Answer:
The true statements are given below.
Explanation:
1 D glucose is a reducing sugar
2 The oxidation of reducing sugar forms a carboxylic acid sugar.
D glucose is a reducing sugar because glucose contain a free hydroxyl group (-OH)in its anomeric carbon.
The oxidation of reducing sugar result in the conversion of -CHO group in case of aldose sugar and -CH2OH group in case of ketose sugar into carboxylic acid(-COOH).
This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and 
and 
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
Learn more:
Answer:N
2
+ 3
H
2
-----> 2N
H
3
Explanation:
N
2
+
H
2
-----> N
H
3
Let us balance this equation by counting the number of atoms on both sides of the arrow.
N
2
+
H
2
-----> N
H
3
N=2 , H=2 N=1, H=3
To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N
H
3
on RHS
N
2
+
H
2
-----> 2N
H
3
N=2 , H=2 N=2 , H= 6
To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of
H
2
on LHS
N
2
+ 3
H
2
-----> 2N
H
3
N=2 , H=6 N=2 , H= 6
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