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>The formula for quantity of energy is given thus;
Q = n × HF
Where n represents number of moles
HF represents heat of fusion
To find the number of moles, we have
235.0 = n × 12.550
number of moles = = 18. 725 moles
Note that molar mass of Gold is 197g/ mol
Let's note that;
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:
The solution's new volume is 1.68 L
Explanation:
Dilution is the procedure to prepare a less concentrated solution from a more concentrated one, and simply consists of adding more solvent. So, in a dilution the amount of solute does not vary, but the volume of the solvent varies.
In summary, a dilution is a lower concentration solution than the original.
The way to do the calculations in a dilution is through the expression:
Ci*Vi=Cf*Vf
where C and V are concentration and volume, respectively; and the i and f subscripts indicate initial and final respectively.
In this case, being:
- Ci= 7 M
- Vi= 0.60 L
- Cf= 2.5 M
- Vf=?
Replacing:
7 M*0.60 L= 2.5 M* Vf
Solving:
Vf= 1.68 L
<u><em>The solution's new volume is 1.68 L</em></u>