Answer:
Similarities: both state the mass of chemical species and they have the same numerical value
Differences: molecular mass refers to one single molecule and molar mass refers to one mole of a molecule
Explanation:
The molecular mass is the value of the mass of each molecule and it is measured in mass units (u). It is calculated adding the mass of each atom of the molecule.
The molar mass is the value of the mass of one mole of molecules, which means the mass of 6.022140857 × 10²³ molecules. The unit is g/mol.
For example, we can consider the methane molecule, which has the chemical formula of CH₄:
Molecular mass CH₄ = C mass + 4 x (H mass)
Molecular mass CH₄ = 12.01 + 4 x (1.01)
Molecular mass CH₄ = 16.05 u
Now to calculate the molar mass we multiply the value of the molecular mass by the Avogadro number and convert the units to g/mol:
Molar mass CH₄: 16.05 x 
x 6.022140857 × 10²³ mol⁻¹
Molecular mass CH₄ = 16.05 g / mol
This is because amino group of p-aminobenzoic acid is an aniline and is less electrophilic than an alkyl amine.
<h3>What is an Aniline?</h3>
This is an aromatic amine which consists of a phenyl group attached to an amino group.
The amino group of p-aminobenzoic acid being an aniline makes it less electrophilic which is why an alkyl amine participates in the reaction instead.
Read more about Aniline here brainly.com/question/9982058
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: 51.9961 g/mol, don't know if it helps :)
Explanation:
