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Chemistry 16 Online
OpenStudy (anonymous):

PLEASE HELP WILL AWARD MEDAL! A pressure of 743 mm Hg is equal to how many atm?

OpenStudy (technodynamic):

\[1 atm = 760 mmHg\] By using dimensional analysis you'll get \[743 mmHg * \frac{ 1atm }{ 760mmHg } = 0.977 atm\]

OpenStudy (anonymous):

THANK YOU SOOO MUCH @Technodynamic

OpenStudy (technodynamic):

I'd round it though to 0.978 atm for more accuracy

OpenStudy (anonymous):

Yeh it asks for three sig figs @Technodynamic

OpenStudy (technodynamic):

You're welcome! Correct answer based on accuracy: 0.978 atm

OpenStudy (anonymous):

Do you know how to do Percentge Yield? @Technodynamic

OpenStudy (technodynamic):

Yes. \[Percent Yield = \frac { actual }{ theoretical } * 100% =\]

OpenStudy (anonymous):

I'm sorry if i'm being such a nuisance but i'm taking a quiz I missed from week, since I was absent to know the material could you help me out?

OpenStudy (anonymous):

@Technodynamic

OpenStudy (anonymous):

Its stuff like this- A sample of gas at 47 ºC and 0.97 atm occupies a volume of 2.20 L. What volume would this gas occupy at 107 ºC and 0.789 atm?

OpenStudy (anonymous):

But there is Temp. Do I avoid that?? @Technodynamic

OpenStudy (anonymous):

Thats what Im confused on. How do I solve it if they give me more than 3

OpenStudy (technodynamic):

Sorry, typo. \[\frac{ P_1V_1 }{ T_1 }= \frac{ P_2V_2 }{ T_2 }\]

OpenStudy (technodynamic):

Plug the first numbers into the equation on the left. For the second sentence, plug those numbers on the right side. Then divide.

OpenStudy (technodynamic):

Don't worry, I'll walk you through the steps...

OpenStudy (anonymous):

OK one moment

OpenStudy (technodynamic):

can you repost your question into a new question box? my tablet won't scroll any farther

OpenStudy (anonymous):

Yeah sure

OpenStudy (anonymous):

A sample of gas at 47 ºC and 0.97 atm occupies a volume of 2.20 L. What volume would this gas occupy at 107 ºC and 0.789 atm?

OpenStudy (anonymous):

Ok so I have this- T1= 320k P1=.97 V1=2.20 T2=380 P2=.789

OpenStudy (technodynamic):

Yes that's correct

OpenStudy (anonymous):

OK so am itrying to find V2?

OpenStudy (technodynamic):

Yep, you're doing this correctly

OpenStudy (technodynamic):

You just plug in the numbers, divide, and boom you're done!

OpenStudy (anonymous):

Ok so (.97)(2.20)/320= .789/380 ?

OpenStudy (technodynamic):

That looks about right

OpenStudy (anonymous):

So did you get 1.9994?

OpenStudy (anonymous):

If you typed it in the calculator

OpenStudy (technodynamic):

My calc is charging but let me get some paper...one sec

OpenStudy (anonymous):

I justw ant to be sure before i submit my answer. Cause i Multiplied the top and divided by the two temps at the bottom

OpenStudy (anonymous):

or do you cross multiply them.. or am i just confusing myself.

OpenStudy (technodynamic):

It's asking for the volume, so in this case you will be dividing

OpenStudy (anonymous):

SO what did i do wrong.

OpenStudy (anonymous):

I divided the temp into P and V

OpenStudy (technodynamic):

You arranged it correctly. After you calculate the left side, and the right side. You will make the left side your numerator and the right side denominator when dividing. Did you do it that way?

OpenStudy (anonymous):

OH no I didnt I juts took the two temps to make my denominator

OpenStudy (technodynamic):

Yes that part is right

OpenStudy (anonymous):

Now im confused

OpenStudy (technodynamic):

(.97*2.20)/(320) = (0.789)/380 =

OpenStudy (anonymous):

OK

OpenStudy (anonymous):

and so i divide them?

OpenStudy (technodynamic):

After you solve that. Divide both the answers

OpenStudy (anonymous):

I got this .0066875/.0020763158=3.211818742

OpenStudy (technodynamic):

Excellent!

OpenStudy (anonymous):

Thank you very much

OpenStudy (technodynamic):

Yw!

OpenStudy (anonymous):

Can I ask you what Law I use for : The pressure exerted on a 243 mL sample of hydrogen gas at constant temperature is increased from 0.428 atm to 0.724 atm. What will the final volume of the sample be? @Technodynamic

OpenStudy (anonymous):

I thought the Combined gas law but it doesnt give me temp. only pressure and volume

OpenStudy (technodynamic):

alright here's how you do it...

OpenStudy (anonymous):

I dont understand why its so complicated, ive been at this for 3 hours..... sigh

OpenStudy (technodynamic):

i answered your initial post from your other post with the worked out answer. Hope that helps!

OpenStudy (technodynamic):

Go check it :)

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