Answer:
im pretty sure its b
Explanation:
im sorry if im wrong i tried my best
Mg + Cl2 ---> MgCl2
<span>Which element is reduced?
</span>answer: chlorine (chlorine is going from 0 on -I)
Answer:
the entropy change of the fluid during the process process is is 1.337 kJ/K, the change for the source is -1.337 kJ/K and the total entropy change is 0
Explanation:
since the Carnot cycle is a reversible cycle, the entropy change is related with the heat exchanged through:
ΔS =∫dQ/T
since the temperature remains constant
ΔS =∫dQ/T=(1/T)*∫dQ = Q/T
Q= heat added to the system
T= absolute temperature = 400°C= 673 K
therefore
ΔS = Q/T = 900 kJ/ 673 K = 1.337 kJ/K
ΔS working fluid = 1.337 kJ/K
since the process is reversible, the entropy change of the universe (total entropy change) is 0 (there is no entropy generation). thus
ΔS universe = ΔS working fluid + ΔS source = 0
ΔS source= -ΔS working fluid = -1.337 kJ/K
Answer:
The answer to your question is: ΔHrxm = -23.9 kJ
Explanation:
Data:
2Fe(s)+3/2O2(g)→Fe2O3(s), ΔH = -824.2 kJ (1)
CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ (2)
Reaction:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
We invert (1) and change the sign of ΔH
Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ
We multiply (2) by 3
3( CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ) (2)
3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ
We add (1) and (2)
Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ
3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ
Fe2O3(s) + 3CO(g)+3/2O2(g) → 2Fe(s)+3/2O2 + 3CO2(g)
Simplify
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) and ΔHrxm = -23.9 kJ