#47 on Chemical Equilibrium 1B Video Module Assessment


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Gisselle Sainz 2F
Posts: 76
Joined: Wed Feb 21, 2018 3:00 am

#47 on Chemical Equilibrium 1B Video Module Assessment

Postby Gisselle Sainz 2F » Mon Jan 14, 2019 11:38 pm

How do you solve for the partial pressures in this problem?
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Charles Hood Disc 1C
Posts: 39
Joined: Fri Sep 28, 2018 12:19 am

Re: #47 on Chemical Equilibrium 1B Video Module Assessment

Postby Charles Hood Disc 1C » Tue Jan 15, 2019 10:00 am

Use PV=nRT, then convert it to P=MRT, (n/v = concentration = M) Then proceed to use the values given in concentrations to solve for the pressures individually. These are the partial pressures.

Ava Kjos 1D
Posts: 65
Joined: Fri Sep 28, 2018 12:19 am

Re: #47 on Chemical Equilibrium 1B Video Module Assessment

Postby Ava Kjos 1D » Thu Jan 17, 2019 2:29 pm

Convert each concentration to a partial pressure using PV=nRT. You are given n/V, R and T so you just need to solve for P.

Gisselle Sainz 2F
Posts: 76
Joined: Wed Feb 21, 2018 3:00 am

Re: #47 on Chemical Equilibrium 1B Video Module Assessment

Postby Gisselle Sainz 2F » Fri Jan 18, 2019 6:41 pm

I got 0.00192 atm, 0.0137 atm, and 0.0237 atm as my answers but they're aren't exactly any of the answer choices listed. Why would my answers be two decimal points off?


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