1.more
2.longer
3.warmer
4.northern
5.less
6.shorter
7.colder
8.southern
MA= output force/ input force
MA= 100N/20N
MA= 50
What was the question here?
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
Answer : The amount of formaldehyde permissible are, 
Explanation : Given,
Density of air = 
First we have to calculate the mass of air.
Now we have to calculate the amount of formaldehyde.
Permissible exposure level of formaldehyde = 0.75 ppm = 
Amount of formaldehyde in 7.2 g of formaldehyde = 
Amount of formaldehyde in 7.2 g of formaldehyde =
Thus, the amount of formaldehyde permissible are,
