C. Increases. Increasing temperature=Increasing Volume
The mass of a sample of alcohol is found to be = m = 367 g
Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.
Explanation:
The Energy of the sample given is q = 4780
We are required to find the mass of alcohol m = ?
Given that,
The specific heat given is represented by = c = 2.4 J/gC
The temperature given is ΔT = 5.43° C
The mass of sample of alcohol can be found as follows,
The formula is c = 
We can drive value of m bu shifting m on the left hand side,
m = 
mass of alcohol (m) = 
m = 367 g
Therefore, The mass of the given sample of alcohol is
m = 367g
It requires 4780 J of heat to raise the temperature by 5.43 C in the process which yields a mass of 367 g of alcohol.
The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

The pH of the solution is 2.54.
Explanation:
pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.
The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.
So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the
![K_{a}=\frac{[H^{+}][HB] }{[reactant]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BHB%5D%20%7D%7B%5Breactant%5D%7D)
[HB] is the concentration of base.


Then
![pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5Bx%5D%20%3D%20-%20log%20%5B%200.283%20%2A%2010%5E%7B-2%7D%5D%5C%5C%20%5C%5CpH%20%3D%202%20%2B%200.548%20%3D%202.54)
So the pH of the solution is 2.54.