Chemistry
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OpenStudy (anonymous):
How many grams of methane gas (CH4) are in a 15.8 liter sample at 1.2 atmospheres and 27°C? Show all work used to solve this problem
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OpenStudy (somy):
PV=nRT
OpenStudy (somy):
can you use it?
OpenStudy (somy):
on your own i mean
OpenStudy (anonymous):
I have no idea what that means
OpenStudy (somy):
its a formula for ideal gases
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OpenStudy (somy):
PV= nRT
p- pressure
V- volume
n- mole
R- constant
T- temperature
OpenStudy (somy):
so what are you given in the question ?
OpenStudy (anonymous):
I don't know, I know we have 15.8 liters
OpenStudy (somy):
so ur volume 15.8 L
OpenStudy (somy):
pressure is also given
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OpenStudy (somy):
and temperature
OpenStudy (somy):
so
P= 1.2 atm
V= 15.8 L
T= 27 °C
OpenStudy (somy):
you need to change units
for pressure change it from atm to Pa
for volume change it from L to m3
for temperature change it from °C to Kelvin
OpenStudy (somy):
can u do that?
OpenStudy (anonymous):
If you tell me how haha
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OpenStudy (somy):
lol okay
OpenStudy (somy):
atm to pascal you need to multiply by 101 325
OpenStudy (somy):
L to m3 you need to divide by 1000
OpenStudy (somy):
and °C to Kelvin you need to add 273
OpenStudy (somy):
now do it and tell me the new values
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OpenStudy (anonymous):
Where did you get these numbers?
OpenStudy (somy):
this is conversion
OpenStudy (somy):
you need to learnt how to change units
OpenStudy (anonymous):
So the new temperature is 300?
OpenStudy (somy):
yes :D
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OpenStudy (anonymous):
and the new volume is 0.0158?
OpenStudy (somy):
thats right :)
OpenStudy (anonymous):
And I don't understand the atm one, you have two numbers next to each other
OpenStudy (somy):
come again? what do u mean 2 numbers next to each other?
OpenStudy (anonymous):
How do I convert atm to whatever I'm supposed to convert it to
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OpenStudy (somy):
what is the value for atm?
OpenStudy (anonymous):
1.2
OpenStudy (somy):
good so if 1 atm = 101 325
1.2 atm= ?
OpenStudy (anonymous):
I don't know what 101 325 means are you missing a comma or something
OpenStudy (somy):
1 atm = 101 325 Pa
its number...
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OpenStudy (gazela1997):
101.325 KPa
Change the Celcius to kelivin
OpenStudy (somy):
she did.
OpenStudy (somy):
so it'll be 1.2*101 325 =
OpenStudy (anonymous):
that's so confusing, 121590?
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OpenStudy (somy):
yes
OpenStudy (somy):
so
now your new values
P= 121590 Pa
V= 0.0158 m3
T= 300 K
OpenStudy (somy):
right?
OpenStudy (anonymous):
Right
OpenStudy (somy):
you have to know that R is constant and the value of R is 8.31
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OpenStudy (somy):
get it?
OpenStudy (anonymous):
Okay
OpenStudy (somy):
cool so the formula we'll be using is
PV=nRT
what you don't have?
OpenStudy (anonymous):
So basically pressure x volume = R x temp?
OpenStudy (somy):
yes but don't forget 'n'
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OpenStudy (somy):
pressure*volume= mole*R*temperature
OpenStudy (anonymous):
n means mole?
OpenStudy (somy):
yes
OpenStudy (somy):
so we don't have mole right?
OpenStudy (anonymous):
Okay so pressure*volume = 18.96?
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OpenStudy (somy):
okay first can u make mole as subject?
OpenStudy (anonymous):
I don't know what that means
OpenStudy (somy):
for example if i want to find V , i'll make it as subject and the formula will look like this
V= nRT/P
OpenStudy (anonymous):
So I'm looking for n = something?
OpenStudy (somy):
yes
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OpenStudy (somy):
so if u make n as subject how the formula will look like?
OpenStudy (anonymous):
n = R*temperature ?
OpenStudy (somy):
how about volume and pressure?
OpenStudy (anonymous):
What about it?
OpenStudy (somy):
your main formula is
PV= nRT
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OpenStudy (anonymous):
Okay
OpenStudy (somy):
so if u make n as subject
it'll be like this
n= PV/RT
OpenStudy (somy):
do u understand?
OpenStudy (anonymous):
Okay
OpenStudy (somy):
now put the values into the formula and get your mole
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OpenStudy (anonymous):
So n = 1921. 122/2493
= 0.770 ?
OpenStudy (somy):
yup :) 0.77
OpenStudy (somy):
good job :)
OpenStudy (somy):
now u found mole
OpenStudy (somy):
can you tell me what is the formula of mole?
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OpenStudy (anonymous):
Mass/molecular mass?
OpenStudy (somy):
good :)
OpenStudy (somy):
what is the question asking you to find?
OpenStudy (anonymous):
how many grams of CH4 are in a 15.8 liter sample at 1.2 atmospheres and 27°C?
OpenStudy (somy):
so they are asking for Mas right?
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OpenStudy (somy):
mass*
OpenStudy (anonymous):
Yeah they are
OpenStudy (somy):
mole= mass/Mr
make mass as subject
OpenStudy (anonymous):
I'm not good at that haha but mass = Mr/n?
OpenStudy (somy):
no
mass= Mr*mole
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OpenStudy (somy):
now what is Mr of CH4?
OpenStudy (anonymous):
That was my next guess, really quick Mr is what?
OpenStudy (somy):
Mr is molecular mass
OpenStudy (anonymous):
Okay where do I find it is it on the periodic table?
OpenStudy (somy):
yes
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OpenStudy (anonymous):
16.04 g/mol
OpenStudy (somy):
yeah basically 16
OpenStudy (anonymous):
Okay so mass = 16 x 0.770
= 12.32
OpenStudy (somy):
yup :D
OpenStudy (somy):
done :)
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OpenStudy (anonymous):
Now what or are we done?
OpenStudy (anonymous):
Yay! Do you mind if i write my answer in paragragh form and than I'll tag you in it and you can tell me if it looks good?
OpenStudy (somy):
okay np :)