Find the ΔG° for the combustion of half a mole of glucose C6H12O6 in kJ
1 answer:
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Answer: D=8.27 g/cm³
Explanation:
Density is mass/volume. Mass is in grams and volume is in liters. In this case, the problem wants our volume to be in cm³. All we need to do is to make some conversions to convert kg/m³ to g/cm³.
With this equation, the m³ and kg cancel out, and we are left with g/cm³.
D=8.27 g/cm³
Answer:
See explanation and image attached
Explanation:
The standard cell potential at 298 K is given by;
E°cathode - E°anode
Hence;
E°cell = 0.34 V - (-0.76 V)
E°cell = 0.34 V + 0.76 V
E°cell = 1.1 V
To reduce Zn^2+ to Zn then Zn must be the cathode, hence;
E°cell = (-0.76 V) - 0.34 V
E°cell = -1.1 V
