6N.3c


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Lindsey Chheng 1E
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am

6N.3c

Postby Lindsey Chheng 1E » Mon Mar 02, 2020 4:32 pm

Predict the potential of:

Pt(s)|Cl2(g, 250 Torr)|HCl (aq, 1.0 mol/L)‖HCl (aq, 0.85 mol/L)|H2(g, 125 Torr)|Pt(s)

How do I find Q for this question when the gases are in units of Torr and the aqueous ions are in units of mol/L?

Ryan Chang 1C
Posts: 105
Joined: Sat Aug 24, 2019 12:17 am

Re: 6N.3c

Postby Ryan Chang 1C » Mon Mar 02, 2020 9:46 pm

You can convert torr to atm by dividing the torr values by 760. The units will then cancel out as you solve the equation.

Daniela Shatzki 2E
Posts: 53
Joined: Sat Aug 24, 2019 12:16 am

Re: 6N.3c

Postby Daniela Shatzki 2E » Tue Mar 03, 2020 3:53 pm

Ryan Chang 1C wrote:You can convert torr to atm by dividing the torr values by 760. The units will then cancel out as you solve the equation.


after this do you implement the pressures into the Q equation? I keep getting the wrong overall answer so I'm not sure what I'm doing wrong.

MAC 4G
Posts: 121
Joined: Wed Sep 18, 2019 12:16 am

Re: 6N.3c

Postby MAC 4G » Tue Mar 03, 2020 4:39 pm

Daniela Shatzki 2E wrote:
Ryan Chang 1C wrote:You can convert torr to atm by dividing the torr values by 760. The units will then cancel out as you solve the equation.


after this do you implement the pressures into the Q equation? I keep getting the wrong overall answer so I'm not sure what I'm doing wrong.



Yes, you would plug in the converted values in the Q equation. So, Q = ((250/760)*(125/760))/((1^2)*(1)) = 0.075. If you got this, you may have a problem in a different part of your overall equation.

Daniela Shatzki 2E
Posts: 53
Joined: Sat Aug 24, 2019 12:16 am

Re: 6N.3c

Postby Daniela Shatzki 2E » Tue Mar 03, 2020 9:15 pm

MAC 4G wrote:
Daniela Shatzki 2E wrote:
Ryan Chang 1C wrote:You can convert torr to atm by dividing the torr values by 760. The units will then cancel out as you solve the equation.


after this do you implement the pressures into the Q equation? I keep getting the wrong overall answer so I'm not sure what I'm doing wrong.



Yes, you would plug in the converted values in the Q equation. So, Q = ((250/760)*(125/760))/((1^2)*(1)) = 0.075. If you got this, you may have a problem in a different part of your overall equation.


how did you get 1^2 and 1 for the reactants?


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