We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.
The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.
Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-
N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)
According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia,
moles = 0.375 moles of nitrogen is required.
Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.
Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.
Ammonium acetate:
Formula C2H7NO<span>2</span>
You have to use density=mass/volume formula:

Solve for x and you get
2282.75 grams
A Brønsted-Lowry base is a base is a proton acceptor.
In the only case where this is done is when HCO3- accepts a proton and becomes H2CO3.
In the other cases, HCO3- is donating a proton which makes it an acid.
Mayonnaise is a heterogeneous mix. The correct answer is colloids.