Answer: option D. triple bond and one lone pair
Explanation:
Raised temperature, decreased volume.
Temperature and Pressure are directly related, when volume increases so does the your pressure.
Volume and Pressure are indirectly related. When volume decreases, your pressure will increase.
Answer : The correct option is, (c) 79.62
Explanation :
The formula used for percent humidity is:
..........(1)
The formula used for relative humidity is:
...........(2)
where,
= partial pressure of water vapor
= vapor pressure of water
p = total pressure
First we have to calculate the partial pressure of water vapor by using equation 2.
Given:


Relative humidity = 80 % = 0.80
Now put all the given values in equation 2, we get:


Now we have to calculate the percent humidity by using equation 1.


Therefore, the percent humidity is 79.62 %
Answer:
Avogadro's number is the number of particles in one mole of anything. In this context, it is the number of atoms in one mole of an element. It's easy to find the mass of a single atom using Avogadro's number. Simply divide the relative atomic mass of the element by Avogadro's number to get the answer in grams.
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