What’s a chemical property
1 answer:
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Answer:
The answers to the questions are;
a. The entropy of sublimation for carbon dioxide (the system) is
134.07 J/Kmol.
b. The entropy of the universe for this reversible process is 376 J/K.
Explanation:
Entropy of sublimation is the entropy change experienced following the transformation of a mole of solid to vapor at the temperature where the sublimation is taking place
a. We note that the mass of the solid CO₂ = 389 g
Molar mass of CO₂ = 44.01 g/mol
Number of moles of CO₂ in the sculpture = Mass/(Molar mass)
= (389 g)/(44.01 g/mol) = 8.84 Moles
Entropy of sublimation is given by
ΔS =
- S
=
Where:
ΔH = 26.1 KJ/mol
T = Temperature = –78.5°C = 194.65 K
Therefore the amount of heat required to cause the 389 g of dry ice to sublime = 26.1 KJ/mol × 8.84 Moles = 230.695 KJ
Therefore the entropy of sublimation = ΔS =
= 1.185 KJ/K
= 1185 J/K = 1185/8.84 J/Kmol = 134.07 J/Kmol
b. The entropy of the universe is given by;
ΔS =
+ ΔS
If the heat absorbed by the system is the same as the heat given off by the surrounding, then we have;
ΔS =
=1.185 KJ/K - = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K
= 376 J/K.
Answer:
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?
A) BrCH2CH2Br
B) CH3CH2CH2Br
C) CH3CHBr2
D) CH3CH2CH2CH3
E) BrCH2CH2CH2CH2Br
Explanation:
The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.
This is an example of free radical substitution.
The structure of ethane and its bromination is shown below:
Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).
Remaining all other products are possisble to form on free radical substitution of ethane.
