Answer:
K = 3.37
Explanation:
2 NH₃(g) → N₂(g) + 3H₂(g)
Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).
2 NH₃(g) → N₂(g) + 3H₂(g)
Initally 4moles - -
React 2moles 2m + 3m
Eq 2 moles 2m 3m
We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)
The expression for K is: ( [H₂]³ . [N₂] ) / [NH₃]²
We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)
K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²
K = 27/8 / 1 → 3.37
Answer:
Total ATP molecules produced = 66 molecules of ATP
Explanation:
A 10-carbon fatty acid when it has undergone complete oxidation will yield 5 acetyl-CoA molecules and 4 FADH₂ and 4 NADH molecules each. Each of the 5 acetyl-CoA molecules enters into the citric acid cycle and is completely oxidized to yield further ATP and FADH₂ and NADH molecules.
The total yield of ATP in the various enzymatic step is calculated below:
Acyl-CoA dehydrodenase = 4 FADH₂
β-Hydroxyacyl-CoA dehydrogenase = 4 NADH
Isocitrate dehydrogenase = 5 NADH
α-Ketoglutarate dehydrogenase = 5 NADH
Succinyl-CoA synthase = 5 ATP (from substrate-level phosphorylation of GDP)
Succinate dehydrogenase = 5 FADH₂
Malate dehydrogenase = 5 NADH
Total ATP from FADH₂ molecoles = 9 * 1.5 = 13.5
Total NADH molecules = 19 * 2.5 = 47.5
Total ATP molecules produced = 13.5 + 47.5 + 5
Total ATP molecules produced = 66 molecules of ATP
Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be 
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio 
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.
