.linkedin.com/company/pecb/
Facebook: https://www.facebook.com/PECBInternational/
Slideshare: http://www.slideshare.net/PECBCERTIFICATION
Main Java[All of the Base Concepts}.docx
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...Nguyen Thanh Tu Collection
https://app.box.com/s/y977uz6bpd3af4qsebv7r9b7s21935vdANATOMY AND BIOMECHANICS OF HIP JOINT.pdf
it describes the bony anatomy including the femoral head , acetabulum, labrum . also discusses the capsule , ligaments . muscle that act on the hip joint and the range of motion are outlined. factors affecting hip joint stability and weight transmission through the joint are summarized.
Hindi varnamala | hindi alphabet PPT.pdf
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
Iván Bornacelly, Policy Analyst at the OECD Centre for Skills, OECD, presents at the webinar 'Tackling job market gaps with a skills-first approach' on 12 June 2024
Recently uploaded (20)
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Reimagining Your Library Space: How to Increase the Vibes in Your Library No ...
Reimagining Your Library Space: How to Increase the Vibes in Your Library No ...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
South African Journal of Science: Writing with integrity workshop (2024)
South African Journal of Science: Writing with integrity workshop (2024)
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
B. Router(config-if)#ip address dhcpThe ip address dhcp command mu.pdf
- 1. 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.