CHEM 120 WEEK 8 FINAL EXAM

CHEM 120 WEEK 8 FINAL EXAM 2023 with A+ Grade Solutions 1. (TCO 8) 35.0 mL of 0.25 M NaOH is neutralized by 23.6 mL of an HCl solution. The molarity of the HCl solution is (show your work): (Points : 5) 2. (TCO 1) How many mL are in 3.5 pints? Show your work. (Points : 5) 3.5 pints is equivalent to 1656.116 1 pint = 473.176 ml 3.5 pints* 473.176mL = 1656.116mL 3. (TCO 3) What is the name of the following compound: Zn3P2? (Points : 5) It's Zinc Phosphide 4. (TCO 3) What is the name of the following compound: AgNO3? (Points : 5) Silver nitrate 5. (TCO 6) Calculate the pressure, in atmospheres, of 2.78 mol CO(g) in a 4.25 L tank at 51 degrees C. (Points : 5) Given that n = 2.78 mol; V = 4.25 L; and temperature [2.78 mol* 0.0821 L -atm/ mol-K*304 K)/4.25 L = 16.3 6. (TCO 6) A gas at a temperature of 95 degrees C occupies a volume of 165 mL. Assuming constant pressure, determine the volume at 25 degrees C. Show your work. (Points : 5) Answer: Using Charles’ Law, (V1/T1) = (V2/T2). First, convert temperature to KELVIN (T1 = t1 +273) Thus, T1 = 95 + 273 = 368. We have V1 (165 mL) & T2 = (25 + 273) = 298. V2 = (V1*T2)/T1 = (165 mL*298)/368 = 133.6 mL. 7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant temperature, determine the volume of helium at 745 mmHg. Show your work. (Points : 5) Answer: 1021mL * 719 mm/745 mmHg = 985.36mL =985mL Using Boyle’s law, P1V1 = P2V2. We have V1 (1021 mL), P1 (719 mmHg) and P2 (745 mmHg).