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.
Answer:
the electricity dissipated, because of the time.
Explanation:
electricity dissipates.
The frequency of a wave represents B. the number of wave cycles that pass through a specific point within a given time.
The distance between two consecutive crests and the length of a wave are the <em>wavelength</em>.
The distance between the highest and lowest points of a wave is <em>twice the amplitude</em>.
Answer:
Size and Temperature or E & B
Explanation:
<h3>
Answer: 386.67 g/mol </h3>
Explanation:
Molar Mass = Mass ÷ Mole
= 0.406 g ÷ 0.00105 mol
= 386.67 g/mol
∴ molar mass of cholesterol = 386.67 g/mol