The molar heat of fusion of gold is 12.550 kJ mol–1. At its melting point, how much mass of melted gold must solidify to release
1 answer:
The mass of melted gold to release the energy would be 3, 688. 8 Kg
<h3>How to determine the mass</h3>We have quantity of energy is;
Q = n × HF
n = number of moles
HF = heat of fusion
Let's find number of moles
235.0 = n × 12.550
number of moles = = 18. 725 moles
Note that molar mass of Gold is 197g/ mol
Number of moles = mass/ molar mass
Mass = number of moles × molar mass
Mass = 18. 725 × 197
Mass = 3, 688. 8 Kg
Thus, the mass of melted gold to release the energy would be 3, 688. 8 Kg
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Answer:
6,78 mL of 12,0 wt% H₂SO₄
Explanation:
The equilibrium in water is:
H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)
The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = , thus,
Thus, you need to add:
[H⁺] = = 5,31x10⁻⁸ M
The total volume of the pool is:
9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L
Thus, moles of H⁺ you need to add are:
5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺
These moles comes from
H₂SO₄ → 2H⁺ +SO₄²⁻
Thus:
1,792x10⁻² moles of H⁺ × = 8,96x10⁻³ moles of H₂SO₄
These moles comes from:
8,96x10⁻³ moles of H₂SO₄ × ×
×
=
6,78 mL of 12,0wt% H₂SO₄
I hope it helps!