Question

http://prntscr.com/ctfp3s
@cuanchi help lol

(idk what exactly this is asking. Tried PV=nRT but got lost halfway)

Answer 1

you have to calculate the number of total moles of gas in the mix. Then you have to built a 2 equations system with 2 unknown x and y such as x=moles of O2 and y moles of He

x+y = 100
x MMO2 + y MMHe = mass of the mix

solve the equation system

Answer 2

Oh, so a system equation

Answer 3

yep

Answer 4

alright thanks;

Can I ask another question? :p

Answer 5

sure

Answer 6

http://prntscr.com/ctfv4w

Answer 7

what book is that problems from?

Answer 8

I know I need to apply partial pressure but I don't know exactly how

*petrucci 10th

Answer 9

Well the site is based on that I think?

Answer 10

OK
I know

Answer 11

calculate the number of moles in the 3.15g of NH4NO3
then multiply by 3 because after decomposition you will have 3 molecules in place of 1.
This will be n, then you have T and V you can calculate the P

PV=nRT

Answer 12

So then...
dimensional analysis, then stoichiometry, then the gas law?

Answer 13

yes you have to convert T to K

Answer 14

okay

and for the first one, sorry I got stuck again --
how would I find the \(n\) (moles) for the gas total?

Answer 15

yea!
assume the V=1L

n=PV/RT

the mass 0.502g is the mass of n moles of the mix because the density is 0.502g/L

Answer 16

I thought about the 1L too but I was paranoid haha

okay thank you!! ☺

Answer 17

I also had a question on this one if it's okay
http://prntscr.com/ctg3gu

Answer 18

with the density and the T and 1L (again) calculate the number of moles -> and the molecular mass of the compound multiplying by the mass

n=m/MM
d=m/V
PV=nRT

PV=(m/MM) RT
P= (m/V)(RT/MM)
MM=d (RT/P)

with the percentage composition C,H you can calculate the empirical formula, then calculate the molecular mass of the empirical formula and the mass of the compound to calculate if you have to multiply for some factor or not

Answer 19

alright O:

that's a lot of information...
thanks !

Answer 20

@cuanchi I solved the system for the first problem and got y=114.

... I think that's wrong.

Answer 21

how many moles did you get?

Answer 22

0.03862 \(mol_{gas}\)

Answer 23

is that wrong?

Answer 24

x+y = 0.03862 mol
x MMO2 + y MMHe = 0.502g
x 32 + 4y = 0.502

Answer 25

this is what I did with the molar-mass equation:\[(100-y)(31.9988)+y(4.0026)=0.502\]

Answer 26

wait... what?

Answer 27

it is not 100 it is the 0.0386 moles

Answer 28

oh

but you said 100 so I was not sure lol

Answer 29

sorry

Answer 30

it's okay, I didn't submit any answers and get point deductions, so I forgive you lol

Answer 31

Oops I submitted on accident;

oh well, I still have more tries... I got this\[y\approx0.02622\text{ mol }He\]

Answer 32

Should I find the grams of \(He\) from this and then find percent mass this way?
\[\frac{\text{grams }He}{0.502\text{ grams total gas}}\]

Answer 33

yes

Answer 34

Okay thank you

Answer 35

times 100

Answer 36

yup thanks

Answer 37

@cuanchi I seem to have gotten an incorrect answer on this problem:
http://prntscr.com/ctg3gu

I put \(C_{2}H_{6}\) and it is listed incorrect.

Answer 38

I think the answer is
C4H10

Answer 39

Oh okay