The required mass of calcium bromide is 35.98 grams.
<h3>What is molarity?</h3>
Molarity is any solution is define as the number of moles of solute present in per liter of solution as;
M = n/V, where
- M = molarity = 4M
- V = volume = 45mL = 0.045L
Moles will be calculated by using the above equation as:
n = (4)(0.045) = 0.18 mole
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
- W = given mass
- M = molar mass
Mass of CaBr₂ = (0.18mol)(199.89g/mol) = 35.98g
Hence required mass of CaBr₂ is 35.98 grams.
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Answer:

Explanation:
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In this case, we write the reaction again:

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

Best regards.
In keeping with the general trends, K-Br will have the smallest bond energy. The bond energy refers to the energy that keeps the atoms in a bond together.
<h3>What is bond energy?</h3>
Bond energy is the energy that is required to hold atoms together in a bond. This energy must also be supplied when the atoms are to be separated.
We have the bond energies of each of the bons in the question, we have to note that the the smallest value of bond energy is Na-Br hence in keeping with the general trends, K-Br will have the smallest bond energy.
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True some explanations are not always based on empirical evidence