Charles' Law/Boyle's Law Gas Problems?
Can anyone help me with these? I got my answer but I don't know if I did it right
An aerosol can with a volume of 325mL contains hair spray at 445kPa and 12 degrees C. What volume would the gas occupy if it was allowed to escape at 101kPa and at 21 degrees C?
I forgot the other one:
For the combustion of pentane:
a) Complete the balanced equation.
b) If 250L of pentane at STP was combusted, what volume of carbon dioxide would be produced at 25 degrees C and 99.9kPa?
- **PiNoY YFC**Lv 71 decade agoFavorite Answer
(V1*P1)/T1 = (V2 *P2)/T2
(325ml * 445kPa)/(273 + 12) = (V * 101kPa)/(273 + 21)
507.46 = (V * 101)/294
101V = 149192.11
V = 1477.15ml
here is to your combustion of pentane:
a. C5H12 + 8 02---------->5 CO2 + 6 H20
b. (V1*P1)/T1 = (V2 *P2)/T2
(250L*101.3kPa)/273 = (V * 99.9kPa)/(273 + 25)
92.77 = (99.9V)/298
27,644.14 = 99.9V
V = 276.72L
- kumorifoxLv 71 decade ago
pV/T is constant, so determine the constant value first, then reverse the equation and determine V (V = (T/p) × constant)
(445 × 325)/(12 + 273) = 144625/285 kPa mL K^-1.
Now plug in the new values.
V = [(273 + 21)/101] × (144625/285) = 1477.150 mL
a) C5H12 + 8O2 -> 5CO2 + 6H2O
b) Use pV/T = constant again. I'm not sure about the STP definition you use, apparently the IUPAC definition has changed since I needed it.
- ErikaLv 44 years ago
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- Anonymous1 decade ago
Taryn, i think you should not do chemistry anymore, seeing as it does not help you in any way aha, jk.
if i could i would help you buttt i cant.
sorry im not a good friend in that categorie, but i do think we could talk about it over a nice baked potato.hmm?Source(s): brain smart.