Answer:
True => ΔH°f for C₆H₆ = 49 Kj/mole
Explanation:
See Thermodynamic Properties Table in appendix of most college level general chemistry texts. The values shown are for the standard heat of formation of substances at 25°C. The Standard Heat of Formation of a substance - by definition - is the amount of heat energy gained or lost on formation of the substance from its basic elements in their standard state. C₆H₆(l) is formed from Carbon and Hydrogen in their basic standard states. All elements in their basic standard states have ΔH°f values equal to zero Kj/mole.
Answer:
Explanation:
Heterogeneous mixtures: Milk, salt and pepper, smog, chocolate chip cookie, oil and water.
Solutions we might use everyday: Gatorade, Apple Juice, most wines and liquor, liquid detergent, coffee etc.
Answer:
22.73s
Explanation:
The reaction is a second order reaction, we know this by observing the unit of the slope.
rate constant = k = 0.056 M-1s-1
the initial concentration of BrO- [A]o = 0.80 M
time = ?
Final concentration [A]t= one-half of 0.80 M = 0.40M
1 / [A]t = kt + 1 / [A]o
1 / 0.40 = 0.056 * t + 1 / 0.80
t = (2.5 - 1.25) / 0.056
t = 22.73s
....oxygen and sulfur this is cuz of the potential energy there in the same properties
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62