Answer:
true
Explanation:
A well designed product will increase in sells and in stock.
100: D, third law of motion
101: D, second law of motion
Answer:
(a) The Final Temperature is 315.25 K.
(b) The amount of mass that has entered 0.5742 Kg.
(c) The work done is 56.52 kJ.
(d) The entrophy generation is 0.0398 kJ/kgK.
Explanation:
Explanation is in the following attachments.
Answer:
0.2 kcal/mol is the value of
for this reaction.
Explanation:
The formula used for is:
![\Delta G_{rxn}=\Delta G^o+RT\ln Q](https://tex.z-dn.net/?f=%5CDelta%20G_%7Brxn%7D%3D%5CDelta%20G%5Eo%2BRT%5Cln%20Q)
![\Delta G^o=-RT\ln K](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Cln%20K)
where,
= Gibbs free energy for the reaction
= standard Gibbs free energy
R =Universal gas constant
T = temperature
Q = reaction quotient
k = Equilibrium constant
We have :
Reaction quotient of the reaction = Q = 46
Equilibrium constant of reaction = K = 35
Temperature of reaction = T = 25°C = 25 + 273 K = 298 K
R = 1.987 cal/K mol
![\Delta G_{rxn}=-RT\ln K+RT\ln Q](https://tex.z-dn.net/?f=%5CDelta%20G_%7Brxn%7D%3D-RT%5Cln%20K%2BRT%5Cln%20Q)
![=-1.987 cal/K mol\times 298 K\ln [35]+1.987 cal/K mol\times 298K\times \ln [46]](https://tex.z-dn.net/?f=%3D-1.987%20cal%2FK%20mol%5Ctimes%20298%20K%5Cln%20%5B35%5D%2B1.987%20cal%2FK%20mol%5Ctimes%20298K%5Ctimes%20%5Cln%20%5B46%5D)
![=-2,105.21 cal/mol+2,267.04 cal/mol=161.82 cal/mol=0.16182 kcal/mol\approx 0.2 kcal/mol](https://tex.z-dn.net/?f=%3D-2%2C105.21%20cal%2Fmol%2B2%2C267.04%20cal%2Fmol%3D161.82%20cal%2Fmol%3D0.16182%20kcal%2Fmol%5Capprox%200.2%20kcal%2Fmol)
1 cal = 0.001 kcal
0.2 kcal/mol is the value of
for this reaction.