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Pictures are not legible, could you plz post it again. Solution Pictures are not legible, could you plz post it again..

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In the First molecule there will be Resonance which will distribute the -ve charge between the two O atoms making it the least Basic. (Increase in concentrated -ve charge increases basicity) The first molecule i.e CH3CH2O-Li+. is the most basic due to the +I (Inductive) Effect of the CH3CH2- group. This group would push e- towards the O atom increasing its charge by delta-. Solution In the First molecule there will be Resonance which will distribute the -ve charge between the two O atoms making it the least Basic. (Increase in concentrated -ve charge increases basicity) The first molecule i.e CH3CH2O-Li+. is the most basic due to the +I (Inductive) Effect of the CH3CH2- group. This group would push e- towards the O atom increasing its charge by delta-..

In the First molecule there will be Resonance whi.pdf

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He down the group, IE decreases. Solution.pdf

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Uncouple agents will never disrupt the electron transport (they will not damage or disrupt the mitochondrial structure), they only inhibit the ATP synthesis and best example for this Uncouple agents are 2,4-dinitrophenol (DNP) and carbonylcyanide-ptrifluoromethoxyphenylhydrazone (FCCP) and these agents are known to bear H+ and are capable of diffusing into the matrix. Like this these two chemicals successfully reduce the electrochemical potential across the inner mitochondrial membrane and as a result you can see an inhibition of ATP synthesis. 2,4-Dinitrophenol, dicumarol and carbonyl cyanidep-trifluorocarbonyl-cyanide methoxyphenyl hydrazone (FCCP) chemicals have hydrophobic character, thus these are easily soluble in the bilipid membrane (this gives them any easy entry into the matrix in its protonated form and thus they releases a proton and can successfully dissipate the proton gradients. Another things through resonance stabilization will delocalize the charge which is there on anionic forms and thus this chemical again will diffuse back across the membrane will try to pick up a proton and will try to repeat the process) and in these chemical decouples by having the dissociable protons will carry protons from the intermembrane space to the matrix, so that the pH gradient existing there can be collapsed and all the potential energy of the proton gradient will be seen losing as mere heat. Research has showed when 2,4-dintrophenol is added to the reaction mixture, it has uncoupled electron transfer from that of ATP synthesis and as a result the oxygen is completely seen to be consumed in the absence of ADP. Solution Uncouple agents will never disrupt the electron transport (they will not damage or disrupt the mitochondrial structure), they only inhibit the ATP synthesis and best example for this Uncouple agents are 2,4-dinitrophenol (DNP) and carbonylcyanide-ptrifluoromethoxyphenylhydrazone (FCCP) and these agents are known to bear H+ and are capable of diffusing into the matrix. Like this these two chemicals successfully reduce the electrochemical potential across the inner mitochondrial membrane and as a result you can see an inhibition of ATP synthesis. 2,4-Dinitrophenol, dicumarol and carbonyl cyanidep-trifluorocarbonyl-cyanide methoxyphenyl hydrazone (FCCP) chemicals have hydrophobic character, thus these are easily soluble in the bilipid membrane (this gives them any easy entry into the matrix in its protonated form and thus they releases a proton and can successfully dissipate the proton gradients. Another things through resonance stabilization will delocalize the charge which is there on anionic forms and thus this chemical again will diffuse back across the membrane will try to pick up a proton and will try to repeat the process) and in these chemical decouples by having the dissociable protons will carry protons from the intermembrane space to the matrix, so that the pH gradient existing there can be collapsed and all the potenti.

Uncouple agents will never disrupt the electron transport (they will.pdf

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The N and O atoms are both sp2 hybridized. The sp2 hybrid orbitals (three each for N and O) either form sigma bonds or contain lone electron pairs. The pi bond is formed by overlap of the remaining unhybridized p orbital of N and the remaining unhybridized p orbital of O. Solution The N and O atoms are both sp2 hybridized. The sp2 hybrid orbitals (three each for N and O) either form sigma bonds or contain lone electron pairs. The pi bond is formed by overlap of the remaining unhybridized p orbital of N and the remaining unhybridized p orbital of O..

The N and O atoms are both sp2 hybridized.The sp2 hybrid orbitals .pdf

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d.Addition of sulfuric acid to copper(II) oxide p.pdf

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The 2nd statement and 3rd statement follow the seed and soil theory of metastasis or simply metastasis. It involves the main step “spreading of cells”. Integrins are responsible for signal transduction and do not move. Seed and soil theory is acceptable to the both statements. Micrometastases even remain dormant but have some growth in an organ. This organ is still work as soil for Micrometastases. So tumor cell are still as seed and organ is soil for them. Solution The 2nd statement and 3rd statement follow the seed and soil theory of metastasis or simply metastasis. It involves the main step “spreading of cells”. Integrins are responsible for signal transduction and do not move. Seed and soil theory is acceptable to the both statements. Micrometastases even remain dormant but have some growth in an organ. This organ is still work as soil for Micrometastases. So tumor cell are still as seed and organ is soil for them..

The 2nd statement and 3rd statement follow the seed and soil theory .pdf

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S2F6SolutionS2F6.pdf

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Quicksort Algorithm: Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort the sub-arrays. The steps are: The base case of the recursion is arrays of size zero or one, which never need to be sorted. The pivot selection and partitioning steps can be done in several different ways; the choice of specific implementation schemes greatly affects the algorithm\'s performance. This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the problem in two parts. And then two parts are solved recursively. The key concept is two count the number of inversion in merge procedure. In merge procedure we pass two sub-list. The element is sorted and inversion is found as follows a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n]. b)Conquer : Conquer the sub-problem by solving them recursively. 1) Set count=0,0,i=left,j=mid. C is the sorted list. 2) Traverse list1 and list2 until mid element or right element is encountered . 3) Compare list1[i] and list[j]. i) If list1[i]<=list2[j] c[k++]=list1[i++] else c[k++]=list2[j++] count = count + mid-i; 4) add rest elements of list1 and list2 in c. 5) copy sorted list c back to original list. 6) return count. void quickSort(int arr[], int left, int right) { int i = left, j = right; int tmp; int pivot = arr[(left + right) / 2]; /* partition */ while (i <= j) { while (arr[i] < pivot) i++; while (arr[j] > pivot) j--; if (i <= j) { tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; i++; j--; } }; /* recursion */ if (left < j) quickSort(arr, left, j); if (i < right) quickSort(arr, i, right); } Solution Quicksort Algorithm: Quicksort is a divide and conquer algorithm. Quicksort first divides a large array into two smaller sub-arrays: the low elements and the high elements. Quicksort can then recursively sort the sub-arrays. The steps are: The base case of the recursion is arrays of size zero or one, which never need to be sorted. The pivot selection and partitioning steps can be done in several different ways; the choice of specific implementation schemes greatly affects the algorithm\'s performance. This algorithm is based on Divide and Conquer paradigm. It is implemented using merge sort. In this approach the time complexity will be O(n log(n)) . Actually in divide step we divide the problem in two parts. And then two parts are solved recursively. The key concept is two count the number of inversion in merge procedure. In merge procedure we pass two sub-list. The element is sorted and inversion is found as follows a)Divide : Divide the array in two parts a[0] to a[n/2] and a[n/2+1] to a[n]. b)Conquer : Conquer the sub-problem by solving them recursively. 1) Set count=0,0,i=left,j=mid. C is the sorted list. 2) Traverse list1 and list2 until mid elem.

Quicksort AlgorithmQuicksort is a divide and conquer algorithm. Q.pdf

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Carbonic acid leaves the soda solution as CO2 (carbon dioxide). When a soda goes flat, all the CO2 leaves, however the H+ (acidic protons) are still left in the solution, leaving the solution acidic but flat. Solution Carbonic acid leaves the soda solution as CO2 (carbon dioxide). When a soda goes flat, all the CO2 leaves, however the H+ (acidic protons) are still left in the solution, leaving the solution acidic but flat..

Carbonic acid leaves the soda solution as CO2 (ca.pdf

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C. III S.pdf

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mass= density volume= 2.330.10.10.01 = 2.3310-4gmsatomic .pdf

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John is suffering from fifth disease. It is caused by an airborne virus called parvovirus B19. It causes a red rash on children\'s arms, legs and cheeks and hence it is also known as \"slapped cheek disease\". It is more common in children, but it can be more severe for pregnant women or anyone with a compromised immune system. For children with healthy immune systems, fifth disease is a common, mild illness that rarely presents lasting consequences. John can go to kindergarden- in the initial stages of this disease the symptoms such as headache, sore throat, running nose etc are seen, most young people develop a red rash that frist appears on cheeks, the rash often spreads to the arms, legs and trunk of the body within a few days. The rash may last for weeks, but usually by the time you see it you\'re no longer contagious. This infection is common among school age children but is rare in infants and adults Measles A paramyxovirus causes the disease, it usually begins with nasal congestion, eye redness, swelling and tearing, cough, lethargy and high fever. On the third or fourth day of the illness, the child will develop a red rash on the face, which dpreads rapidly and lasts about 7 days. A measles rash, which appears as red, itchy bumps, commonly develops on the head and slowly spreads to other parts of the body. Solution John is suffering from fifth disease. It is caused by an airborne virus called parvovirus B19. It causes a red rash on children\'s arms, legs and cheeks and hence it is also known as \"slapped cheek disease\". It is more common in children, but it can be more severe for pregnant women or anyone with a compromised immune system. For children with healthy immune systems, fifth disease is a common, mild illness that rarely presents lasting consequences. John can go to kindergarden- in the initial stages of this disease the symptoms such as headache, sore throat, running nose etc are seen, most young people develop a red rash that frist appears on cheeks, the rash often spreads to the arms, legs and trunk of the body within a few days. The rash may last for weeks, but usually by the time you see it you\'re no longer contagious. This infection is common among school age children but is rare in infants and adults Measles A paramyxovirus causes the disease, it usually begins with nasal congestion, eye redness, swelling and tearing, cough, lethargy and high fever. On the third or fourth day of the illness, the child will develop a red rash on the face, which dpreads rapidly and lasts about 7 days. A measles rash, which appears as red, itchy bumps, commonly develops on the head and slowly spreads to other parts of the body..

John is suffering from fifth disease.It is caused by an airborne v.pdf

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Initial concentration of IO4- = moles of NaIO4/final volume = 0.025 x 0.915/0.250 = 0.0915 M IO4- (aq) + 2 H2O (l) <=> H4IO6- (aq) I 0.0915 0 C -a +a E 0.0915-a a Kc = [H4IO6-]/[IO4-] = a/(0.0915 - a) = 3.5 x 10-2 [H4IO6-] = a = 3.09 x 10-3 M Solution Initial concentration of IO4- = moles of NaIO4/final volume = 0.025 x 0.915/0.250 = 0.0915 M IO4- (aq) + 2 H2O (l) <=> H4IO6- (aq) I 0.0915 0 C -a +a E 0.0915-a a Kc = [H4IO6-]/[IO4-] = a/(0.0915 - a) = 3.5 x 10-2 [H4IO6-] = a = 3.09 x 10-3 M.

Initial concentration of IO4- = moles of NaIO4final volume= 0.025.pdf

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atomic elements, Hg, Au, molecular elements, H2,.pdf

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In Drosophila the gene Tra encodes a protein that is expressed only in females. Individuals homozygous for this mutation be male but they are sterile.this is mainly because In males, the Sex lethal splice factor is not made; consequently, the Transformer message is not properly processed, and Transformer proteins are not produced. This causes an alternative splicing of Doublesex mRNA to an mRNA that codes for a male version of Doublesex protein. Solution In Drosophila the gene Tra encodes a protein that is expressed only in females. Individuals homozygous for this mutation be male but they are sterile.this is mainly because In males, the Sex lethal splice factor is not made; consequently, the Transformer message is not properly processed, and Transformer proteins are not produced. This causes an alternative splicing of Doublesex mRNA to an mRNA that codes for a male version of Doublesex protein..

In Drosophila the gene Tra encodes a protein that is expressed only .pdf

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Ecell = E0 - (0.0591n)log[H+]0.207 = 0.0591log[H+]1Therefore.pdf

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Explanation:- The processes that occur within the nucleus are Replication Transcription Post transcriptional modification Nucleolus synthesis Ribosome synthesis The histones are basic proteins that help in DNA packaging through nucleosme model .These proteins are synthesised during S phase of cell cycle at the time DNA replication . The nucleoporins are help in nuclear transport and are synthesised in the cytoplasm and then localised in the nuclear membrane. Solution Explanation:- The processes that occur within the nucleus are Replication Transcription Post transcriptional modification Nucleolus synthesis Ribosome synthesis The histones are basic proteins that help in DNA packaging through nucleosme model .These proteins are synthesised during S phase of cell cycle at the time DNA replication . The nucleoporins are help in nuclear transport and are synthesised in the cytoplasm and then localised in the nuclear membrane..

Explanation- The processes that occur within the nucleus areRepli.pdf

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Dominant strategy for Jane EarlyDominant strategy for Bob Early.pdf

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Critical infrastructures : Compose the asset or systems or network related , whether physical or virtual network, so vital to the US(united states) that their destruction would have a very bad effect on security, national economic security, national public health or safety. Elite hackers : the name used by the community that have the aim of identifying those individuals who are considered to be as experts in their line of work Hacker : a person who uses his own or other public computers to gain unauthorized access to data. Hacking : it is an attempt to make a computer system or a private network inside a computer vunerable or exploit it . it is basically the unauthorised access to or control over others computers or network, hacktivist : a person who takes unauthorized access to computer files or networks in order to use it for social or political ends highly structured threat : these are basically an Organized efforts to attack a specific target , it can be a organization specific or related to anything but that is also specific. information warfare : is a concept that involves the use and management of information and communication technology in avocation of a competitive advantage over an opposing party. Ping sweep :ping sweep is a method which establishes a range of IP addresses that can map to live hosts. The tool used for ping sweeps is fping Port scan : It is basically a attack in which the attacker sends packets to your machine, varying the destination port and find out the sevices you are running. Script kiddies : person who uses existing computer scripts or codes to hack into computers, as he lacks the expertise to write his own code or scripts Structured threat : Explained above in highly structured threat Unstructured threat : these are basically an Organized efforts to attack a un specific target , it can be a organization specific or related to anything but not specific. Solution Critical infrastructures : Compose the asset or systems or network related , whether physical or virtual network, so vital to the US(united states) that their destruction would have a very bad effect on security, national economic security, national public health or safety. Elite hackers : the name used by the community that have the aim of identifying those individuals who are considered to be as experts in their line of work Hacker : a person who uses his own or other public computers to gain unauthorized access to data. Hacking : it is an attempt to make a computer system or a private network inside a computer vunerable or exploit it . it is basically the unauthorised access to or control over others computers or network, hacktivist : a person who takes unauthorized access to computer files or networks in order to use it for social or political ends highly structured threat : these are basically an Organized efforts to attack a specific target , it can be a organization specific or related to anything but that is also specific. information warfare : .

Critical infrastructures Compose the asset or systems or network r.pdf

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Code: import java.util.Random; import java.util.Scanner; public class Test{ public static void main(String args[]){ Scanner input = new Scanner(System.in); String bin = \"1\"; int n; System.out.print(\"Enter the number of binary digits : \"); n = input.nextInt(); for(int i = 0; i < n-1; i++){ bin += randomBin(); } System.out.println(bin); } public static String randomBin() { Random rg = new Random(); int n = rg.nextInt(2); return Integer.toBinaryString(n); } } Output: Enter the number of binary digits : 5 10110 Solution Code: import java.util.Random; import java.util.Scanner; public class Test{ public static void main(String args[]){ Scanner input = new Scanner(System.in); String bin = \"1\"; int n; System.out.print(\"Enter the number of binary digits : \"); n = input.nextInt(); for(int i = 0; i < n-1; i++){ bin += randomBin(); } System.out.println(bin); } public static String randomBin() { Random rg = new Random(); int n = rg.nextInt(2); return Integer.toBinaryString(n); } } Output: Enter the number of binary digits : 5 10110.

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