How many grams of water vapor (H2O) are in a 10.2 liter sample of 0.98 atmosphere and 26 C
1 answer:
Hey there!
<span>Use the equation of Clapeyron:
</span>
T in kelvin :
26 + 273.15 => 299.15 K
R = 0.082
V = 10.2 L
P = 0.98 atm
number of moles :
P *V = n * R * T
0.98 * 10.2 = n * 0.082 * 299.15
9.996 = n * 24.5303
n = 9.996 / 24.5303
n = 0.4074 moles
Therefore:
Molar mass H2O = 18.01 g/mol
1 mole H2O ------------- 18.01 g
0.4074 moles ----------- m
m = 0.4074 * 18.01 / 1
m = 7.339 g of H2O
You might be interested in
97 mols I think approximately
Thjjhccfhijjhhbvvbjjhv bob. Hi. H.
these are the correct answers
The answer would be, it would be warmer.
Answer:104.6cm^3
Explanation:
V of a cylinder =πr2h
V = pi((1.045cm)^2)(30.48)
V = 104.57cm^3
Explanation: