I think it enters a new material
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
Answer:
<em> 14, 508J/K</em>
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
<em>m = mass of substance (g) = 0.1184g</em>
1 mole of Mg - 24g
<em>n</em> moles - 0.1184g
<em>n = 0.0049 moles.</em>
Also, q = m × c × ΔT
<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>
<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>
<em>= 14, 508 J/K/kg</em>
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>
<em>= 1,7117.7472 J °C-1 g-1</em>
<em />
<em>∴ ΔHrxn = q/n</em>
<em>=1,7117.7472 ÷ 0.1184 </em>
<em>= 14, 508J/K</em>
B. 538 j hoped this helps
