1 / 6

Agenda

Agenda. Attendance Hand in your Labs Matching Gas Laws Challenge Excepted….. Pass the problem on Hot air balloon project Super Duper Easy Homework. Match the Gas Laws. PV = nRT. Challenge Excepted .

hiroko
Download Presentation

Agenda

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Agenda Attendance Hand in your Labs Matching Gas Laws Challenge Excepted….. Pass the problem on Hot air balloon project Super Duper Easy Homework

  2. Match the Gas Laws PV = nRT

  3. Challenge Excepted Group A. How many grams of H2 is in a 3.1 L sample of H2 measured at 300 kPa and 20°C? Group B. How many grams of O2 are in a 315 mL container that has a pressure of 12 atm at 25°C?

  4. (300 kPa)(3.1 L) (8.31 kPa•L/K•mol)(293 K) (1215.9 kPa)(0.315 L) (8.31 kPa•L/K•mol)(298 K) Sample problems How many grams of H2 is in a 3.1 L sample of H2 measured at 300 kPa and 20°C? PV = nRT P = 300 kPa, V = 3.1 L, T = 293 K (300 kPa)(3.1 L) = n (8.31 kPa•L/K•mol)(293 K) = n = 0.38 mol 0.38g x 2.02g/mol= 0.7676g How many grams of O2 are in a 315 mL container that has a pressure of 12 atm at 25°C? PV = nRT P= 1215.9 kPa, V= 0.315 L, T= 298 K = n = 0.1547 mol 0.1547 mol x 32 g/mol = 4.95 g

  5. Pass It On • How many moles of CO2(g) is in a 5.6 L sample of CO2 measured at STP? • a) Calculate the volume of 4.50 mol of SO2(g) measured at STP. b) What volume would this occupy at 25°C and 150 kPa? (solve this 2 ways)

  6. (1 atm)(5.6 L) ( 0.08206L·atm·mol-1·K-1 )(273 K) (101.3 kPa) • Moles of CO2 is in a 5.6 L at STP? P= 1atm, V= 5.6 L, T= 273 K PV = nRT (1 atm)(5.6 L) = n ( 0.08206 L·atm·mol-1·K-1)(273 K) • a) Volume of 4.50 mol of SO2 at STP. P= 101.3 kPa, n= 4.50 mol, T= 273 K PV=nRT = n = 0.25 mol (101.3kPa)(V)=(4.5mol)(8.31kPa•L/K•mol)(273K) (4.50 mol)(8.31 kPa•L/K•mol)(273 K) V = = 100.8 L

More Related