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# Thread: a couple chemistry questions

1. ## a couple chemistry questions

Now i nkow that some of you must be better at chem than I am...

A 1.0L sample of unknown gas at 25 degrees C and 211mmhg is found to weigh 0.727g. the gas is He, NO, CO2, SO2??

Consider the combustions of CH4:

CH4(g) + 2O2 (gas) = CO2(g) + 2H2O(L). If the reaction takes place at STP, how many liters of CH4 is needed to produce 1 mol of H20 (water).

a. 22.4L
b. 11.4L
c. 44.8L
d. 0.5L

also if you know the anwer that is great but I to have the answers but am not quite sure how to get there so if you could explain a little that would be great..

2. atomic weights
SO2 64.07
CO 44.01
NO 30.01
HE 4.00

Sorry the second part of the question the answer is 11.2L or B.

3. 211mmHg is the atmospheric pressure...

4. haha it seems like my batteries always die when I need to use something...

also i think 1 mol = 1 L

5. Can't you just convert pressure to atm and use ideal gas law?

PV=nRT, and then convert the moles that you got into the atomic weight of one of the elements.

6. Originally Posted by Jenseno9
haha it seems like my batteries always die when I need to use something...

also i think 1 mol = 1 L
1 mol at standard temperature and pressure = 22.4 L

7. 22.4L CH4/1 * 1 mol CH4/1 Mol CH4 * 1 mol H20/ 2 Mol H20

= 11.2 L

Although i wouldnt you would do this b/c water is a pure liquid and normally pure liquids and solids arent counted but oh well....

8. For the second one you need to put everything in standard state so:

V1/V2 = T1/T2 * P2/P1 the combined law

1/ x = 298 K /273 K * 1 atm/ (211/760)
x = 3.932 L @ STP

then use pv=nrt

1 * 3.932 = (.727/x) * .0821 * 273
x = 4 g/mol

If you dont convert it to STP you get
211/760 * 1 = (.727/x) * .0821 * 298
x = 64 g/mol

I think the first one (4 g is correct tho)....

9. Originally Posted by AbsolutelyLethal
22.4L CH4/1 * 1 mol CH4/1 Mol CH4 * 1 mol H20/ 2 Mol H20

= 11.2 L

Although i wouldnt you would do this b/c water is a pure liquid and normally pure liquids and solids arent counted but oh well....
****...this part i actually figured out...was proud of myself until i saw someone had already answered it...haha...but the second post is way beyond me at this point...i dumped chemistry back in undergrad...no more i say!