Answer:
ΔG°′ = 1.737 KJ/mol
Explanation:
The reaction involves the transfer of two electrons in the form of hydride ions from reduced coenzyme Q, CoQH₂ to fumarae to form succinate and oxidized coenzyme Q, CoQ.
The overall equation of reaction is as follows:
fumarate²⁻ + CoQH₂ ↽⇀ succinate²⁻ + CoQ ; ΔE∘′=−0.009 V
Using the equation for standard free energy change; ΔG°′ = −nFΔE°′
where n = 2; F = 96.5 KJ.V⁻¹.mol⁻¹; ΔE°′ = 0.009 V
ΔG°′ = - 2 * 96.5 KJ.V⁻¹.mol⁻¹ * 0.009 V
ΔG°′ = 1.737 KJ/mol
B, D and E
Explanation:
conversion factors
1c = 8oz
1pt = 2c
1qt = 2pt
For A and B
ounces to cup = 160/8 = 20c
cup to pints = 20c / 2c = 10pt
pint to quarts = 10pt/2pt = 5qt
B applies as four 1-quart and two 1-pt = 5 1-quart
Considering C
4 8oz = 32oz
160-32 = 128oz /8 = 16c/2 = 8pints C does not apply
considering D
8 * 8 = 64oz
160 - 64 = 96oz/8 = 12c/2 = 6pt/2 = 3qt
So D applies
E applies
1bonding and 3non-bonding
(Refer to the attachment for structure
Answer:
There are 0.93 g of glucose in 100 mL of the final solution
Explanation:
In the first solution, the concentration of glucose (in g/L) is:
15.5 g / 0.100 L = 155 g/L
Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.
- 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)
The concentration of the second solution is:
So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:
1 L --------- 9.3 g
0.1 L--------- Xg
Xg = 9.3 g * 0.1 L / 1 L = 0.93 g
1) Temperature (heat) of the solution
2) Concentration (amount) of both solvent (usually water) and solute (substance being dissolved by solvent)
3) Movement (kinetic energy) of the solution, as in shaking/stirring
