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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6.023*1023 He atoms=1 mole He
1.23×1024 helium atoms=1.23×1024/6.023*1023 =2.042 mole He
one mole He=4.0 gm
2.042 mole He=2.042*4.0=8.168 gm
answer in grams of Helium is 8.168 gm
Explanation:
The total mass is:
100 g + 200 g
= 300 g
The mass of salt is:
0.20 (100 g) + 0.10 (200 g)
= 20 g + 20 g
= 40 g
So the concentration is:
40 g / 300 g
≈ 13.33%
Round as needed.
