Answer : The final temperature of the solution in the calorimeter is, 
Explanation :
First we have to calculate the heat produced.

where,
= enthalpy change = -44.5 kJ/mol
q = heat released = ?
m = mass of
= 1.52 g
Molar mass of
= 40 g/mol

Now put all the given values in the above formula, we get:


Now we have to calculate the final temperature of solution in the calorimeter.

where,
q = heat produced = 1.691 kJ = 1691 J
m = mass of solution = 1.52 + 35.5 = 37.02 g
c = specific heat capacity of water = 
= initial temperature = 
= final temperature = ?
Now put all the given values in the above formula, we get:


Thus, the final temperature of the solution in the calorimeter is, 
2.25739773716275. I used a calculator during class today to get this answer, and I am pretty sure it is right, hope it helps.
PbO2
You have to take the mass of lead in the problem, and divide by the molar mass.
When you do the same with oxygen, you get a number about twice as large as when you divide the mass of lead by the molar mass of lead. This means that the simplest formula would be PbO2
Answer:

Explanation:
Given , molarity of glycerol= 
Volume= 1 L.
Therefore, No of moles of glycerol= 
Now, volume of water needed, V=998.8 mL.
Density is given as= 0.9982 g/mL.
Therefore, mass of water = 
Now, molality=
Hence, this is the required solution.