Chem 1045 - Study guides, Class notes & Summaries

CHEM 120 Week 8 Final Exam A+ GRADED 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) 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 ...

CHEM 120 WEEK 8 FINAL EXAM - QUESTION ONLY CHEM 120 WEEK 8 FINAL EXAM - QUESTION ONLY Page 1 1. Question: (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): 2. Question: (TCO 1) How many mL are in 3.5 pints? Show your work. 3. Question: (TCO 3) What is the name of the following compound: Zn3P2? 4. Question: (TCO 3) What is the name of the following compound: AgNO3? 5. Question: (TCO 6) Calculate the pressure...

CHEM 120 Week 8 Final Exam + Review+ Study Guide (Collection) Page 1 Question: (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): Question: (TCO 1) How many mL are in 3.5 pints? Show your work. Question: (TCO 3) What is the name of the following compound: Zn3P2? Question: (TCO 3) What is the name of the following compound: AgNO3? Question: (TCO 6) Calculate the pressure, in atmospheres, of 2.78 mol CO(g) in a 4.25...

CHEM 120 Week 8 Final Exam (Solved Q & A) Page 1 Question: (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): Questio n: (TCO 1) How many mL are in 3.5 pints? Show your work. Question: (TCO 3) What is the name of the following compound: Zn3P2? Question: (TCO 3) What is the name of the following compound: AgNO3? Question: (TCO 6) Calculate the pressure, in atmospheres, of 2.78 mol CO(g) in a 4.25 L tank at 51 degrees ...