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Chemistry 19 Online
OpenStudy (anonymous):

How many grams of Oxygen gas will occupy 1.12L space at STP?

OpenStudy (anonymous):

@Somy

OpenStudy (somy):

STP is standard temperature and pressure temp = 273 K perssure = 1 atm

OpenStudy (somy):

PV= nRT u know P , V and T R is constant value, and u also know it

OpenStudy (somy):

R = 0.0821

OpenStudy (somy):

find n

OpenStudy (anonymous):

How do you figure out R?

OpenStudy (somy):

well its a constant value http://www.csupomona.edu/~lllee/gasconstant.pdf

OpenStudy (anonymous):

oh oh ideal gas, got it

OpenStudy (somy):

because we are using P in atm, v in L and T in K thats why K = 0.0821

OpenStudy (somy):

i personally use SI system units which is P= pa V= m3 T= K thus R = 8.31

OpenStudy (somy):

but i don't want to confuse u so lets do it ur way

OpenStudy (anonymous):

so I have (1atm)(1.12L)=n(0.08206L*atm/mol*K)(273K)

OpenStudy (somy):

yup

OpenStudy (anonymous):

I get 0.049995

OpenStudy (somy):

yup

OpenStudy (anonymous):

The answer is 1.6g

OpenStudy (somy):

so now u have mole use formula mole= mass/Mr Mr will be 16 mole u have mass= X and im getting 0.79992g

OpenStudy (somy):

coz they are asking for Oxygen gas they are not saying O2 gas, i've seen this happening a lot hhhh so i don't get if the question is asking for O2 or just O usually when they don't specifically tell u O2, we take it as O

OpenStudy (somy):

but what is the answer that u have?

OpenStudy (somy):

i mean the right answer

OpenStudy (anonymous):

The correct answer is 16g

OpenStudy (somy):

alright then they mean O2 so yeah u got it

OpenStudy (anonymous):

would this also work?

OpenStudy (anonymous):

\[1.12LO * \frac{ 1mol O }{ 22.4L O } * \frac{ 16g O }{ 1mol O } = .8 g <-- *2 = 16g O\]

OpenStudy (anonymous):

lol

OpenStudy (somy):

no don't overdo it lol u'll kill the formula that exists

OpenStudy (anonymous):

but but my formula worked lol

OpenStudy (somy):

lol thats not your formula there is another way of doing this lmao

OpenStudy (anonymous):

I know is not mine, but still worked :o

OpenStudy (somy):

volume= mole* 22.4 mole= volume/22.4 mole= mass/ Mr mass= mole*Mr mass= (volume/22,4) * Mr

OpenStudy (somy):

this is what u did lol

OpenStudy (anonymous):

so it work the way I did it :o

OpenStudy (anonymous):

BUT it only works if pressure stays as 1 and temperature at 273K

OpenStudy (somy):

yes but u better play with formulas like i did just now rather then 'as u think' making up a formula from nowhere well if u did consider those formulas then u are on the right way

OpenStudy (somy):

yes it only works as long as the condition is STP

OpenStudy (anonymous):

do you think I will remember all if I fall sleep for 3 hours?

OpenStudy (somy):

if its RTP its same but value 22.4 will change to 24

OpenStudy (somy):

lol u will if its test before the test DON'T LOOK THROUGH STUFF YOU DID ur brain will be messed up

OpenStudy (somy):

so look through everything u did before u sleep that way it'll get fixed lol

OpenStudy (anonymous):

I somehow remember every single digit from constants and stuff... even though im almost sleeping

OpenStudy (somy):

lol well do ur best :)

OpenStudy (anonymous):

I have another question. Which gases behave as a perfect ideal gas?

OpenStudy (anonymous):

I just learned half a book of a material in 8 hours buh

OpenStudy (somy):

hmm gas at high temp and low pressure behaves kinda ideally noble gasses are one's that are closest to behaving ideally

OpenStudy (anonymous):

so between Methane and Ammonia, which one behaves closer to an ideal gas? and I would like to know how you figure this out.

OpenStudy (somy):

hmmm i'd say methane coz i remember Ammonia not behaving ideally lol

OpenStudy (somy):

let me check just a sec

OpenStudy (somy):

oh i think i understand its not ammonia because of H bonding intermolecular forces are just too strong so we cannot ignore it in case of methane its just van der Waal's forces and are not much thats why if we compare these 2 the one close to ideal gas behavior would be methane rather then ammonia

OpenStudy (anonymous):

ok I understand it.

OpenStudy (somy):

:)

OpenStudy (anonymous):

I have one last question before I die

OpenStudy (anonymous):

What is the volume of 56 g of Nitrogen at STP? I did it but I forgot how I did it -.-?

OpenStudy (somy):

lol use the awesome formula u made

OpenStudy (anonymous):

I tried =.=

OpenStudy (anonymous):

\[56g N*\frac{ 1mol N }{ 14g N }*\frac{ 22.4L N }{ 1mol N } = 89.6L \]

OpenStudy (anonymous):

The answer is half of that. I am missing something..

OpenStudy (somy):

\[mass = \frac{ volume }{22.4 }\times Mr\] \[mass = \frac{ volume \times Mr }{22.4 } \] \[mass \times 22.4 = volume \times Mr \] so

OpenStudy (somy):

\[volume = \frac{ mass \times 22.4 }{ M r }\]

OpenStudy (anonymous):

You will get 89.6L too..

OpenStudy (somy):

Nitrogen is N2 lol

OpenStudy (anonymous):

so is divided by 28g

OpenStudy (somy):

its not 14 xD its 28 the Mr

OpenStudy (anonymous):

How am I suppose to know if there is just 1N or 2N? D;

OpenStudy (anonymous):

N or N2*

OpenStudy (somy):

idk im always confused in this in previous question when they say Oxygen gas, they meant O2

OpenStudy (somy):

so i thought here they mean N2 so i guess they will specify if its O or N but when they don't specify then its O2 or N2

OpenStudy (anonymous):

I suppose it is to be seen as Nitrogen Gas since it says Nitrogen at STP

OpenStudy (anonymous):

I will do the work as if the professor is asking O2 or N2 or any other gas to subscript of 2

OpenStudy (somy):

i guess thats how it is

OpenStudy (anonymous):

I have to remember that now. I suppose this must be the reason why so many had low grade on this quiz of this chapter =.=

OpenStudy (somy):

lol well good luck ))

OpenStudy (anonymous):

since it is N2 it is still 1 mole of N right

OpenStudy (anonymous):

1 mole of N2*

OpenStudy (somy):

hmm yeah tho it'll depend on the question

OpenStudy (anonymous):

aaaaaaaaaaaaaa chem make it all so difficult lol

OpenStudy (somy):

chem is interesting tho lol

OpenStudy (somy):

take a look at this http://openstudy.com/study#/updates/53a5b02ae4b0a819ab109a05

OpenStudy (anonymous):

all to start over -.-. is same as calculus :D

OpenStudy (anonymous):

I once took 3 times longer than that solving something with a friend

OpenStudy (anonymous):

took us around 2-3 hours for one problem. is fun, I can imagine that being as fun as calculus lol

OpenStudy (anonymous):

Instead of 8 hours I actually spent around 10 hours non stop learning half the book

OpenStudy (somy):

lol that's cool i mean doing this on ur own must be tough

OpenStudy (anonymous):

that's why I came here for any doubt I had =.=

OpenStudy (somy):

lol that's good :) i can help out as long as its my level chem lol

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