# Chemistry

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Test corrections :) They're calculations/short answers. I'd just like something to compare answers to or help me get the full answer (all partial credits). Thanks!

#9. When water is electrolyzed it produces hydrogen and oxygen. If 10.0% of 150.0g of water is electrolyzed, wht is the final volume of gas if the temperature is 17.2 degrees C and the pressure is 98.9kPa?

#11. Caffeine has 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen and 16.49% oxygen. Its molar mass is 194.2g/mol. Its molecular formula is?

#14. The compound X2O is 63.7% X (mystery element). What is the identity of X?

#15. Carbon disulfide burns in oxygen. Complete combustion gives the reaction CS2 + 3O2 -> CO2 + 2SO2. [16.352 L SO2; 8.064 L CO2; Excess CS2] What was the pressure in the container after the reaction if the container is 3.86 L and the temperature was 392K?

• Chemistry -

I'm having trouble understanding what you want. Have you worked parts of the problems and you want us to work all of it in detail so you can compare your answers and see where you went wrong? I have a better idea. You show your work on each and I shall be happy to check the results and show corrections where needed.

• Chemistry -

Alright, sure.

#9.
2H2O -> 2H2 + O2

15g H2O/1 x 1mol/18.02 = 0.832 mol H2O
PV=nRT
(98.9kPa)(v)=(0.832mol)(8.31 kPaxL/molxK)(290.2K)
v=20.29L

#11.

12.01g/gC x 100=49.48%
24.27gC/1 x 1mol/12.01g=2.02mol C

1.01g/gH x 100=5.15%
19.61gH/1 x 1mol/1.01g=19.42mol H

14.01g/gN x 100=28.87%
48.53g/1 x 1mol/14.01g=3.46mol N

16g/gO x 100=16.49%
97.03g/1 x 1mol/16g=6.06mol O

2.02/2.02=1C
19.42/2.02=9H
3.46/2.02=2N
6.06/2.02=3O

N2CO3H9

#14.

O=36.3%
X=63.7%

16g/gO x 100=36.3%
44.08gO
44.08gO/1 x 1mol/16g= 2.755mol O
[That's as far as I got.]

#15.

(P)(3.86L)=(.39mol)(0.0821 atmxL/molxK)(392K)
P=3.25 atm

(P)(3.86)=(1.09mol)(0.0821)(392K)
P=9.09 atm

3.25+9.09=12.34 atm

• Chemistry -

#9. I get 20.32 using your numbers (I used 290.4 for T) but that should be rounded to 20.3. I think three s.f. are allowed (from the 10.0%). For oxygen, it will be half of that.

#11. First I think you have calculated g of the elements and not mols and second I think you've rounded too much. I do them this way.
Take a 100 g sample which give you
49.48 g C
5.15 g H
28.87 g N
16.49 g O
Then convert to mols.
49.48/12.01 = 4.11 mols C.
5.15/1 = 5.15 H
28.87/14.008 = 2.06
16.49/16 = 1.03

Divide all by the smallest and I get
3.99 C which rounds to 4.0o
5.00 H
2.00 N
1.00 O
empirical formula is C4H5N2O and the empircal mass is 97.1
Then empirical formula x Number = molar mass and solve for N. That is 2 and the molecular formula is C8H19N4O2.

#14 is #11 in reverse.
63.7 g X
36.3 g O
mols X = 63.7/atomic mass X = ?
mols O = 36.3/16 = 2.27
If the empirical formula is X2O that makes the 2.27 = 1; therefore X must be twice that or 4.54. That means
63.7/4.54 =14.04 which I would guess to be nitrogen. Check it out.
N2O = (14+14)/(14+14+16) = (28/44)*100 = 63.6% (You can get 63.7 if you use 14.04)

#15. I don't understand the 8.064 and 16.352

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