O Sounds like the best answer
The amount of heat that could be removed by 20.0 g of ethyl chloride is 8.184 kJ.
<h3>How do we calculate required heat?</h3>
Required amount of heat which can be removed for the vaporization will be calculated as:
Q = (n)(ΔHv), where
- n = moles of ethyl chloride
- ΔHv = heat of vaporization = 26.4 kj/mol
Moles will be calculated as:
n = W/M, where
- W = given mass of ethyl chloride = 20g
- M = molar mass of ethyl chloride = 64.51 g/mol
n = 20 / 64.51 = 0.31 mol
On putting all these values in the above equation, we get
Q = (0.31)(26.4) = 8.184 kJ
Hence involved amount of heat is 8.184 kJ.
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Answer:
50 g
Explanation:
d= m/v
rearranging the above equation
m = d x v
m = 2.5 g x 20 g/cm3
m = 50 g
Is 2.0 x 10 to the -9th power an option
