Answer:
(a) 
(b) 
(c) 
Explanation:
Hello,
(a) In this case, since entropy remains unchanged, the constant 
should be computed for air as an ideal gas by:
Next, we compute the final temperature:
Thus, the work is computed by:
(b) In this case, since 
is given, we compute the final temperature as well:
And the isentropic work:
(c) Finally, for isothermal, final temperature is not required as it could be computed as:
Regards.
Answer: The factor that lead to cyclopropane being less stable than the other cycloalkanes is the presence of a RING STRAIN.
Explanation:
In organic chemistry, the end carbon atoms of an open aliphatic chain can join together to form a closed system or ring to form cycloalkanes. Such compounds are known as cyclic compounds. Examples include cyclopropane, cyclobutane, cyclopentane and many among others.
Cyclopropane is less stable than other cycloalkanes mentioned above because of the presence of ring strain in its structural arrangement. The ring strain is the spatial orientation of atoms of the cycloalkane compounds which tend to give off a very high and non favourable energy. The release of heat energy which is stored in the bonds and molecules cause the ring to be UNSTABLE and REACTIVE.
The presence of the ring strain affects mainly the structures and the conformational function of the smaller cycloalkanes. cyclopropane, which is the smallest cycloalkane than the rest mentioned above, contains only 3 carbons with a small ring.
The answer is B, Let me know if you get it right!
Answer:
increase
Explanation:
because in endothermic reaction heat is absorbed
Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
