Answer is: enthalpy is -1276,8 kJ.
Chemical reaction: 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂<span>O(g).
</span>ΔH(CH₃OH) = -238,7 kJ/mol.
ΔH(H₂O) = -241,8 kJ/mol.
ΔH(CO₂) = -<span>393,5 kJ/mol.
</span>ΔH(O₂) = 0 kJ/mol.
ΔHreaction = ∑ΔHproducts - ∑ΔHreactants
ΔHreaction = (2·(-393,5) + 4·(-241,8) - (2·(-238,7))
ΔH = (-787 - 967,2) + 477,4
ΔH = -1276,8 kJ.
Answer:
![M_{base}=0.957M](https://tex.z-dn.net/?f=M_%7Bbase%7D%3D0.957M)
Explanation:
Hello,
This problem could be solved in terms of normality as follows:
- The normality of phosphoric acid (triporitic) is:
![N_{acid}=0.314\frac{mol}{L}*\frac{3eq-g}{1mol}=0.942N](https://tex.z-dn.net/?f=N_%7Bacid%7D%3D0.314%5Cfrac%7Bmol%7D%7BL%7D%2A%5Cfrac%7B3eq-g%7D%7B1mol%7D%3D0.942N)
Thus, the normality of barium hydroxide turns out:
![N_{base}=\frac{N_{acid}V_{acid}}{V_{base}}=\frac{0.942N*32.7mL}{16.1mL} =1.91N](https://tex.z-dn.net/?f=N_%7Bbase%7D%3D%5Cfrac%7BN_%7Bacid%7DV_%7Bacid%7D%7D%7BV_%7Bbase%7D%7D%3D%5Cfrac%7B0.942N%2A32.7mL%7D%7B16.1mL%7D%20%3D1.91N)
Finally, the molarity:
![M_{base}=1.91\frac{eq-g}{L}*\frac{1mol}{2eq-g}=0.957M](https://tex.z-dn.net/?f=M_%7Bbase%7D%3D1.91%5Cfrac%7Beq-g%7D%7BL%7D%2A%5Cfrac%7B1mol%7D%7B2eq-g%7D%3D0.957M)
Best regards.
Answer:
there is 1 leftover because your adding to 1... I think
Explanation:
Answer:
Fermentation
Explanation:
If your body were to break down glucose with the use of oxygen then it would be aerobic, but since oxygen isn't being used in this case, the process would be anaerobic or fermentation.
Hope this helps :)
I suck at explaining biology questions, but I tried.