It is typically formed by the evaporation of salty water (such as sea water) which contains dissolved Na+ and Cl- ions. One finds rock salt deposits ringing dry lake beds, inland marginal seas, and enclosed bays and estuaries in arid regions of the world
Answer:
Products are AgBr and KNO3
Answer:
No
Explanation:
<u>In order for glucose to be produced inside the mixture, photosynthesis has to take place</u>. The photosynthetic process requires a series of steps within an organelle called the <u>chloroplast</u>. The chloroplast contains the chlorophyll and other enzymes that are necessary for photosynthesis.
<em>Once the chlorophyll is isolated, it becomes separated from the enzymes necessary for the completion of photosynthesis, and the process is truncated. </em>When light is shined on the mixture, the majority would instead be lost as heat while some cause the chlorophyll molecules to glow.
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
