the heat gained by the cooler sample is equal to the heat lost by sample 1 (Q2=-Q1).

Q1Q 1 and Q2Q 2 are found from: Q1 = (specific heat of water) × (100 g)×(T2 - 70),are found from Q 1 equals the specific heat of water times the mass m times the change in temperature which equals the specific heat of water, times 100 grams times parenthesis, T 2 minus 70, close parenthesis Q2 = (specific heat of water) × (200 g) × (T2 - 40)Q 2 equals specific heat of water times 200 grams times parenthesis, T 2 minus 40 close parenthesis. Since Q2 = -Q1Q 2 equals Q 1, and since the specific heat of water cancels out, the following expression results: (200 g)×(T2 - 40) = (-100 g)×(T2 - 70)200 grams times parenthesis, T 2 minus 40, close parenthesis equals negative 100 grams times parenthesis, T 2 minus 70, close parenthesis or 2(T2 - 40) = (70 - T2)2 times parenthesis, T 2 minus 40, close parenthesis equals parenthesis, 70 minus T 2, close parenthesis.. Solving for T2T 2, T2T 2 is found to be 50°C50 degrees Celsius. The temperature decrease of the 100 g100 grams sample was 20ºC20 degrees Celsius, while the temperature increase of the 200 g200 grams sample was only 10°C10 degrees Celsius.` The balanced equation is: H2SO3 + 2 Mn + 4 H+

→→

S + 2 Mn2+ + 3 H2O. H 2 S O 3 plus, 2 M n, plus 4 H plus, react to produce, S plus 2 M n positive 2 ion, plus 3 H 2 OSince this is a redox reaction, the number of electrons lost in the oxidation process must equal the number of electrons gained in the reduction process. Also, the number of atoms of each element in the reactants must equal the number in the products. Sulfur has an oxidation number of +4plus 4 in H2SO3H 2 S O 3 and is reduced to SoS oxidation number of zero, while MnoM n oxidation number of zero is oxidized to Mn2+M n positive 2 ion. S4+ + 4e-

→→

SoS positive 4 ion, plus 4 electrons, react to produce S oxidation number of zero and 2 Mn

→→

2 Mn2+ + 4e-.2 M n react to produce 2 M n positive 2 ion, and 4 electrons Then the H and O atoms must be balanced. Based on the assumption that gasoline is about 80% carbon by mass, a mass of about 1,200 kg1,200 kilograms of carbon (0.80×1,500 kg = 1,200 kg)parenthesis, zero point eight zero times 1,500 kilograms equals 1,200 kilograms, close parenthesis is consumed in a combustion reaction with oxygen that produces CO2C O 2. If a complete combustion reaction occurs, then 12 g12 grams of carbon produces 44 g44 grams of carbon dioxide. Hence, if 1,200 kg1,200 kilograms of carbon is consumed, approximately 4,400 kg4,400 kilograms of CO2C O 2 is released per year by the automobile. Based on the proposed mechanism, the second step is the slow elementary reaction A + C

→→

BA reacts with C to produce B and is the rate-determining step. Hence, rate = k2[A][C]rate equals k 2, times the concentration of A times the concentration of C, where k2k 2 is the rate constant for the step. C is an intermediate and [C]the concentration of C can be found from K = [B][C]/[A],K equals, the concentration of B times the concentration of C, divided by the concentration of A where K is the equilibrium constant for the fast equilibrium step. Hence, [C] = K[A]/[B]the concentration of C equals K, times the concentration of A divided by the concentration of B and can be substituted into the rate expression yielding, rate = k2[A]K[A]/[B] = k2K[A]2/[B]rate equals k 2, times the concentration of A, times K times the concentration of A, divided by the concentration of B, equals k 2 K, times the concentration of A squared, divided by the concentration of B or k[A]2/[B]. 3rd EditionJames Girard371 solutions

6th EditionJanice Gorzynski Smith2,029 solutions

8th EditionDaniel C. Harris921 solutions

3rd EditionDavid Klein3,049 solutions