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17 CHAPTER 2 ATOMS AND THE ATOMIC THEORY PRACTICE EXAMPLES 1A The total mass must be the same before and after reaction. mass before reaction = 0.382 g magnesium + 2.652 g nitrogen = 3.034 g mass after reaction = magnesium nitride mass + 2.505 g nitrogen = 3.034 g magnesium nitride mass = 3.034 g 2.505 g = 0.529 g magnesium nitride 1B Again, the total mass is the same before and after the reaction. mass before reaction = 7.12 g magnesium +1.80 g bromine = 8.92 g mass after reaction = 2.07 g magnesium bromide + magnesium mass = 8.92 g magnesium mass = 8.92 g 2.07 g = 6.85 g magnesium 2A In Example 2-2 we are told that 0.500 g MgO contains 0.301 g of Mg. With this information, we can determine the mass of magnesium needed to form 2.000 g magnesium oxide. 0.301 g Mg mass of Mg = 2.000 g MgO = 1.20 g Mg 0.500 g MgO The remainder of the 2.00 g of magnesium oxide is the mass of oxygen mass of oxygen = 2.00 g magnesium oxide 1.20 g magnesium = 0.80 g oxygen 2B In Example 2-2, we see that a 0.500 g sample of MgO has 0.301 g Mg, hence, it must have 0.199 g O 2 . From this we see that if we have equal masses of Mg and O 2 , the oxygen is in excess. First we find out how many grams of oxygen reacts with 10.00 g of Mg. 2 oxygen 2 2 2 0.199 g O mass =10.00 g Mg 6.61 g O (used up) 0.301 g Mg Hence, 10.00 g - 6.61 g = 3.39 g O unreacted. Mg is the limiting reactant. MgO(s) mass = mass Mg + Mass O = 10.00 g + 6.61 g = 16.6  2 1 g MgO. There are only two substances present, 16.61 g of MgO (product) and 3.39 g of unreacted O 3A Silver has 47 protons. If the isotope in question has 62 neutrons, then it has a mass number of 109. This can be represented as 109 47 Ag . 3B Tin has 50 electrons and 50 protons when neutral, while a neutral cadmium atom has 48 electrons. This means that we are dealing with Sn 2+ . We do not know how many neutrons tin has. so there can be more than one answer. For instance, 116 2+ 117 2+ 118 2+ 119 2+ 120 2+ 50 50 50 50 50 Sn , Sn , Sn , Sn , and Sn are all possible answers. 4A The ratio of the masses of 202 Hg and 12 C is: 202 12 Hg 201.97062u = =16.830885 C1 2 u
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Chapter 2: Atoms and the Atomic Theory 18 4B Atomic mass is 12 u × 13.16034 = 157.9241 u. The isotope is 158 64 Gd . Using an atomic mass of 15.9949 u for 16 O, the mass of 158 64 Gd relative to 16 O is 157.9241 u relative mass to oxygen-16 = 9.87340 15.9949 u 5A The average atomic mass of boron is 10.811, which is closer to 11.0093054 than to 10.0129370. Thus, boron-11 is the isotope that is present in greater abundance. 5B The average atomic mass of indium is 114.818, and one isotope is known to be 113 In. Since the weighted- average atomic mass is almost 115, the second isotope must be larger than both In-113 and In-114. Clearly, then, the second isotope must be In-115 ( 115 In). Since the average atomic mass of indium is closest to the mass of the second isotope, In- 115, then 115 In is the more abundant isotope. 6A Weighted-average atomic mass of Si = (27.9769265325 u × 0.9223) 25.80 u (28.976494700 u × 0.04685) 1.358 u (29.973377017 u × 0.03092) 0.9268 u 28.085 u We should report the weighted-average atomic mass of Si as 28.08 u. 6B We let x be the fractional abundance of lithium-6.
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Full Answers to Questions - CHAPTER 2 ATOMS AND THE ATOMIC...

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