#58 A Compound has the same elementX:elementY ratio no matter where it comes from. You can always use proportions: [2.00 g] O/ 1.50 g C = 1.33 < ratio of amounts from synthetic form. [6.35 g] O X 1.33 g C = 6.47g of O < amount of O from natural form. #63 First let's get the number of grams: 40.0 lb of peat X [453.6 g/lb] > 18144 grams of peat For the volumes, there are conversion factors for the cubic units but it is best to play it safe and start with the linear units. For the volume of peat: 14.00 in X [2.54 cm/in] = 35.56 cm 20.00 in X [2.54 cm/in] = 50.80 cm 30.00 in X [2.54 cm/in] = 76.20 cm [35.56 cm]X [50.80 cm]X [76.20 cm] = 137651.3376 cm^{3} ρ_{peat} = 18144 grams / 137651.3376 cm^{3} = 0.13181 g/cm^{3} Some conversion factors are written in items you see often, 1.00 gal = 3.7854 L < where have you seen this before? (rounded off to 3.8) Now, for the volume of soil: 1.9 gal X [3.7854 L] = 7.19226 = 7.19226 L 7.19226 L X [1000 cm^{3}/L] = 7192.26 cm^{3} < 1 mL = 1 cm^{3} ρ_{soil} = 18144 grams / 7192.26 cm^{3} = 2.5227 g/cm^{3} The density of top soil is larger than that of peat moss, so a volume of peat moss will be lighter than an equivalent volume of top soil. Answering this last part of the question in the way the author would consider correct requires a clear understanding of the English language, but we are studying chemistry. #64 We just look at the density formula, ρ = Mass / Volume *With the volume constant, mass is directly proportional to the density. *With the mass constant, volume is inversely proportional to the density. And volume of a sphre is given by, V_{sphere} = [4/3] π r^{3} = [1/16] π d^{3} The spheres (drawn to scale) would look like this, The formulas are not actually required, just know that volumes high density materials are heavier than equivalent volumes of low density materials. Because the density of silver (10.5 g/cm^{3}) is comparable to the density of lead (11.3 g/cm^{3}), the two spheres have nearly the same size. Since the density of aluminum is a lot smaller than both silver and lead, the aluminum sphere is a lot larger. #68 Mass of toluene added to sample = 58.58 g  32.65 g = 25.93 g volume of toluene = V = Mass / ρ = 25.93 g / [0.864 g / cm^{3}] = 30.01 g/cm^{3} Volume of solid = 50.00 cm^{3}  30.01 cm^{3} = 19.99 cm^{3} ρ_{solid} = 32.65 g / [19.99 cm^{3}] = 1.633 g/cm^{3} #69 The G scale: There are 209.1 ^{0}C between 0 ^{0}G and 100 ^{0}, so the conversion factor in terms of magnitude will be: 1.00 ^{0}G = 100 / 209.1 = 2.091 ^{0}C / ^{0}G The ^{0}C scale is 11.5 at 0 ^{0}G, then, ^{0}C to ^{0}G < add 11.5 ^{0}G to ^{0}C < substract 11.5 For the freezing point of water, [100 / 209.1] X [0^{0}C + 11.5^{0}G] = 5.50 ^{0}G #74a Volume = π r^{2} h Volume = &pi X [3.55 cm]^{2} X 75.3 cm = 2.98 X 10^{3} cm^{3} #74b Radius = diameter / 2 Radius = 12.9 in / 2 = 6.45 in Radius = 6.45 in X [2.54 cm/in] = 16.383 cm Height = 22.5 in X [2.54 cm/in] = 57.15 cm Volume in cm^{3} = π X [16.383 cm]^{2} X 57.15 cm = 48189 cm^{3} or 48189 X 10^{6} m^{3} Volume in m^{3} = π X [16.383 X 10^{2} m]^{2} X 57.15 ^{2} m = 0.048189 m^{3}, which is the same as adding the 10^{6} on front of the cm^{3} answer. #74c Volume X ρ = Mass = 48189 cm^{3} X 13.6 g/cm^{3} = 655370 grams or 655370 X 10^{3} kg < there are 1000 grams in 1.000 kg #76a Mass of gold = Ψ Mass of silver = 9.85  Ψ Volume of alloy = 0.675 cm^{3} = [Ψ grams / 19.3 g/cm^{3}] + [(9.85  Ψ) grams / 10.5 g/cm^{3}] Remember Lowest Common Denominator? Here the LCD is [19.3 g/cm^{3} X 10.5 g/cm^{3}], but we should remove the units for the moment, 0.675 X [19.3 X 10.5] = {[19.3 X 10.5] X [Ψ / 19.3]} + {[19.3 X 10.5] X [(9.85  Ψ) / 10.5]} 136.79 = {10.5Ψ} + {19.3 X (9.85  Ψ)} 136.79 = 10.5Ψ + 190.105  19.3Ψ Put Ψ's on one side, plain numbers on the other, 136.79  190.105 = 10.5Ψ  19.3Ψ 53.315 = 8.8Ψ < both sides should have the same sign 6.06 = Ψ < grams of gold 9.85  Ψ = 9.85  6.06 = 3.79 < grams of silver % gold in jewelry = [6.06 g of gold / 9.85 g of jewelry] X 100 = 61.52 %
