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Topic: {gas volume}  (Read 13776 times)

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Offline peterpan1372

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{gas volume}
« on: September 08, 2017, 03:52:03 PM »
Can someone help me out with this one?

I have 105cm^3(2*100+10)/2)
« Last Edit: September 08, 2017, 06:01:12 PM by Arkcon »

Offline Borek

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Re: Help please
« Reply #1 on: September 08, 2017, 04:14:51 PM »
Per forum rules you have to show your attempts at solving the problem.
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Offline peterpan1372

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Re: Help please
« Reply #2 on: September 08, 2017, 04:16:40 PM »
I already have?

Offline Arkcon

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Re: Help please
« Reply #3 on: September 08, 2017, 05:59:44 PM »
I already have?

Ok then.  Why does:

Can someone help me out with this one?

I have 105cm^3(2*100+10)/2)

give the correct answer, or why doesn't it?  You add a bunch of numbers, 2 100 10 slash 2 -- where did those come from?  Why are they there, instead of not there, or somewhere else?

Are you guessing?  What good will that do the next person who uses this thread?

Or do you know what you're doing, and just looking for us to check, to insure you get the bestesterest possible grade number?

Don't do those things on our forum, that's not what where here for.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline peterpan1372

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Re: Help please
« Reply #4 on: September 08, 2017, 06:00:28 PM »
All I know is that 1mole of any gas occupies 24dm^3. We would have then 48dm^3 + 24dm^3 ... however, from here on i do not know further...

Offline XeLa.

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Re: {gas volume}
« Reply #5 on: September 08, 2017, 06:43:40 PM »
All I know is that 1mole of any gas occupies 24dm^3.

Remember that 1 mole of gas doesn't always occupy 24 L of space, however, as long as we keep this assumption consistent throughout our calculations (as temperature and pressure don't change) we should be fine.

We would have then 48dm^3 + 24dm^3

I don't understand your logic here. The question specifically says that we initially have 100 mL of X and 10 mL of y... You also know that 1 mole of gas occupies 24 L of space (from your assumption)...

EDIT
« Last Edit: September 08, 2017, 07:07:59 PM by XeLa. »

Offline peterpan1372

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Re: {gas volume}
« Reply #6 on: September 08, 2017, 06:56:23 PM »
well, obviously Y is the limiting factor since it is less available than X, however i do not know how to calculate the rest..

Offline peterpan1372

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Re: {gas volume}
« Reply #7 on: September 08, 2017, 09:28:23 PM »
I seriously do not know why they state in the solutions that 100cm^cm would be the right answer...

Offline Borek

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Re: {gas volume}
« Reply #8 on: September 09, 2017, 03:18:33 AM »
Do you know Avogadro's hypothesis?

With what volume of X will 10 mL of Y react?

What will be the volume of Z produced?

By how much will the volume of the mixture change?
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Offline peterpan1372

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Re: {gas volume}
« Reply #9 on: September 09, 2017, 12:17:05 PM »
well, 10ml of y react with a volume of 10ml of x or not?. but i do not know how much they produce.. i first thought of adding them up, but 110ml is obviously wrong... since the answer is 100ml, i  thought of multiplying them, but do not know the sense behind it since you have to add them up...
« Last Edit: September 09, 2017, 12:52:29 PM by peterpan1372 »

Offline Borek

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Re: {gas volume}
« Reply #10 on: September 09, 2017, 01:55:18 PM »
well, 10ml of y react with a volume of 10ml of x or not?

No, you are guessing, it won't work.

Do you know Avogadro's hypothesis?
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Offline peterpan1372

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Re: {gas volume}
« Reply #11 on: September 09, 2017, 05:16:11 PM »
we have 100ml of x reacting with 10ml of y,  giving 110ml of z... thats for me the only logical way, since we have to add them up...

Offline Borek

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Re: {gas volume}
« Reply #12 on: September 10, 2017, 03:28:41 AM »
well, ive looked up what it says, but do not know how to adapt it into our context...

Can you state what it says?

Do you remember that things react in molar ratio?

we have 100ml of x reacting with 10ml of y,  giving 110ml of z... thats for me the only logical way, since we have to add them up...

This logic is wrong, as you have to follow stoichiometry.
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Offline peterpan1372

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Re: {gas volume}
« Reply #13 on: September 10, 2017, 10:56:10 AM »
equal volumes of gases at the same temperature and pressure contain equal numbers of molecules...

the ratio is 2:1...

Offline peterpan1372

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Re: {gas volume}
« Reply #14 on: September 10, 2017, 12:43:19 PM »
and no, I do not know how to transfer it into our context...

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