Answer:
C would be the right answer
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
I think the answer for this is 4702.5 J/g*k Depending on if it is water as a solid liquid or gas. I used water as a liquid when I solved it. J=(75g)(4.18 J/g*k)(15K)
Answer:
Explanation:
Step 1
The question is based on the concept of PH and pOH calculations.
pH is defined as negative logarithmic of hydronium Ion concentration.
while pOH is defined as negative logarithmic of hydroxide ion concentration of the solution.
Step 2
[H+] = 7.7*10-7 M
pH = -log[H+]
= -log ( 7.7*10-7 )
= 6.12
Step 3
pOH = 14 - pH
= 14 - 6.11
= 7.89