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Offline uns33n

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help me please?
« on: October 19, 2008, 05:53:44 PM »
I'm doing a sample midterm as to review for my midterm tomorrow but for the life of me I cannot get the answer to this question correctly..


a 250.0 L cylinder contains 65.0% He(g) and 35.0% Kr (g) by mass at 25.0 C and 1.35 atm total pressure. What is the partial pressure of He in this container?

a) 0.473atm
b)0.675atm
c)0.878atm
d)1.32atm
e)1.35 atm


I keep getting C. The answer according to the Key is D.

can someone help me please?

Offline Mitch

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Re: help me please?
« Reply #1 on: October 19, 2008, 06:07:18 PM »
Need to show what you've done and how.
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Offline uns33n

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Re: help me please?
« Reply #2 on: October 19, 2008, 06:08:42 PM »
Need to show what you've done and how.

i used the formula pv= nrt to find for n.. moles.

n = pv/rt

Since I was given total pressure, (1.35) I used that as my P value.

once I found that i multiplied it by .65 to get He's moles.. then I used the formula again... for pressure p = nrt/v



I get C.. aways.

Offline Borek

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Re: help me please?
« Reply #3 on: October 19, 2008, 06:12:32 PM »
d seems to be OK.

0.65*total moles is not number of moles of He. 65% by MASS.
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Offline uns33n

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Re: help me please?
« Reply #4 on: October 19, 2008, 06:21:35 PM »
d seems to be OK.

0.65*total moles is not number of moles of He. 65% by MASS.
can you elaborate?


so im not supposed to multiply my n value by .65 because of mass?


so what do i do? 200.6g (according to the table) x .65 = 130.39 g He...

then what?

Offline Borek

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Re: help me please?
« Reply #5 on: October 19, 2008, 06:29:01 PM »
Assume numbers of moles are nHe and nKr.

What is sum of these numbers?

What is mass of the helium?

What is mass of the krypton?

What is ratio of their masses?

How many equations and in how many unknows do you have?
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Offline uns33n

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Re: help me please?
« Reply #6 on: October 19, 2008, 06:33:02 PM »
wow i totally do NOT recall any of this.

I feel like an idiot


im gonna do sooo bad tomorrow.. this is the ONLY question im stuck on and the midterm has one very similar to it in written form.

Offline Borek

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Re: help me please?
« Reply #7 on: October 19, 2008, 06:41:57 PM »
That's not difficult.

You have already calculated number of moles of both gases, so:

nHe + nKr = 13.795

Now, mass of helium is 4*nHe - molar mass times number of moles.

Do the same for krypton.
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Offline uns33n

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Re: help me please?
« Reply #8 on: October 19, 2008, 08:57:13 PM »
I have now officially spent 2 hours on a question that took you abou 3 minutes to do and still havent gotten it right.


Please can you show me step by step what exactly it is you did? Because giving me nHe and so on isn't doing anything for me. :|

Offline nj_bartel

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Re: help me please?
« Reply #9 on: October 20, 2008, 01:03:16 AM »
that just stands for moles of helium

Offline Borek

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Re: help me please?
« Reply #10 on: October 20, 2008, 02:50:14 AM »
Please can you show me step by step what exactly it is you did? Because giving me nHe and so on isn't doing anything for me. :|

nHe and nKr ar just names of unknowns - as nj already wrote, they denote unknown number of moles of helium and unknown number of moles of krypton.

From PV=nRT we can calculate n, thus it is obvious what is sum of number of moles of helium and number of moles of krypon.

If you have nHe moles of helium its mass is 4*nHe - that's just molar mass times number of moles, nothing fancy.

Do the same for krypton, follow questions from my earlier post.
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