Q = m x c x ΔT
2500 = 0.135 x C x 80.5
2500 = 10.8765 x C
C = 230.043 J/Kg.K
hope this helps
Cl2(g) -------> Cl-(aq) + ClO-(aq)
2e- + Cl2(g) -------> 2Cl-(aq) [reduction]
4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
______________________________________...
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
(P1)(V1)=(P2)(V2)
(1.50)(5.00)=(1240/760)(V2)
(7.5)/(1240/760)=V2
V2=4.596774194 L
When sodium carbonate is dissolved in water, the equation is
.
When carbon dioxide is placed in water, aqueous carbon dioxide is formed: 
<h3>Dissolution of compounds in water</h3>
Some compounds are water-soluble, some are just partially soluble, while others are insoluble in water. Some soluble or partially soluble substances dissociate in water into their component ions. These substances are said to be ionic.
Sodium carbonate, like every other sodium salt, is soluble in water. It dissolves in water to form an aqueous solution of sodium carbonate.
While in solution, sodium carbonate dissociates into its component ions according to the following equation:

Carbon dioxide, on the other hand, does not dissociate in water. Instead, it dissolves in water where most of it remains as aqueous carbon dioxide in equilibrium with a small amount of hydronium ion and hydrogen carbonate ion.
Since the hydronium and hydrogen carbonate ions formed are so minute, the equation of the reaction can be written as: 
More on the dissolution of substances can be found here: brainly.com/question/28580758
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Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of
=
= 0.0458 M
Concentration of
=
= 0.0521 M
GIven that :
Ka = 
Thus; it's pKa = 4.72




pH ≅ 4.80