The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.
The balanced chemical equation for the reaction can be represented as,
Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL
Heat of the reaction, q = 
Δ
m is mass of the solution = 151.8 mL * 
C is the specific heat of solution = 4.18 
ΔT is the temperature change = 
q = 
Moles of NaOH = 
NaOH
Moles of 
= 
Enthalpy of the reaction = 
I answered all of them except 2 for you to do
Hope this helps :))
The solution would be like this for this specific problem:
Given:
pH of a 0.55 M hypobromous
acid (HBrO) at 25.0 °C = 4.48
[H+] = 10^-4.48 = 3.31 x
10^-5 M = [BrO-] <span>
Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>
To add, Hypobromous Acid does not require acid
adjustment, which is necessary for chlorine-based product and is stable and
effective in pH ranges of 5-9.<span>
</span>Hypobromous Acid combines with organic
compounds to form a bromamine. Chlorine also combines with the same organic
compounds to form a chloramine. <span>It is also
one of the least expensive intervention antimicrobial compounds available.</span>
