1. On average, human lungs can contain 5.8 g of air at sea level at room tempera
ID: 927380 • Letter: 1
Question
1. On average, human lungs can contain 5.8 g of air at sea level at room temperature (1.00 atm and 25 °C). Assuming air is mostly N2, what is the average volume of a pair of human lungs? (a) 0.42 L (b) 0.85 L (c) 5.1 L (d) 10. L (e) 140 L
2. Given the data below, calculate H° for the reaction: BaCO3(s) BaO(s) + CO2(g) 2Ba(s) + O2(g) 2BaO(s) H° = –1107.0 kJ Ba(s) + CO2(g) + ½O2(g) BaCO3(s) H° = –822.5 kJ (a) –1929.5 kJ (b) –1376.0 kJ (c) –284.5 kJ (d) 269.0 kJ (e) 537 kJ
3. Consider the following reaction: 3ClO– (aq) ClO3 – (aq) + 2Cl– (aq) K = 3.2×103 The concentrations present are: [Cl– ] = 0.50 M, [ClO3 – ] = 0.32 M, [ClO– ] = 0.24 M. Which of the following statements is true? (a) The system is at equilibrium (b) The reaction will proceed to the right (c) The reaction will proceed to the left (d) The system cannot reach equilibrium because the ClO3 – and Cl– concentrations are not at the stoichiometric ratio. (e) There is not enough information to determine whether the system is at equilibrium
4. Use the following data to calculate the average bond energy of an N–H bond. NH3(g) + HCl(g) NH4Cl(s) H° = –177 kJ N2(g) + 3H2(g) 2NH3(g) H° = –92.2 kJ N2(g) + 4H2(g) + Cl2(g) 2NH4Cl(s) H° = –629 kJ Bond Type Bond Energy (kJ/mol) NN 945 H–Cl 432 Cl–Cl 243 H–H 436 (a) 360 kJ/mol (b) 376 kJ/mol (c) 391 kJ/mol (d) 766 kJ/mol (e) 1173 kJ/mol
5. Consider the following reaction at some constant temperature: H2(g) + I2(g) 2HI(g) Kp = 16 In an experiment, H2(g) at 1.00 atm and I2(g) at 1.00 atm are allowed to react to reach equilibrium. At equilibrium, what is the partial pressure of HI(g)? (a) 1.5 atm (b) 0.67 atm (c) 1.3 atm (d) 0.75 atm (e) 1.0 atm
Explanation / Answer
1. On average, human lungs can contain 5.8 g of air at sea level at room temperature (1.00 atm and 25 °C). Assuming air is mostly N2, what is the average volume of a pair of human lungs? (a) 0.42 L (b) 0.85 L (c) 5.1 L (d) 10. L (e) 140 L
m = 5.8 g of Air
MW air = 29
mol = mass/MW = 5.8/28 = 0.2 mol
PV = nRt
V =nRT/P = 0.2*0.082*298/1= 5.06 L
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