Answer:
The original concentration of ethanol was 18 % (v/v)
Explanation:
For the analysis of a sample, mostly dilutions are made of the original concentrated sample. The analysis results obtained from the diluted sample are then calculated for the concentrated samples.
According to the given data, the wine sample was diluted 75 times. This means that the initial concentration of ethanol was 75 times more than the final concentration i.e. 0.24 % (v/v). So mathematically:
original concentration (v/v)= final concentration (v/v) x times diluted
original concentration (v/v)= 0.24 % x 75
original concentration (v/v) = 18 %
Answer:
2 NO (g) → N2 (g) + O2 (g)
2 NOCl (g) → 2 NO (g) + Cl2 (g)
____________________________
2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)
ΔH = [90.3 kJ x 2 x -1] + [-38.6 kJ x -1 x 2] = -103.4 kJ
The ΔH for the reaction is -103.4 kJ
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