2) Answer is The cloud storage provider ensures that you have a bac.pdf
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FOR ACIDIC MEDIUM: First separate into half-reactions. Second balance all elements other than \"O\" and \"H\" Third balance \"O\" with \"H2O\" Fourth balance \"H\" with \"H+\" ions Fifth balance charge with electrons then repeat for the second half reaction. make sure the electrons are on opposite sides of you half reactions. (one reaction should have them in the reactants, the other the products) make the electrons \"cancel\" each other by multiplying the other half reaction by the number of the electrons. then add the two half reactions together by combining like terms. Solution FOR ACIDIC MEDIUM: First separate into half-reactions. Second balance all elements other than \"O\" and \"H\" Third balance \"O\" with \"H2O\" Fourth balance \"H\" with \"H+\" ions Fifth balance charge with electrons then repeat for the second half reaction. make sure the electrons are on opposite sides of you half reactions. (one reaction should have them in the reactants, the other the products) make the electrons \"cancel\" each other by multiplying the other half reaction by the number of the electrons. then add the two half reactions together by combining like terms..
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a.) Topoisomerase-Topoisomerases are enzymes used in the overwindin.pdf
NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound
(composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water,
therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium
equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq)
Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ ..........
0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x
Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb
= 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a
CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller
than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN-
] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = -
log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2
Solution
NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound
(composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water,
therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium
equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq)
Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ ..........
0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x
Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb
= 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a
CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller
than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN-
] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = -
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a.) Topoisomerase-Topoisomerases are enzymes used in the overwindin.pdf
NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound
(composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water,
therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium
equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq)
Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ ..........
0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x
Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb
= 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a
CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller
than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN-
] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = -
log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2
Solution
NaCN(aq) ? Na+(aq) + CN-(aq) Assuming that NaCN is an ionic compound
(composed of a monatomic cation and a polyatomic anion) it will completely dissociate in water,
therefore [NaCN] = [CN-] Then, since HCN (Prussic Acid) is a weak acid, write the equilibrium
equation between the acid and its conjugate base: ..... H2O(l) + CN-(aq) ? HCN(aq) + OH-(aq)
Pure solids and liquids are ignored in equilibrium reactions, so water is ignored. I ........ ..........
0.1 . . | . . . . 0 . . . . . 0 C ...... ............ -x . . | . . . .+x . . . . +x E ...... .......... 0.1 . . | . . . . x . . . . . x
Since Ka was given, change it to Kb by using: Ka * Kb = Kw 4.9*10^(-10) * Kb = 10^(-14) Kb
= 2.0*10^(-5) Then set up the base dissociation constant equation (since it was written with a
CN- (base) on the left and HCN (acid) on the right). It is assumed that x is significantly smaller
than [CN-], just to make the calculations easier (avoiding quadratics). Kb = [HCN] [OH-] / [CN-
] 2.0*10^(-5) = x² / 0.1 x² = 2.0*10^(-6) x = [OH-] = 1.4*10^(-3) M pOH = -log[OH-] = -
log(1.4*10^(-3)) = 2.8 pH = 14 - pOH = 14 - 2.8 = 11.2.
A) There are several factors taken into consideration before giving .pdf
Tanzania is seeking a credit rating to raise capital for infrastructure projects through bond issues. It has shown strong economic growth of 7% in 2014, reducing poverty and increasing incomes, allowing it to advance from low- to middle-income status. Tanzania recently issued a $500 million bond with a 5-year maturity to be repaid in semi-annual installments, though the price faced speculation. Further bond issues may have higher coupons given Tanzania's continued growth, with proceeds still intended for infrastructure development.
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here heat capacity of calorimeter = (50 x 4.186 x ( 37.92 - 29.11 ) - 50 x 4.186 x (
29.11 - 20.91 ))/dt = 127.673/dt where dt is the change in temp .. it is not given so i have
mentioned it as dt
Solution
here heat capacity of calorimeter = (50 x 4.186 x ( 37.92 - 29.11 ) - 50 x 4.186 x (
29.11 - 20.91 ))/dt = 127.673/dt where dt is the change in temp .. it is not given so i have
mentioned it as dt.
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gases do not have a fixed volume as they are to be compressed.
Solution
gases do not have a fixed volume as they are to be compressed..
3.) 2 methyl 2 penteneSolution3.) 2 methyl 2 pentene.pdf
H2CrO4 ? 2 H+ + CrO42- Ka = 5.83 x 10-8 Initial .0375 2(.150) 0 Change - x +x
+x Eq. .0375 -x .3 + x x Ka = [H+][A-]/[HA] 5.83E-8 = (.3 + x)(x)/(.0375 - x) Let\'s assume
that the value of [H+] is within +- 5%, so we can simplify the above eq. to make calculations
simpler. 5.83E-8 = (.3x + x^2)/(.0375) x = [CrO4^2-] = 7.3E-9 mol/L
Solution
H2CrO4 ? 2 H+ + CrO42- Ka = 5.83 x 10-8 Initial .0375 2(.150) 0 Change - x +x
+x Eq. .0375 -x .3 + x x Ka = [H+][A-]/[HA] 5.83E-8 = (.3 + x)(x)/(.0375 - x) Let\'s assume
that the value of [H+] is within +- 5%, so we can simplify the above eq. to make calculations
simpler. 5.83E-8 = (.3x + x^2)/(.0375) x = [CrO4^2-] = 7.3E-9 mol/L.
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Fe2O3(s)+3CO(g)=======>2Fe(s)+3CO2(g) 50.0g Fe2O3*(1 mol Fe2O3/
159.69 g Fe2O3)*(2 mol Fe/1 mol Fe2O3)*(55.85 g Fe/1 mol Fe) 34.97 g Fe [ theoretical yield]
% of yield = practical/theoretical * 100 = 32.8/34.97 *100 = 93.80 %
Solution
Fe2O3(s)+3CO(g)=======>2Fe(s)+3CO2(g) 50.0g Fe2O3*(1 mol Fe2O3/
159.69 g Fe2O3)*(2 mol Fe/1 mol Fe2O3)*(55.85 g Fe/1 mol Fe) 34.97 g Fe [ theoretical yield]
% of yield = practical/theoretical * 100 = 32.8/34.97 *100 = 93.80 %.
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a. Sn2/E2 mixture (secondary; strong nucleophile) b. Mostly Sn2 with some E2
(primary; strong nucleophile) c. E2 (secondary and very hindered, blocking Sn2) d. Sn1/E1
mixture (tertiary; water indicates it\'s going Sn1/E1) a. cyanide b. alcohol c. ester
Solution
a. Sn2/E2 mixture (secondary; strong nucleophile) b. Mostly Sn2 with some E2
(primary; strong nucleophile) c. E2 (secondary and very hindered, blocking Sn2) d. Sn1/E1
mixture (tertiary; water indicates it\'s going Sn1/E1) a. cyanide b. alcohol c. ester.
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