Answer:
351.43mL
Explanation:
To calculate the original volume of hydrogen gas in this question, the Boyle's law equation will be used. Boyle's law equation is:
P1V1 = P2V2
Where; P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
According to this question, the P1= 1.56atm, V1 = ?, P2 = 0.73atm, V2 = 751mL
Hence;
P1V1 = P2V2
1.56 × V1 = 0.73 × 751
1.56 V1 = 548.23
V1 = 548.23/1.56
V1 = 351.43mL
Therefore, the original volume of hydrogen gas is 351.43 mL.
Answer:- 335 kcal of heat energy is produced.
Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.
We could solve this using dimensional analysis as:

= 1401.5 kJ
Now, let's convert kJ to kcal.
We know that, 1kcal = 4.184kJ
So, 
= 335 kcal
Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.
Answer:
136.63 °C
Explanation:
ΔTb=Tb solution - Tb pure
Where; Tb pure = 133.60°C
molar mass of solute = 121.14 g/mol
number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles
molality = 0.431 moles/350 * 10^-3 = 1.23 molal
Then;
ΔTb = Kb * m * i
Kb = 2.46°C kg mol^-1
m = 1.23 molal
i = 1
ΔTb = 2.46 * 1.23 * 1
ΔTb = 3.03 °C
Hence;
Tb solution = ΔTb + Tb pure
Tb solution = 3.03 °C + 133.60°C
Tb solution = 136.63 °C