Answer:
The base must be ethylamine.
Explanation:
The pH of solution of a weak base gives us an idea about the Kb of the base.
![pOH=-log[OH^{-}]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E%7B-%7D%5D)
![[OH^{-}]=0.0062M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D0.0062M)
The relation between Kb and hydroxide ion concentration is:
![Kb=\frac{[OH^{-}]^{2}}{[base]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BOH%5E%7B-%7D%5D%5E%7B2%7D%7D%7B%5Bbase%5D%7D)
Thus the weak base must be ethylamine.
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
Answer:
1 por 10 y otro por el culo
Think it though and always go back to the cause of the problem
