A) [Sn2+]^3 [Fe3+]^2 (dont include solids) B).pdf

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A) [Sn2+]^3 / [Fe3+]^2 (don\'t include solids) B) [OH-]^4 [Cu2+]^2 / [O2] (no solids or liquids) C) [Br-]^2 / [I-]^2 Solution A) [Sn2+]^3 / [Fe3+]^2 (don\'t include solids) B) [OH-]^4 [Cu2+]^2 / [O2] (no solids or liquids) C) [Br-]^2 / [I-]^2.

A) [Sn2+]^3 / [Fe3+]^2 (don't include solids) B) [OH-]^4 [Cu2+]^2 / [O2] (no
solids or liquids) C) [Br-]^2 / [I-]^2
Solution
A) [Sn2+]^3 / [Fe3+]^2 (don't include solids) B) [OH-]^4 [Cu2+]^2 / [O2] (no
solids or liquids) C) [Br-]^2 / [I-]^2

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HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06 Solution HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06.

HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf

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Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction. Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide. The product formed is 3-bromo 1-iodo 3 methylpentane Solution Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction. Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide. The product formed is 3-bromo 1-iodo 3 methylpentane.

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NaCl : It\'s the coulombic interaction between Na+ and Cl- due to which crystal exists. It\'s not ionic intereaction. It\'s ionic interaction. (very strong). NaCl dissolves in water. It\'s not due to hydrogen bonding. It\'s due to strong ion-dipole interactions. Initially, the positive and negative ion are only attracted to each other. The water molecules are hydrogen bonded to each other. If the crystal is to dissolve, these bonds must be broken. Negative chloride ions on the surface are attracted by neighboring positive sodium ions and by the partially positive hydrogen atom in the polar water molecule (See the graphic on the left). Similarly, the positive sodium ions are attracted by both chloride ions and the partially negative oxygen atom in the polar water molecule. A \"tug-of-war\" occurs for the positive and negative ions between the other ions in the crystal and the water molecules. Several water molecules are attracted to each of the ions. Whether the crystal dissolves is determined by which attractive force is stronger. If the internal ionic forces in the crystal are the strongest, the crystal does not dissolve. This is the situation in reactions where precipitates form. If the attractions for the ions by the polar water molecules are the strongest, the crystal will dissolve. This is the situation in sodium chloride. Once the ions are released from the crystals, the ions are completely surrounded by water molecules. Note that the proper atom in the water molecule must \"point\" toward the correct ion. The charge principle and the partial charges in the polar molecule determine the correct orientation. Partially negative oxygen atoms in the water molecule interact with the positive sodium ion. Partially positive hydrogen atoms in the water molecule interact with the negative chloride ion. Sugar : The interaction between sugar molecule and water is greater than between sugar molecules. So When sugar is put into water, it likes to form hydrogen bonds with water. So it breaks glycosidic linkages and forms hydrogen bonds with water. Initially it separates into polymer sugar chains, but later it dissociates into individual molecules. Solution NaCl : It\'s the coulombic interaction between Na+ and Cl- due to which crystal exists. It\'s not ionic intereaction. It\'s ionic interaction. (very strong). NaCl dissolves in water. It\'s not due to hydrogen bonding. It\'s due to strong ion-dipole interactions. Initially, the positive and negative ion are only attracted to each other. The water molecules are hydrogen bonded to each other. If the crystal is to dissolve, these bonds must be broken. Negative chloride ions on the surface are attracted by neighboring positive sodium ions and by the partially positive hydrogen atom in the polar water molecule (See the graphic on the left). Similarly, the positive sodium ions are attracted by both chloride ions and the partially negative oxygen atom in the polar water molecule. A \"tug-of-war\" occurs for.

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Here: Jupiter is the fifth planet from the Sun and the largest planet within the Solar System.It is a gas giant with a mass slightly less than one-thousandth of the Sun but is two and a half times the mass of all the other planets in our Solar System combined. Jupiter is classified as a gas giant along with Saturn, Uranus and Neptune. Together, these four planets are sometimes referred to as the Jovian or outer planets. The planet was known by astronomers of ancient times and was associated with the mythology and religious beliefs of many cultures. The Romans named the planet after the Roman god Jupiter.[14] When viewed from Earth, Jupiter can reach an apparent magnitude of -2.94, making it on average the third-brightest object in the night sky after the Moon and Venus. (Mars can briefly match Jupiter\'s brightness at certain points in its orbit.) Jupiter is primarily composed of hydrogen with a quarter of its mass being helium; it may also have a rocky core of heavier elements. Because of its rapid rotation, Jupiter\'s shape is that of an oblate spheroid (it possesses a slight but noticeable bulge around the equator). The outer atmosphere is visibly segregated into several bands at different latitudes, resulting in turbulence and storms along their interacting boundaries. A prominent result is the Great Red Spot, a giant storm that is known to have existed since at least the 17th century when it was first seen by telescope. Surrounding the planet is a faint planetary ring system and a powerful magnetosphere. There are also at least 63 moons, including the four large moons called the Galilean moons that were first discovered by Galileo Galilei in 1610. Ganymede, the largest of these moons, has a diameter greater than that of the planet Mercury. Jupiter has been explored on several occasions by robotic spacecraft, most notably during the early Pioneer and Voyager flyby missions and later by the Galileo orbiter. The most recent probe to visit Jupiter was the Pluto-bound New Horizons spacecraft in late February 2007. The probe used the gravity from Jupiter to increase its speed. Future targets for exploration in the Jovian system include the possible ice-covered liquid ocean on the moon Europa. Jupiter\'s structure: Jupiter is composed primarily of gaseous and liquid matter. It is the largest of four gas giants as well as the largest planet in the solar system with a diameter of 142,984 km at its equator. The density of Jupiter, 1.326 g/cm³, is the second highest of the gas giant planets. However, the density is lower than any of the four terrestrial planets. Solution Here: Jupiter is the fifth planet from the Sun and the largest planet within the Solar System.It is a gas giant with a mass slightly less than one-thousandth of the Sun but is two and a half times the mass of all the other planets in our Solar System combined. Jupiter is classified as a gas giant along with Saturn, Uranus and Neptune. Together, these four planets are sometimes refe.

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a. Times Interest Earned Ratio = EBIT / interest Expense For 2014: $8,799 / $600 = 14.7 times For 2015: $9,454 / $838 = 11.3 times For 2016: $10,338 / $1,058 = 9.8 times The TImes Interest Earned (TIE) Ratio measures the operating income level that can be used for interest payments in the future, i.e. how many times the Company can make the interest payments. It is a solvency ratio since it measures the Company\'s ability to service it\'s debt interest obligations. Hence, the higher the ratio, the more favorable is the company\'s liquidity position. In the given question, we can observe that the ratio has consistently decreased year on year. This implies that the Company\'s debt has increased while the increase in operating income has not been at the same pace. The Company can afford to service interest fewer times each year from 2014 to 2016. This implies that the Company has become comparatively more risky than before from the Creditor\'s standpoint. b. Current Ratio = Current Assets / Current Liabilities For 2014: $25,983 / $19,027 = 1.4 For 2015: $29,158 / $23,169 = 1.3 For 2016: $31,042 / $26,250 = 1.2 The Current Ratio is an indicator of how efficiently the Company manages it\'s operating cycle. A higher Current Ratio implies that the COmpany is in a better position to discharge all it\'s short term obligations without affecting or jeopardising it\'s working capital requirements. On the flipside, a ratio that is too high indicates that the Company\'s current assets are lying idle or that it is not efficiently using short term funds. In the Given case, the Company\'s current ratio shows a steady decline. However, the lowest is 1.2 times, which implies that the Company is still sufficiently liquid. This is becase even at the lowest, the Company\'s Current Assets can pay off the Current Liabilities 1.2 times without affecting working capital cycle for regular business needs. Hence, as per the options given, we can say that CVS’s current ratio has decreased over the past three years, however, it is greater than 1 indicating CVS is liquid. c. Total Liabilities to Equity Ratio = Total Liabilities / Total Equity For 2014: $36,224 / $37,963 = 0.95 For 2015: $55,234 / $37,203 = 1.5 For 2016: $57,628 / $36,834 = 1.6 Total Liabilities include all the claims, long term and short term, against the Company\'s Assets. Total Liabilities to Equity Ratio indicates how much of the Company\'s Assets are funded by creditors versus how much is funded by the equity holders. Broadly, a 1:1 ratio is considered acceptable for most industries. Creditors would prefer a lower ratio as this implies their money is at less risk. However, equity holders would like to benefit from the risk taken by them due to the Company using debt, and would prefer a higher ratio. In the given case, this ratio is increasing year on year. This implies that the COmpany is taking on more debt financing. This implies that the Company has more risk overall, although it benefits the equity .

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the bacteria isolated from the dead mouse is smooth (S) S. pneumoniae bacteria. and the DNA was radio-labeled. the Griffith experiment proves that bacteria are capable of transforming genetic information. in the above experiment Griffith grown the smooth (S) S. pneumoniae bacteria with radio-labelled phosphate (32P), radiolabelled 32P, which is incorporated into the DNA and he had also separately grown rough (R) bacteria in regular (not radio-labelled) phosphate. while injecting the heat killed S bacteria to live R bacteria, the R bacteria had been transformed into lethal S bacteria. the DNA had survived in Heating process of S strain, and this DNA transform into the R strain and synthesizes the polysaccharide capsule that protects form host immune response in the R bacteria and transform the R strain into the S strain which results Mouse dies. in the first step we are labeled the DNA of S Strain with 32P, so the final DNA which is isolating from the died mouse contains both radiolabeled DNA and non radiolabeled DNA in different proportions. Solution the bacteria isolated from the dead mouse is smooth (S) S. pneumoniae bacteria. and the DNA was radio-labeled. the Griffith experiment proves that bacteria are capable of transforming genetic information. in the above experiment Griffith grown the smooth (S) S. pneumoniae bacteria with radio-labelled phosphate (32P), radiolabelled 32P, which is incorporated into the DNA and he had also separately grown rough (R) bacteria in regular (not radio-labelled) phosphate. while injecting the heat killed S bacteria to live R bacteria, the R bacteria had been transformed into lethal S bacteria. the DNA had survived in Heating process of S strain, and this DNA transform into the R strain and synthesizes the polysaccharide capsule that protects form host immune response in the R bacteria and transform the R strain into the S strain which results Mouse dies. in the first step we are labeled the DNA of S Strain with 32P, so the final DNA which is isolating from the died mouse contains both radiolabeled DNA and non radiolabeled DNA in different proportions..

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H2SO4 = 2H+ + SO42-[H+] = 2[H2SO4] = 0.0092 MpH = -log[H+] = 2..pdf

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here reaction is occurring when Sn and I2 are refluxed together reduction and oxidation is takes place(called redox ) he would ask the oxidation potential &its reduction potential is asking . Sn + 2I2------>SnI4 here Sn oxidised to Sn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0= -0.136, I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Sn & it can be reduced. Zn(s) + I2(s) -----> Zn2+ (aq) + 2I- (aq) here Zn oxidised to Zn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0=-0.763 I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Zn & it can be reduced Sn(s)+2HCl (aq) -----> Sn2+ (aq) +2Cl- (aq) + H2 (g) here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136, cl- reduction potential is =+1.36 ,its indicates reduction potential of chlorine is more so it can oxidise the Sn & it(Cl) can be reduced. Sn2+ (aq) + 2I- (aq) -----> SnI2 (s) I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is less than the Sn2+ so it can reduce the Sn2+ & it(I-) can be oxidised. Solution here reaction is occurring when Sn and I2 are refluxed together reduction and oxidation is takes place(called redox ) he would ask the oxidation potential &its reduction potential is asking . Sn + 2I2------>SnI4 here Sn oxidised to Sn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0= -0.136, I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Sn & it can be reduced. Zn(s) + I2(s) -----> Zn2+ (aq) + 2I- (aq) here Zn oxidised to Zn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0=-0.763 I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Zn & it can be reduced Sn(s)+2HCl (aq) -----> Sn2+ (aq) +2Cl- (aq) + H2 (g) here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136, cl- reduction potential is =+1.36 ,its indicates reduction potential of chlorine is more so it can oxidise the Sn & it(Cl) can be reduced. Sn2+ (aq) + 2I- (aq) -----> SnI2 (s) I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is less than the Sn2+ so it can reduce the Sn2+ & it(I-) can be oxidised..

here reaction is occurring when Sn and I2 are refluxed together redu.pdf

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An undirected graph is acyclic (i.e., a forest) iff a DFS yields no back edges. Since back edges are those edges (u, v) connecting a vertex u to an ancestor v in a depth-first tree, so no back edges means there are only tree edges, so there is no cycle. So we can simply fun DFS. If find a back edge, there is a cycle. The complexity is O(V ) instead of O(E + V ). Since if there is a back edge, it must be found before seeing |V | distinct edges. This is because in a acyclic (undirected ) forest, |E| |V | + 1 Solution An undirected graph is acyclic (i.e., a forest) iff a DFS yields no back edges. Since back edges are those edges (u, v) connecting a vertex u to an ancestor v in a depth-first tree, so no back edges means there are only tree edges, so there is no cycle. So we can simply fun DFS. If find a back edge, there is a cycle. The complexity is O(V ) instead of O(E + V ). Since if there is a back edge, it must be found before seeing |V | distinct edges. This is because in a acyclic (undirected ) forest, |E| |V | + 1.

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Given LOSS Sells a variety of equpiment including the executive office chair Particulars Current No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Current manufacturing costs in $ per unit 175 Current Fixed cost 3,60,000 Proposed Changes No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Alternative manufacturing costs in $ per unit 75 Alternative Fixed cost 9,45,000 Sl Particulars Current Proposed A Sales 15,00,000 15,00,000 B Variable cost 10,50,000 4,50,000 C Contribution ( A - B ) 4,50,000 10,50,000 D Fixed cost 3,60,000 9,45,000 E Profit ( C - D ) 90,000 1,05,000 F price Volume Ratio % ( C/A * 100 ) 30% 70% G Break even Point Volume ( D/F) 12,00,000 13,50,000 H Contribution Per Unit ( C/6000 ) 75 175 I Break even Point Units ( D/H) 4800 5400 J Margin of safety sales ( A -G ) 3,00,000 1,50,000 K Margin of safety in Units ( J / selling Price pu ) 1200 600 L Margin of safety in Percentage ( J / A )*100 20% 10% Indiffernce point between two alternatives Indifference cost = point where there is no difference in cost between two alternatives Indifference point Units Assumptions Q = Quantity / units Cpu 1 = Contribution per unit for current alternative Cpu 2= Contribution per unit for proposed alternative FC 1 = fixed cost for current alternative FC 2 = fixed cost for Proposed alternative alternative Indifference point equation Cpu1 * Q - FC1 = Cpu2 * Q - FC2 = 75 * Q - 360,000 = 175 * Q -945,000 = 945,000 - 360,000 = Q ( 175 - 75 ) = 100 * Q = 585,000 = Q = 585,000/100 = Q = 5850 Units Therefore Indifference point is 5850 Units Given LOSS Sells a variety of equpiment including the executive office chair Particulars Current No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Current manufacturing costs in $ per unit 175 Current Fixed cost 3,60,000 Proposed Changes No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Alternative manufacturing costs in $ per unit 75 Alternative Fixed cost 9,45,000 Sl Particulars Current Proposed A Sales 15,00,000 15,00,000 B Variable cost 10,50,000 4,50,000 C Contribution ( A - B ) 4,50,000 10,50,000 D Fixed cost 3,60,000 9,45,000 E Profit ( C - D ) 90,000 1,05,000 F price Volume Ratio % ( C/A * 100 ) 30% 70% G Break even Point Volume ( D/F) 12,00,000 13,50,000 H Contribution Per Unit ( C/6000 ) 75 175 I Break even Point Units ( D/H) 4800 5400 J Margin of safety sales ( A -G ) 3,00,000 1,50,000 K Margin of safety in Units ( J / selling Price pu ) 1200 600 L Margin of safety in Percentage ( J / A )*100 20% 10% Indiffernce point between two alternatives Indifference cost = point where there is no difference in cost between two alternatives Indifference point Units Assumptions Q = Quantity / units Cpu 1 = Contribution per unit for current alternative Cpu 2= Contribution per unit for.

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Answer for question 1 DuPont analysis explains what is affecting company\'s ROE. DuPont analysis can be explained with formula as below ROE= Profit Margin× Total asset turnover× Leverage factor Profit Margin= Net Income/ Total Revenue Total Asset turnover=Revenue/ Total assets Leverage factor= Total assets/ Shareholder equity DuPont analysis points out operating efficiency in form of profit margin, Asset efficiency in form of asset turnover and leverage impact with the help of leverage factor as how much of the equity is used to finance total assets Solution Answer for question 1 DuPont analysis explains what is affecting company\'s ROE. DuPont analysis can be explained with formula as below ROE= Profit Margin× Total asset turnover× Leverage factor Profit Margin= Net Income/ Total Revenue Total Asset turnover=Revenue/ Total assets Leverage factor= Total assets/ Shareholder equity DuPont analysis points out operating efficiency in form of profit margin, Asset efficiency in form of asset turnover and leverage impact with the help of leverage factor as how much of the equity is used to finance total assets.

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