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.
Ion, any atom or group of atoms that bears one or more positive or negative electrical charges. Positively charged ions are called cations; negatively charged ions, anions.
B. Heat is a form of energy so boiling it would increase energy. (I guess)
Answer:
Mass = 51 g
Explanation:
Given data:
Mass of nitrogen = 41.93 g
Mass of ammonia formed = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 41.93 g/ 28 g/mol
Number of moles = 1.5 mol
now we will compare the moles of nitrogen and ammonia.
N₂ : NH₃
1 : 2
1.5 : 2/1×1.5 = 3 mol
Mass of ammonia formed:
Mass = number of moles × molar mass
Mass = 3 mol × 17 g/mol
Mass = 51 g
Answer:
It is because water is liquid and it can be weathered easily. While weathering by solid is a bit dangerous because of its hardness.
Explanation:
