11.89

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Hyein Cha 2I
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11.89

Postby Hyein Cha 2I » Sun Dec 03, 2017 2:49 pm

Can someone please explain both parts of the question 11.89? I first dont get how you would write balanced equation when you don’t know the composition of A B and C, and also why is partial pressures all over 100 in the answer sheet for part b? For example, why is partial pressure for C not 10 and 10/100?
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Srbui Azarapetian 2C
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Re: 11.89

Postby Srbui Azarapetian 2C » Sun Dec 03, 2017 4:16 pm

Although you are no explicitly told to ratio of reactants to products, you can deduce this information from the provided graph. Because use the C and B start from 0, you can assume that they are both the products. The A stats from a positive number, which means it was previously present, so it is the reactant. To figure out their relative ratios, you can tell from the scale on the left that B ends up at 5 and C ends up at 10, so C is twice as much as B. The concentration of A is a little harder to figure out numerically, but if you count by boxes, it decreases by 2 boxes, which is the same as the increase of C. This way, A also is double B and the same as C. The resulting equation is 2A(g) <-> B(g) + 2C(g).
As for why the 5, 10, and 18 are over 18, I thin it is because the scale is in kPa.

Kendall Schemmer 1I
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Re: 11.89

Postby Kendall Schemmer 1I » Sun Dec 03, 2017 4:56 pm

I'm still not sure why the 5, 10, and 18 are over 100 in the equilibrium concentration calculation. A kPa equals 1000 Pa, so it would make sense if it were over 1000, not 100??

Kyra LeRoy 1E
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Re: 11.89

Postby Kyra LeRoy 1E » Thu Dec 07, 2017 1:26 am

100 kPa=1 bar, so to go from 1kPa you multiply each by 1/100 bar.

Hyein Cha 2I
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Re: 11.89

Postby Hyein Cha 2I » Thu Dec 07, 2017 10:06 am

Kyra LeRoy 1F wrote:100 kPa=1 bar, so to go from 1kPa you multiply each by 1/100 bar.


ohh so we always need to have pressure in terms of 'bar?'


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