The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted - 0.126 mol/L x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol
then 1000 mL contains - 0.00109 mol / 25 mL x 1000 mL = 0.0436 mol/L
therefore molarity of H₂SO₄ is 0.0436 M
H is the answer :)
Hope that helps
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
