Answer:
The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
Explanation:
As HCl is a strong acid and hence a strong electrolyte, it will dissociate as
HCl ⟶ H⁺ + Cl⁻
So, The concentration of H⁺ will be 2.5 M (same as HCl)
Thus, The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
<u>-TheUnknownScientist</u><u> 72</u>
<span>write out the balance equation
3NaOh+H3PO4->Na3PO4+3H2O
You are given everything needed to calculate
q=heat transfer=2.2*10^2,
H3PO4 moles= 1.5*10^-3,
NaOH moles=5.0*10^-3
equation is deltaHneutraliztion=q/Moles of limiting reagent
H3PO4 is limiting reagent because lowest moles, and is used up first
Now plug in variables
DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole
Notice we had to convert J to kj,</span>
E = mct
Energy = (mass) x (specific heat capacity of water) x (change in temp)
585.24 = 53.2 x 4.2 x (X-24.15)
585.24 divided by 53.2 divided by 4.2 = X - 24.15
2.62 = X - 24.15
X= 26.77degrees C
(Specific heat capacity for water is 4.2 but is different for other liquids)
Answer:
25.7 kJ/mol
Explanation:
There are two heats involved.
heat of solution of NH₄NO₃ + heat from water = 0
q₁ + q₂ = 0
n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃
∴ n = 0.100 mol NH₄NO₃
q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln
m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g
q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J
q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0
ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol