Answer:- 21.4 grams of 
are formed.
Solution:- The balanced equation is:
From this equation, lithium and nitrogen reacts in 6:1 mol ratio. Limiting reactant gives the theoretical yield of the product. We will calculate the grams of the product for the given grams of both the reactants and see which one of them gives the limited amount of the product. This limited amount of the product will be the theoretical yield.
The molar mass of Li is 6.94 gram per mol and for 
It is 28.02 gram per mol. The molar mass of is 34.83 gram per mol. The calculations for the grams of the product for given grams of both the reactants are shown below:
= 
= 
From above calculations, Li gives least amount of the product. So, 21.4 g of are formed.
Word equation: magnesium chloride + bromine -> magnesium bromide + chlorine
balance equation: MgCl2 + 2Br -> MgBr2 + 2Cl
It is non-polar molecule.
CF₄ - dipole moment = 0D
1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
= 
(1)
and we also know that:
n = M x 
If we replace this expression in (1) we have:
x 
= 
x
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is :
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
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