Assuming that the contents of the chamber ar ideal gases. We can use the relation PV=nRT. At a constant
temperature and number of moles of the gas the product of PV is equal to some
constant. At another set of condition of temperature, the constant is still the
same. Calculations are as follows:
P1V1 =P2V2
P2 = (1)(450)/ 48
P2 = 9.375 atm
Answer:
I dont know the answer for that question it's hard question isn't it
n = PV/RT
p = 1.6 atm
v = 12.7L
R = 0.0821
T = 24°C which is equivalent to 297.15 degrees k
n = (16 × 12.7) / (0.0821 × 297.15)
n = 20.32 / 24.39
n = 0.83 mol
C = 12.90
H = 1.0079
C2 = 12.010 × 2 = 24.02
H6 = 1.0079 × 6 = 6.0474
C2H6 = 30.0674
Ethane times n which is 30.0674 × 0.83mol
= 24.95 grams of C2H6. Which is Ethane.
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