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Entry Task: Nov 2 nd FridayPowerPoint Presentation

Entry Task: Nov 2 nd Friday

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Entry Task: Nov 2 nd Friday

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Question:

What is the rms speed of NO2at 45°C?

You have 5 minutes!!

- Discuss Graham’s Law ws
- Graham’s Law Lab
- HW: Pre-Lab Molar Mass of a volatile Liquid

10.61 Provide the following gases in order of increasing average molecular speed at 300K : CO2, N2O, HF, F2, H2. B) Calculate and compare the rms speeds of H2 and CO2 at 300K.

Convert g to kg

3*8.314 kg-m2/s2-mol-K * 300K

0.044 kg/mol

= 7482.6

0.044

= 412 or 4.12 x 102m/s of CO2

10.61 Provide the following gases in order of increasing average molecular speed at 300K : CO, N2O, HF, F2, H2. B) Calculate and compare the rms speeds of H2 and CO2 at 300K.

Convert g to kg

3*8.314 kg-m2/s2-mol-K * 300K

0.044 kg/mol

= 7482.6

0.044

= 412 or 4.12 x 102m/s of N2O

10.61 Provide the following gases in order of increasing average molecular speed at 300K : CO, N2O, HF, F2, H2. B) Calculate and compare the rms speeds of H2 and CO2 at 300K.

Convert g to kg

3*8.314 kg-m2/s2-mol-K * 300K

0.020 kg/mol

= 7482.6

0.020

= 612 or 6.12 x 102m/s of HF

10.61 Provide the following gases in order of increasing average molecular speed at 300K : CO, N2O, HF, F2, H2. B) Calculate and compare the rms speeds of H2 and CO2 at 300K.

Convert g to kg

3*8.314 kg-m2/s2-mol-K * 300K

0.0378 kg/mol

= 7482.6

0.0378

= 445 or 4.45 x 102m/s of F2

Convert g to kg

3*8.314 kg-m2/s2-mol-K * 300K

0.002 kg/mol

= 7482.6

0.002

= 1934 or 1.93 x 103 m/s of H2

= 1934 or 1.93 x 103 m/s of H2

= 412 or 4.12 x 102m/s of CO2

CO2 is WAY slow because its molecule is HUGE compared to hydrogen gas

A sample of hydrogen gas effuse through a porous container 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.

Set up problem so x is in the numerator on the right hand side of equation

Square both sides to get rid of the square root sign.

If CO2has an effusion rate of 0.00601moles/sec. What would the effusion rate for Xe be in moles/sec?

Set up problem so x is in the numerator on the left hand side of equation

3.48 x 10 -3 mols/sec

- NH3 has an effusion rate of 0.00314mol/sec. Which of the gases (listed in 10. 61) below would have an effusion rate of 0.00245mol/sec?

Set up problem so x is in the numerator on the right hand side of equation

Square both sides to get rid of the square root sign.

28.0 molar mass

In a given period of time, 0.214 moles of Ne effuses. How many moles of CO would effuse in to the same period of time?

Set up problem so x is in the numerator on the right hand side of equation

Square both sides to get rid of the square root sign.

0.182 moles of CO

- At a given temperature, the speed at which a gas molecule moves depends on its mass. This relationship, first discovered by Thomas Graham, is a result of the definition of kinetic energy. Graham studied two properties of gases that are related to the molecular speed: the rate of effusion (the escape of gas particles through a small hole) and diffusion (the spread of gas particles from an area of high concentration to low concentration).

- His experiments showed that these properties were related to the molecular mass of the gas molecules. What type of relation did he find? How is it related to the formula for kinetic energy? In this lab we will find the answers.

Convert g to kg

8.314kg-m2/s2-mol-K * Temp in K

molar mass

- Effusion is the movement of gas through a tiny hole.
- Diffusion is the movement of gas from a high concentration to a low concentration.
- We are studying the rate of diffusion

- All gases have the same average kinetic energy at the same temperature.

- HCl has a molar mass of 36.45 g/mol
- NH3 has a molar mass of 17.0 g/mol

Concentrated HCl

Concentrated NH4OH

DO NOT TOUCH OR breathe IT IN !!!