1.2mole•20.17g/1mole= 24.20g
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.
To know more about heat of vaporization, visit the below link:
brainly.com/question/13106213
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The solution of which contains of PH 3 called acidic medium which is more stronger acid called HCl when reacts to the stronger base base NaOH forms the acid base reaction .
According to acid base equation you consider ph5 as PH 9 if you understand it then tell me below
