partial pressure- chemistry

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shreya_2003
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#1
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#1
how would i work out partial pressure with no mol value.
i rearranged to get pressure on its own so
p(cl2)= 3.5x10-4/ p(SbCl5) and them /divide by p(SbCl3) and got 9.3582 x10-3
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shreya_2003
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#2
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(Original post by Daddry)
This is a small task pm. I can share with you all the steps on how to calculate this. reach out to me through my whatsApp +254797004634
Do you mind explaining it instead?
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charco
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(Original post by shreya_2003)
how would i work out partial pressure with no mol value.
i rearranged to get pressure on its own so
p(cl2)= 3.5x10-4/ p(SbCl5) and them /divide by p(SbCl3) and got 9.3582 x10-3
The partial pressure is directly proportional to the moles.

It is the mole fraction x the total pressure.

The sum of the partial pressures = the total pressure.
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shreya_2003
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#4
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(Original post by charco)
The partial pressure is directly proportional to the moles.

It is the mole fraction x the total pressure.

The sum of the partial pressures = the total pressure.
Hi i forgot to post the link to the question and just did so, its given no mol so how would i do mole fraction x total p?
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charco
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(Original post by shreya_2003)
Hi i forgot to post the link to the question and just did so, its given no mol so how would i do mole fraction x total p?
Like I said the partial pressure is directly proportional to the moles.

K = pp(SbCl3) * pp(Cl2)/pp(SbCl5)

just rearrange to get pp(Cl2)
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