Answer : The normal boiling point of ethanol will be,
or 
Explanation :
The Clausius- Clapeyron equation is :

where,
= vapor pressure of ethanol at
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:


Hence, the normal boiling point of ethanol will be,
or 
Answer:
This question is incomplete
Explanation:
This question is incomplete because the result of the described experiment would have better determined the type of scientific explanation to profer. However, the type of material that will preserve the relative hotness or temperature of the hot coffee for the longest time will be a material than can resist heat transfer. These materials tend to keep hot substances hot by not allowing the heat of the coffee to be conducted or pass through it. These materials are mostly insulators or made by placing an insulator between two heat conductors.
Generally, heat is usually transferred from a region of higher concentration to a region of lower concentration, hence when the heat is denied of this transfer, the heat will remain trapped in the "heat-donor" substance (in this case the hot coffee). Thus, the material chosen (A, B or C) will be the material that resists heat transfer the most based on the explanation above.
1 property
2 procedure
Results
Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.