a) It is given E is the event of getting 2. We can observe that 2 is.pdf

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a) It is given E is the event of getting 2. We can observe that 2 is an even number. It is given F is the event of getting an even number. Event E is giving an even number and event F is also giving an even number.That means both are occuring at the same time. so they are not mutually exclusive. b)The events E and F to be complementary their probabilities should add up to 1. P(E) = 1/6 P(F) = 3/6 P(E)+P(F) = 1/6 + 3/6 = 4/6 = 2/3 2/3 is not equal to one. so events E and F are not complementary. Solution a) It is given E is the event of getting 2. We can observe that 2 is an even number. It is given F is the event of getting an even number. Event E is giving an even number and event F is also giving an even number.That means both are occuring at the same time. so they are not mutually exclusive. b)The events E and F to be complementary their probabilities should add up to 1. P(E) = 1/6 P(F) = 3/6 P(E)+P(F) = 1/6 + 3/6 = 4/6 = 2/3 2/3 is not equal to one. so events E and F are not complementary..

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E is simply hexane...no polarity at all...nothing to speak of has extra electron density. A is a benzene ring, and if you play with the resonance you will see that sometimes the electrons can be in many different places which will make it polar at those times. :) nate Solution E is simply hexane...no polarity at all...nothing to speak of has extra electron density. A is a benzene ring, and if you play with the resonance you will see that sometimes the electrons can be in many different places which will make it polar at those times. :) nate.

E is simply hexane...no polarity at all...nothing.pdf

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Ferroxyl contains the chemical compound Potassium ferricyanide which reacts with the ferrous ions to produce the color \"Prussian blue\". Iron is being oxidized at the anode and creating Ferrous (Fe+2) ions. Therefore, with the introduction of Ferroxyl, the anode will turn blue as the Ferroxyl reacts with the Ferrous ions. It is also worth noting that this is a acidic solution and the blue pigment is insoluble in water. This is useful in detecting metal oxidation that leads to rust. Same thing basically for the OH. Solution Ferroxyl contains the chemical compound Potassium ferricyanide which reacts with the ferrous ions to produce the color \"Prussian blue\". Iron is being oxidized at the anode and creating Ferrous (Fe+2) ions. Therefore, with the introduction of Ferroxyl, the anode will turn blue as the Ferroxyl reacts with the Ferrous ions. It is also worth noting that this is a acidic solution and the blue pigment is insoluble in water. This is useful in detecting metal oxidation that leads to rust. Same thing basically for the OH..

Ferroxyl contains the chemical compound Potassium.pdf

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There are two different types of mechanisms. There are elimination and substitution reactions. The elimination reaction is when there is a proton transfer and loss of leaving group taking place. The substitution reaction is when there is a loss of leaving group and nucleophlic attack. From what we see in this reaction, we see that there is a nucleophilic attack with the MeI and OH. So this is a substitution reaction. Now to determine whether it is SN1 or SN2, we would have to look at the reaction. There is a tertary compound, which favors SN1 reactions better than SN2. The final answer would be SN1. I hope this makes sense. Let me know if anything is unclear. Solution There are two different types of mechanisms. There are elimination and substitution reactions. The elimination reaction is when there is a proton transfer and loss of leaving group taking place. The substitution reaction is when there is a loss of leaving group and nucleophlic attack. From what we see in this reaction, we see that there is a nucleophilic attack with the MeI and OH. So this is a substitution reaction. Now to determine whether it is SN1 or SN2, we would have to look at the reaction. There is a tertary compound, which favors SN1 reactions better than SN2. The final answer would be SN1. I hope this makes sense. Let me know if anything is unclear..

There are two different types of mechanisms. Ther.pdf

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What must be true about a set of data when its standard deviation is.pdf

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When one drives from sea level to trailhead in high Sierras, altitude increases. At higher altitudes, atmospheric pressure drops and so does the oxygen content of the air. We run out of enough oxygen in blood for our brain and body that leads to dizziness and heavy breathing. This is detrimental as it can even stop our brain due to lack of oxygen. These symptoms are collectively called Altitude sickness which reduces, as one spends some time in such region as our body gets acclimatized to such environment. Solution When one drives from sea level to trailhead in high Sierras, altitude increases. At higher altitudes, atmospheric pressure drops and so does the oxygen content of the air. We run out of enough oxygen in blood for our brain and body that leads to dizziness and heavy breathing. This is detrimental as it can even stop our brain due to lack of oxygen. These symptoms are collectively called Altitude sickness which reduces, as one spends some time in such region as our body gets acclimatized to such environment..

When one drives from sea level to trailhead in high Sierras, altitud.pdf

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the reaction as follows- 14N7 + 0n1 -- 4He2 + 1.pdf

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The important thing here is to understand the Standard Electrode Potential for each of our metals involved. Whichever of the metals has a larger negative number is the one that will be our anode, the other will be our cathode. Therefore, for this reaction, we know that Aluminum is going to be our anode. With this reasoning, we know that where the anode is, so will oxidation occur. If we think about it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and feeding it to the other side. Since we know the salt bridge is there to balance out the lost electrons from the oxidation side, we know that the NO3- are going to flow to the left hand side to make up for the electrons that are going from the left hand side to the right hand side. Your salt bridge sends anions to the left hand side! Hope this helps! Solution The important thing here is to understand the Standard Electrode Potential for each of our metals involved. Whichever of the metals has a larger negative number is the one that will be our anode, the other will be our cathode. Therefore, for this reaction, we know that Aluminum is going to be our anode. With this reasoning, we know that where the anode is, so will oxidation occur. If we think about it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and feeding it to the other side. Since we know the salt bridge is there to balance out the lost electrons from the oxidation side, we know that the NO3- are going to flow to the left hand side to make up for the electrons that are going from the left hand side to the right hand side. Your salt bridge sends anions to the left hand side! Hope this helps!.

The important thing here is to understand the Standard Electrode Pot.pdf

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The pathway would be overactive.The GTPase activity of the alpha s.pdf

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The metal ion is Pb2+.The white precipitate PbCl2 is insoluble in .pdf

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The electron carriers that feed into the ETC are NADH and FADH2.NA.pdf

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The developmental potential, or potency of a cell, describes the range of different cell types it can become. Therefore it involves. The determination means a cell restricts itself choosing a particular fate. Differentiation follows determination and the cell becomes a specific cell type ultimately. Solution The developmental potential, or potency of a cell, describes the range of different cell types it can become. Therefore it involves. The determination means a cell restricts itself choosing a particular fate. Differentiation follows determination and the cell becomes a specific cell type ultimately..

The developmental potential, or potency of a cell, describes the ran.pdf

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//Tested on linux,ubuntu #include #include #include #include #include /*Main function start*/ int main(int argc, char *argv[]) { /*variable for buffer*/ struct stat statbuf; /*pointer variable declaration*/ DIR *dp; struct dirent *dirp; /*if we are not providing file path then it will take current directory as name*/ if(argc==1) { dp = opendir(\"./\"); } else { dp = opendir(argv[1]); } while ( (dirp=readdir(dp)) != NULL) { /*lstat - get symbolic link status * Please readabout lstat more*/ lstat(dirp->d_name, &statbuf); /*checking it is directory or not*/ if(S_ISDIR(statbuf.st_mode)) { /*it is printing directory name with /*/ printf(\"%s/ \ \", dirp->d_name); } else { /*If it is file then we are printing only name of file*/ printf(\"%s\ \", dirp->d_name); } } /*cosing directory pointer*/ closedir(dp); exit(0); } /*main function end*/ /**********output**********/ gopal@gopal:~/Desktop/chegg$ gcc directory.c gopal@gopal:~/Desktop/chegg$ ./a.out dLinkedList.c random.c jasmin/ LinkedStack.cpp VideoGame.java LinkedStack.cpp Client.class web.xml ../ TemplateMain.cpp Energy.py PrecondViolatedExcept.h CSA.cpp BoxofProduce.cpp ./ Unanswere/ Tree.c target/ readFile.c inventory.txt goodQuestion interest.c VideoGameManager.java Thanks a lot Solution //Tested on linux,ubuntu #include #include #include #include #include /*Main function start*/ int main(int argc, char *argv[]) { /*variable for buffer*/ struct stat statbuf; /*pointer variable declaration*/ DIR *dp; struct dirent *dirp; /*if we are not providing file path then it will take current directory as name*/ if(argc==1) { dp = opendir(\"./\"); } else { dp = opendir(argv[1]); } while ( (dirp=readdir(dp)) != NULL) { /*lstat - get symbolic link status * Please readabout lstat more*/ lstat(dirp->d_name, &statbuf); /*checking it is directory or not*/ if(S_ISDIR(statbuf.st_mode)) { /*it is printing directory name with /*/ printf(\"%s/ \ \", dirp->d_name); } else { /*If it is file then we are printing only name of file*/ printf(\"%s\ \", dirp->d_name); } } /*cosing directory pointer*/ closedir(dp); exit(0); } /*main function end*/ /**********output**********/ gopal@gopal:~/Desktop/chegg$ gcc directory.c gopal@gopal:~/Desktop/chegg$ ./a.out dLinkedList.c random.c jasmin/ LinkedStack.cpp VideoGame.java LinkedStack.cpp Client.class web.xml ../ TemplateMain.cpp Energy.py PrecondViolatedExcept.h CSA.cpp BoxofProduce.cpp ./ Unanswere/ Tree.c target/ readFile.c inventory.txt goodQuestion interest.c VideoGameManager.java Thanks a lot.

Tested on linux,ubuntu#include stdlib.h #include systypes..pdf

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Since HEPES is a zwitterionic organic chemical buffering agent. At pH<7.0 it exists as cation (a positive ion , as negative charge is compensated by H+ present in the solution) to bring its pH to 7.45 (basic region) we must add NaOH. pH will increase when [H+] decreases. To decrease [H+] we must add NaOH ( a base). Solution Since HEPES is a zwitterionic organic chemical buffering agent. At pH<7.0 it exists as cation (a positive ion , as negative charge is compensated by H+ present in the solution) to bring its pH to 7.45 (basic region) we must add NaOH. pH will increase when [H+] decreases. To decrease [H+] we must add NaOH ( a base)..

Since HEPES is a zwitterionic organic chemical buffering agent.At .pdf

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public static class LinkedDeque implements DequeADT{ private int size; private Node first; private Node last; public LinkedDeque() { } // checking empty method or not public boolean isEmpty() { return first == null; } // checking size public int size() { return size; } // enqueing front public void enqueueFront(T num) { checkNull(num); Node tempNode = new Node(); tempNode.num = num; if (first != null) { // check root is not null tempNode.next = first; first.previous = tempNode; } first = tempNode; if (last == null) last = first; size++; } // removing front public Item dequeueFront() { CheckEmpty(); Node oldFirst = first; first = first.next; if (first == null) last = null; else first.previous = null; size--; return oldFirst.item; } // inseeting element at end of list public void enqueueBack(Item item) { checkNull(item); Node newLast = new Node(); newLast.item = item; if (last != null) { newLast.previous = last; last.next = newLast; } last = newLast; if (first == null) first = last; size++; } // removing element from end of the list public Item dequeueBack() { CheckEmpty(); Node oldLast = last; last = oldLast.previous; if (last == null) first = null; else last.next = null; size--; return oldLast.item; } private void CheckEmpty() { if (first == null) throw new NoSuchElementException(); } private void checkNull(Item item) { if (item == null) throw new NullPointerException(); } } Solution public static class LinkedDeque implements DequeADT{ private int size; private Node first; private Node last; public LinkedDeque() { } // checking empty method or not public boolean isEmpty() { return first == null; } // checking size public int size() { return size; } // enqueing front public void enqueueFront(T num) { checkNull(num); Node tempNode = new Node(); tempNode.num = num; if (first != null) { // check root is not null tempNode.next = first; first.previous = tempNode; } first = tempNode; if (last == null) last = first; size++; } // removing front public Item dequeueFront() { CheckEmpty(); Node oldFirst = first; first = first.next; if (first == null) last = null; else first.previous = null; size--; return oldFirst.item; } // inseeting element at end of list public void enqueueBack(Item item) { checkNull(item); Node newLast = new Node(); newLast.item = item; if (last != null) { newLast.previous = last; last.next = newLast; } last = newLast; if (first == null) first = last; size++; } // removing element from end of the list public Item dequeueBack() { CheckEmpty(); Node oldLast = last; last = oldLast.previous; if (last == null) first = null; else last.next = null; size--; return oldLast.item; } private void CheckEmpty() { if (first == null) throw new NoSuchElementException(); } private void checkNull(Item item) { if (item == null) throw new NullPointerException(); } }.

public static class LinkedDequeT implements DequeADTT{ priv.pdf

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No matter. you can write either the H (on the C with the Br) and Cl (on the C with the H) are on the wedges or the Br and H instead of the H and Cl, be drawn on wedges. Both will be same that is trans isomers. If he asks about R, S configuration in that case they may be different. Solution No matter. you can write either the H (on the C with the Br) and Cl (on the C with the H) are on the wedges or the Br and H instead of the H and Cl, be drawn on wedges. Both will be same that is trans isomers. If he asks about R, S configuration in that case they may be different..

No matter. you can write either the H (on the C with the Br) and Cl .pdf

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Moles of N2O5 = number of moleculesAvogadros number= 4.04 x 1020.pdf

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LOOPSFor LoopFor loop executes group of Java statements as long.pdf

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Intermediate levels of disturbance often lead to higher diversity because: Both species that do well at low disturbance and those that do well at high disturbance are usually present. Ecological succession refers to: Fairly predictable changes in species composition over time in a particular location. Solution Intermediate levels of disturbance often lead to higher diversity because: Both species that do well at low disturbance and those that do well at high disturbance are usually present. Ecological succession refers to: Fairly predictable changes in species composition over time in a particular location..

Intermediate levels of disturbance often lead to higher diversity be.pdf

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In the life cycle of a sexually reproducing organism, regulation of .pdf

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Initial volume = 10 L Decrease in Volume = 25% of 10 = 10/4 = 2.5 L Final Volume = 10 - 2.5 = 7.5 L Initial Pressure = 1.5 bar PV = nRt PV = Constant {Because Temperature remains Constant} Let the final pressure is P 1.5/7.5 = P/10 P = 2 bar So %age increase = (2-1.5)/1.5 * 100 = 33.33 % Solution Initial volume = 10 L Decrease in Volume = 25% of 10 = 10/4 = 2.5 L Final Volume = 10 - 2.5 = 7.5 L Initial Pressure = 1.5 bar PV = nRt PV = Constant {Because Temperature remains Constant} Let the final pressure is P 1.5/7.5 = P/10 P = 2 bar So %age increase = (2-1.5)/1.5 * 100 = 33.33 %.

Initial volume = 10 LDecrease in Volume = 25 of 10 = 104 = 2.5.pdf

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