Chemistry Chapter 2 Study
Phosphorus forms several compounds with oxygen, including diphosphorus trioxide and diphosphorus pentoxide. # A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen. # The decomposition of a sample of diphosphorus pentoxide forms 0.775 g phosphorus to every 1.00 g oxygen. Show that these results are consistent with the law of multiple proportions.
Answer: (mass of phosphorus to 1.00 oxygen in diphosphorus trioxide) / (mass of phosphorus to 1.00 oxygen in diphosphorus pentoxide) = (1.29 g) / (0.775 g) = 1.66
An automobile gasoline tank holds 22 kg of gasoline. When the gasoline burns, 82 kg of oxygen is consumed, and carbon dioxide and water are produced. What is the total combined mass of carbon dioxide and water that is produced?
Answer: 104 kg (Simply add 22 and 82) (The law of conservation of mass states that matter is neither created nor destroyed in a chemical reaction. The mass of CO2 and H2O should be equal to the mass of gasoline and oxygen.)
There are two different compounds of sulfur and fluorine. # In SF6, the mass of fluorine per gram of sulfur is 3.55 gF/gS # In the other compound, SFx, the mass of fluorine per gram of sulfur is 1.18 gF/gS. What is the value of X for the second compound?
Answer: 2 (For compound one, you can divide 6 by 3.55 to get 1.6901... Using this ratio, you can multiply 1.6901... by 1.18 to get 1.9944... which can be rounded to 2)
Charges in Each Drop: -------------------------- A: 3.20 * 10^-19 B: 4.80 * 10^-19 C: 8.00 * 10^-19 D: 9.60 * 10^-19 -------------------------- Based on the data, how many electrons did drop C contain?
Answer: 5 (First find the smallest increment between any of these droplets. We can first see that A and B have the closest comparison. 4.80 * 10^-19 - 3.20 * 10^-19 = 1.60 * 10^-19 Now that we have 1.60 * 10^-19 we can now use this as a theoretical ratio for a single electron. To find the number of electrons in drop C, divide the charge of C by this 1 electron to find the total amount of electrons. 8.00 * 10^-19 / 1.60 * 10^-19 = 5.00 * 10^-19 5.00 * 10^-19 -> 5)
Three different compounds were analyzed and the masses of each element were determined. N = Nitrogen, O = Oxygen Compound A: Mass N (g) = 5.6, Mass O (g) = 3.2 Compound B: Mass N (g) = 3.5, Mass O (g) = 8.0 Compound C: Mass N (g) = 1.4, Mass O (g) = 4.0 If you have never heard of what a mole is or means, which of the following formulas for the three compounds would be the most likely correct formula? A: NO, NO2, NO4 B: NO2, NO8, NO10 C: NO4, NO10, NO5 D: N2O, N2O4, N2O5
Answer: D (For each compound, find what the mass is of oxygen per gram of nitrogen. A: 3.2 / 5.6 = 0.5714... B: 8.0/3.5 = 2.2857... C: 4.0/1.4 = 2.8571... By then comparing these ratios to the ratio in compound A (rounded to 0.57) we can figure out how much larger compound B and C to A by dividing. Comparing A and B: 2.3 (rounded) is around 4 times larger than 0.57 Comparing A and C: 2.9 (rounded) is around 5 times larger than 0.57. Thus, the relative amounts of oxygen in compounds A,B, and C, is a 1:4:5 ratio.
