The molar heats of fusion and vaporization for silver are 11.3 kJ/mol and 250. kJ/mol, respectively. Silver's normal melting point is 962°C, and its normal boiling point is 2212°C. What quantity of heat is required to melt 13.0 g of silver at 962°C?

To find the quantity of heat required to melt 13.0 g of silver at 962°C, we need to use the equation:

Q = m × ΔHf

Where:
Q is the heat required (in joules),
m is the mass of the substance (in grams), and
ΔHf is the molar heat of fusion (in J/mol).

First, we need to calculate the number of moles of silver:

moles = mass / molar mass

The molar mass of silver (Ag) is approximately 107.87 g/mol.

moles = 13.0 g / 107.87 g/mol
moles = 0.1204 mol

Next, we can calculate the heat required to melt the silver:

Q = moles × ΔHf

The molar heat of fusion for silver is 11.3 kJ/mol, which is equivalent to 11,300 J/mol.

Q = 0.1204 mol × 11,300 J/mol
Q = 1,361.72 J

Therefore, the quantity of heat required to melt 13.0 g of silver at 962°C is approximately 1,361.72 J.

To calculate the quantity of heat required to melt 13.0 g of silver at 962°C, we need to use the molar heat of fusion and the molar mass of silver.

1. Calculate the moles of silver:
- Molar mass of silver (Ag) = 107.87 g/mol
- Moles of silver = Mass of silver (g) / Molar mass of silver (g/mol)
- Moles of silver = 13.0 g / 107.87 g/mol

2. Use the molar heat of fusion to calculate the heat required to melt the moles of silver:
- Heat required to melt the moles of silver = Moles of silver × Molar heat of fusion

Now, let's substitute the values into the equations:

1. Moles of silver:
Moles of silver = 13.0 g / 107.87 g/mol
≈ 0.1205 mol

2. Heat required to melt the moles of silver:
Heat required to melt the moles of silver = 0.1205 mol × 11.3 kJ/mol
≈ 1.36 kJ

Therefore, approximately 1.36 kJ of heat is required to melt 13.0 g of silver at 962°C.

heat to melt= mass*latentheatfusion

= 13.0g*11.3kJ/mol *1mol/xxxgrams

so look up xxxg as the molemass of Ag