10. The push button is pushed to the second stop when we pipette out.pdf

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10. The push button is pushed to the second stop when we pipette out or transfer the sample, and to completely void the sample without leaving any sample in the pipette tips 11. The button is to be released after it is withdrawn from the solution/liquid. If it is released prior to withdrawal, the solution reenters the pipette tip. 12. If the push button is pressed to the second stop and is released even before the pipette tip is withdrawn from the solution, the solution enters the pipette. 13. When concentrations or dilutions of the sample that are to be transferred changes, when different solutions or chemicals are transferred, when containment is required, tips need to be changed with each transfer. 14. The fragments obtained will have blunt ends if the restriction enzyme cuts at the center. Yes. They are still useful because linkers or adapters can be attached to the blunt end to make them useful. Moreover, new restriction sites can be created by attaching linkers or adapters. 15. Yes it can be done. The restriction enzyme creates the same overhanging ends since it recognizes the same restriction site in both the DNA. Therefore, the two different DNA can basepair at the sticky ends. Solution 10. The push button is pushed to the second stop when we pipette out or transfer the sample, and to completely void the sample without leaving any sample in the pipette tips 11. The button is to be released after it is withdrawn from the solution/liquid. If it is released prior to withdrawal, the solution reenters the pipette tip. 12. If the push button is pressed to the second stop and is released even before the pipette tip is withdrawn from the solution, the solution enters the pipette. 13. When concentrations or dilutions of the sample that are to be transferred changes, when different solutions or chemicals are transferred, when containment is required, tips need to be changed with each transfer. 14. The fragments obtained will have blunt ends if the restriction enzyme cuts at the center. Yes. They are still useful because linkers or adapters can be attached to the blunt end to make them useful. Moreover, new restriction sites can be created by attaching linkers or adapters. 15. Yes it can be done. The restriction enzyme creates the same overhanging ends since it recognizes the same restriction site in both the DNA. Therefore, the two different DNA can basepair at the sticky ends..

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moles NH4NO3 = 2.57 g 80.04 gmol=0.0321 moles H.pdf

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$mole fraction of He = mole of He total mole now.pdf$ $mole fraction of He = mole of He total mole now.pdf$

mole fraction of He = mole of He total mole now.pdf

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It is a molecular process involving the exchange of bonds between the two reacting chemical species, which results in the creation of products with similar or identical bonding affiliations. This is represented by the general reaction scheme: AX + BY ? BX + AY These chemical species can either be ionic or covalent. When referring to precipitation reactions between solutions of ions in inorganic chemistry, these were formerly referred to as double displacement or double replacement reactions, and these terms are still encouraged. It seems that the term double decomposition is more specifically used when at least one of the substances does not dissolve in the liquor, as the ligand or ion exchange takes place in the solid state of the reactant, i.e. AX(aq) + BY(s) ? AY(aq) + BX(s). Types: Neutralization A neutralization reaction is a specific type of double displacement reaction. Neutralization occurs when an acid reacts with an equal amount of a base. A neutralization reaction creates a solution of a salt and water. For example, hydrochloric acid reacts with sodium hydroxide to produce salt and water: HCl (aq) + NaOH (aq) ? NaCl (aq) + H2O (l) Aqueous metathesis (precipitation) Metathesis reactions can occur between two inorganic salts when one product is insoluble in water, driving the reaction forward. For example, the precipitation of silver chloride from a mixture of silver nitrate and sodium chloride causes sodium nitrate to be left in solution: AgNO3 (aq) + NaCl (aq) ? AgCl (s) + NaNO3 (aq) The formation of an insoluble gas that bubbles out of the solution, or a molecular compound such as water, also drives the reaction to completion. Therefore, a solubility chart (or general knowledge of solubility rules) can be used to predict whether two aqueous solutions will react. HSAB theory can also be used to predict the products of a metathesis reaction. Aqueous metathesis (double decomposition) The reactants need not to be dissolved for metathesis reactions to take place. An example of this is the formation of barium thiocyanate when boiling a mixture of copper(I)thiocyanate and barium hydroxide in water: Ba(OH)2 (s) + 2 CuCNS (s) ? Ba(CNS)2 (aq) + 2CuOH (s) Acid and carbonates A subcategory of aqueous metathesis reactions is the reaction of an acid with a carbonate or bicarbonate. Such a reaction always yields carbonic acid as a product, which spontaneously decomposes into carbon dioxide and water. The release of carbon dioxide gas from the reaction mixture drives the reaction to completion. For example, a common, science-fair \"volcano\" reaction involves the reaction of acetic acid with sodium bicarbonate: HCH3COO (aq) + NaHCO3 (s) ? NaCH3COO (aq) + CO2 (g) + H2O (l) Solution It is a molecular process involving the exchange of bonds between the two reacting chemical species, which results in the creation of products with similar or identical bonding affiliations. This is represented by the general reaction scheme: AX + BY ? BX + AY These che.

It is a molecular process involving the exchange .pdf

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Helium contains a 1s^2 orbital. .pdf

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From the LCAO method: The overlap of the 1s AO on one H atom with the is AO on a second gives one bonding MO (s) at lower energy to the constituent AOs and one antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO \"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5 so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H. (H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4 in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 - gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the charge will be +1. Solution From the LCAO method: The overlap of the 1s AO on one H atom with the is AO on a second gives one bonding MO (s) at lower energy to the constituent AOs and one antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO \"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5 so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H. (H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4 in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 - gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the charge will be +1..

From the LCAO method The overlap of the 1s AO on.pdf

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condense .pdf

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Cis-trans isomers occur because it is impossible to rotate the double bond without breaking it. Thus, isomers that are related by \"rotation\" about the double bond are actually separate compounds - cis/trans isomers. In the example you have drawn, keep the -CH2CH3 group and the right-hand H in place and reverse the position of the -CH2CH2CH3 group and the other H (so that the H\'s are on \"opposite\" sides of the double bond). This is the trans isomer of the compound you drew (you have the cis isomer). Now, imagine that instead of the CH2CH2CH3 group you had another H. In that case, switching the position of the two H\'s would result in the same compound, and cis-trans isomers would not be possible. In general, cis/trans isomers are possible only if each carbon that is part of the double bond is bonded to two DIFFERENT groups. This is only true for I and III. In II, the right-hand carbon of the double bond is bonded to two H\'s. And in IV, the left hand carbon of the double bond is bonded to two - CH3 groups. Thus neither of these has cis-trans isomers. So the correct answer is (b), I and III. Solution Cis-trans isomers occur because it is impossible to rotate the double bond without breaking it. Thus, isomers that are related by \"rotation\" about the double bond are actually separate compounds - cis/trans isomers. In the example you have drawn, keep the -CH2CH3 group and the right-hand H in place and reverse the position of the -CH2CH2CH3 group and the other H (so that the H\'s are on \"opposite\" sides of the double bond). This is the trans isomer of the compound you drew (you have the cis isomer). Now, imagine that instead of the CH2CH2CH3 group you had another H. In that case, switching the position of the two H\'s would result in the same compound, and cis-trans isomers would not be possible. In general, cis/trans isomers are possible only if each carbon that is part of the double bond is bonded to two DIFFERENT groups. This is only true for I and III. In II, the right-hand carbon of the double bond is bonded to two H\'s. And in IV, the left hand carbon of the double bond is bonded to two - CH3 groups. Thus neither of these has cis-trans isomers. So the correct answer is (b), I and III..

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When a heterozygote (Aa) is not an exact intermediate between the two homozygotes (AA and aa), then not all of the genetic variance in the population can be accounted for by differences in the average effect of the A and a alleles. In this case, the total genetic variance of the population can be subdivided into two components: Additive Genetic Variance (sa2): Genetic variation attributed to average differences between different allelic characters. Dominance Genetic Variance (sd2): Variance due to heterozygotes not being exact midpoints of homozygote parents. So that, Sg2 = Sa2 + Sd2 And now, total phenotypic variance can be rewritten as: Sp2 = Se2 + Sg2 = Se2 + Sa2 + Sd2 From this, we can define a new kind of heritability: Heritability in the narrow sense (h2):Proportion of phenotypic variance due to additive genetic variance alone h2 = Sa2 / Sp2 = Sa2 / Se2 + Sg2 = Sa2 / Se2 + Sd2 + Sa2 The greater h2 is for a population, the greater the fraction of the difference between selected parents and the population as a whole that will be preserved in the offspring of those parents. ______________________________________________________________ To see why it is h2 that matters for response to selection, consider an extreme case: 2 homozygotes are equal in phenotype, say with a value of 20. (AA=aa=20) But the heterozygote is overdominant, maybe with a value of 24. (Aa=24) So for this population, there is only dominance variance, and no additive genetic variance because there is no average difference between the carriers of a alleles and the carriers of A alleles. Now suppose that the population mates randomly and the genotype proportions are: If only individuals with the highest values were selected to breed, all parents would be heterozygotes. Offspring of heterozygotes would exactly reproduce the 1:2:1 distribution of genotypes seen in the parental generation. Thus, there would be no change in genetic composition of the F1 generation, despite the fact that there is genetic variation in the population. So, selection proves to be ineffective when all variance is due to dominance and there is no additive variance. It is important to consider that dominance variance arises from more than just the deviation of heterozygotes from the exact midpoint of homozygotes. Dominance variance (sd2) is considered to be any genetic variation that cannot be explained by average allelic differences, or additive genetic variance (sa2) For example, if a trait is affected by multiple loci then any epistatic interactions between the loci will appear as variance not associated with additive variance. In theory, this interaction variance (si2) could be separated from dominance variance (sd2). In practice, however, it is not possible to achieve accurate measures of separating these two parameters so all nonadditive variance is attributed to dominance variance (sd2). _____________________________________________________________________________ ________________ Important Mess.

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Astatine is a halogen. Column VII(7) on the periodic table. Therefore, the elements that would have similar properties to it are the other halogens: Iodine(I), Bromine(Br), Chlorine(Cl), Fluorine(F) Solution Astatine is a halogen. Column VII(7) on the periodic table. Therefore, the elements that would have similar properties to it are the other halogens: Iodine(I), Bromine(Br), Chlorine(Cl), Fluorine(F).

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The use of a generic data type is preferred over using Object as a general class in a collection because it ensures that all the members of the collection are objects related by inheritance. Using generic data type, the immediate benefit can be obtained without derivation from a base collection and implementation of type specific members Solution The use of a generic data type is preferred over using Object as a general class in a collection because it ensures that all the members of the collection are objects related by inheritance. Using generic data type, the immediate benefit can be obtained without derivation from a base collection and implementation of type specific members.

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RainFall.java public class RainFall { private double months[]; String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\", \"July\", \"August\", \"September\",\"October\",\"November\",\"December\"}; public RainFall(double d[]){ months = d; } public void displayTotalRainFall(){ double total = 0; for(int i=0; i months[i]){ min = months[i]; minIndex = i; } } System.out.println(\"The minimum rainfall is :\"+min); System.out.println(\"The minimum rainfall month is :\"+monthStr[minIndex]); } } RainFallTest.java import java.util.Scanner; public class RainFallTest { static String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\", \"July\", \"August\", \"September\",\"October\",\"November\",\"December\"}; public static void main(String[] args) { double month[] = new double[12]; Scanner scan = new Scanner(System.in); for(int i=0; i Solution RainFall.java public class RainFall { private double months[]; String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\", \"July\", \"August\", \"September\",\"October\",\"November\",\"December\"}; public RainFall(double d[]){ months = d; } public void displayTotalRainFall(){ double total = 0; for(int i=0; i months[i]){ min = months[i]; minIndex = i; } } System.out.println(\"The minimum rainfall is :\"+min); System.out.println(\"The minimum rainfall month is :\"+monthStr[minIndex]); } } RainFallTest.java import java.util.Scanner; public class RainFallTest { static String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\", \"July\", \"August\", \"September\",\"October\",\"November\",\"December\"}; public static void main(String[] args) { double month[] = new double[12]; Scanner scan = new Scanner(System.in); for(int i=0; i.

RainFall.java public class RainFall { private double months.pdf

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Proximal convoluted tubule performs absorption of majority of water and solutes in the kidney. It is lined by cuboidal epithelium. Urethral orifice is the region which is the external opening of the urethra. Solution Proximal convoluted tubule performs absorption of majority of water and solutes in the kidney. It is lined by cuboidal epithelium. Urethral orifice is the region which is the external opening of the urethra..

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Please find my implementation: public KeyedItem remove() { //DO THIS //remove the smallest item // since Linked List is in sorted order, so head is the samllest element if(head == null) return null; // return head element and move head one step forward KeyedItem item = head.getItem(); head = head.getNext(); return item; } private Node locateNodeAdd(KeyedItem item) { //DO THIS //find the insertion location (remember FIFO for duplicates) Node curr = head; Node prev = null; while(curr != null){ if((current.getItem().getKey()).compareTo(item.getKey) >= 0){ return prev; } prev = curr; curr = curr.getNext(); } return prev; } Solution Please find my implementation: public KeyedItem remove() { //DO THIS //remove the smallest item // since Linked List is in sorted order, so head is the samllest element if(head == null) return null; // return head element and move head one step forward KeyedItem item = head.getItem(); head = head.getNext(); return item; } private Node locateNodeAdd(KeyedItem item) { //DO THIS //find the insertion location (remember FIFO for duplicates) Node curr = head; Node prev = null; while(curr != null){ if((current.getItem().getKey()).compareTo(item.getKey) >= 0){ return prev; } prev = curr; curr = curr.getNext(); } return prev; }.

Please find my implementationpublic KeyedItem remove() { DO.pdf

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Oxidation means loss of electrons. Metals are electropositive and lose electrons easily to form positively charged ions with stable octet structures, thus they tend to be oxidized. Reduction means gain of electrons. Non-metals are electronegative and gain electrons easily to form negatively charged ions with stable octet structures, thus they tend to be reduced. Solution Oxidation means loss of electrons. Metals are electropositive and lose electrons easily to form positively charged ions with stable octet structures, thus they tend to be oxidized. Reduction means gain of electrons. Non-metals are electronegative and gain electrons easily to form negatively charged ions with stable octet structures, thus they tend to be reduced..

Oxidation means loss of electrons.Metals are electropositive and l.pdf

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Mutualism : 1. Two organisms live together and both gain from that association Chemotrophs : 2. Use chemicals to produce their own food Commensalism : 3. One organism benefits from the association and the other one is unaffected Autotrophs : 4. Produce their own food Photoautotrophs : 5. Use light to produce their own food Heterotrophs : 6. Feed on others Parasitism : 7. Affects others negatively Solution Mutualism : 1. Two organisms live together and both gain from that association Chemotrophs : 2. Use chemicals to produce their own food Commensalism : 3. One organism benefits from the association and the other one is unaffected Autotrophs : 4. Produce their own food Photoautotrophs : 5. Use light to produce their own food Heterotrophs : 6. Feed on others Parasitism : 7. Affects others negatively.

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a) The area of a disk of radius 1 is Pi. We need the integral over the unit disk of f(x,y) to be 1 so int_(x^2+y^2<=1) f(x,y) dy dx = c*Pi = 1 So c=1/Pi b) Fr(r)=P(R <= r) = P(X^2+Y^2 <= r^2) = int_(x^2+y^2<=r^2) 1/Pi dydx = Pir^2/pi ( since the disk of radius r^2 has area pir^2 ) = r^2 c) since x^2+y^2<=1 then -sqrt(1-x^2) <= y <= sqrt(1-x^2) fX(x) = int(-sqrt(1-x^2)<=y<=sqrt(1+x^2)) f(x,y) dy = 2/Pi*sqrt(1-x^2) for -1<=x<=1 Since we have a clear symmetry of variables here then fY(y) =2/pi*sqrt(1-y^2) Solution a) The area of a disk of radius 1 is Pi. We need the integral over the unit disk of f(x,y) to be 1 so int_(x^2+y^2<=1) f(x,y) dy dx = c*Pi = 1 So c=1/Pi b) Fr(r)=P(R <= r) = P(X^2+Y^2 <= r^2) = int_(x^2+y^2<=r^2) 1/Pi dydx = Pir^2/pi ( since the disk of radius r^2 has area pir^2 ) = r^2 c) since x^2+y^2<=1 then -sqrt(1-x^2) <= y <= sqrt(1-x^2) fX(x) = int(-sqrt(1-x^2)<=y<=sqrt(1+x^2)) f(x,y) dy = 2/Pi*sqrt(1-x^2) for -1<=x<=1 Since we have a clear symmetry of variables here then fY(y) =2/pi*sqrt(1-y^2).

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molality of ions in solution = van\'t Hoff factor x molality of NaCl = 1.9 x 1.55 = 2.945 m Freezing point of water - freezing point of NaCl solution = Kf x molality of ions freezing point of NaCl solution = 0 - 1.86 x 2.945 = -5.48 degrees C. Solution molality of ions in solution = van\'t Hoff factor x molality of NaCl = 1.9 x 1.55 = 2.945 m Freezing point of water - freezing point of NaCl solution = Kf x molality of ions freezing point of NaCl solution = 0 - 1.86 x 2.945 = -5.48 degrees C..

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