Answer:
A. since Nitrogen has 7 electrons and when it gains 3 electrons it will have 10 electrons. using short hand rule it is [Ne]
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:
Water molecules pull the sodium and chloride ions apart, breaking the ionic bond that held them together. After the salt compounds are pulled apart, the sodium and chloride atoms are surrounded by water molecules, as this diagram shows. Once this happens, the salt is dissolved, resulting in a homogeneous solution.
To figure out questions related to reacting moles/masses, the first step is always to write a complete balanced equation.
2Fe (s) + 3Cl2 (g) → 2FeCl3 (s)
Since Cl2 is the excess reactant, and Fe is the limiting reactant, we can simply find the number of moles of the product by comparing the mole ratio of the limiting reactant to the mole ratio of the product from the equation.
From the equation, mole ratio of Fe:FeCl3 = 2:2 = 1:1, the number of moles of product is exactly the same as the number of moles of the limiting reactant, which makes it 8 moles.
Note that if the mole ratio is not 1:1, you have to do some calculations to make sure the no. of moles is balanced at the end. Which means, if the mole ratio happened to be 1:2, the no. of moles of the product would be 8x2=16 instead.
So, your answer is 8 moles.
