The molar heat of fusion of iron is #14.9# kJ/mol. What is the energy (in kJ) needed to melt #4.50g# of iron?

Answer 1

The energy needed is #1.20# kJ for #4.50# gram iron.

To determine the energy that is needed to melt the iron, we have to calculate the amount of mol #Fe# (iron) we have. This can be done with the molar mass of iron, which is #55.85 u# (look up in the periodic table!)
Now we use the formula: #mol=("mass" color(white)(a) color(blue)((gram)) )/("molar mass"color(white)(a)color(blue)(((gram)/(mol)))#
We fill the numbers in and we obtain: #color(white)(a) color(red)("mol Fe")=(4.5color(white)(a)color(red)(cancel(color(blue)(gram))))/(55.85color(white)(a)color(red)(cancel(color(blue)(gram)))/color(blue)(mol))=0.081#
We use the molar heat of iron (#14.9# kJ/mol) to calculate the amount of energy that is needed to melt it.

We make a table:

#color(red)("mol Fe") color(white)(aaaaa)color(red)("Energy needed (kJ)"# #color(white)(aa)1color(white)(aaaaaa):color(white)(aaaaaaa)14.9# #color(white)(aaaaaaaaa)cancel(color(black)(\))# #color(white)(aa)0.081color(white)(aaa):color(white)(aaaaaaa)?#
To calculate the energy needed we use the ratio (the cross) and calculate: #(0.081xx14.9)/1=1.20# kJ

Therefore to melt down the 4.50 gram of iron we need 1.20 kJ energy.

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Answer 2

To calculate the energy needed to melt 4.50 grams of iron, you can use the formula:

Energy = (mass * molar heat of fusion) / molar mass

First, convert the mass of iron from grams to moles using the molar mass of iron (Fe), which is approximately 55.85 g/mol.

Number of moles = mass / molar mass Number of moles = 4.50 g / 55.85 g/mol ≈ 0.0805 mol

Now, use the molar heat of fusion of iron (14.9 kJ/mol) to calculate the energy needed:

Energy = (0.0805 mol * 14.9 kJ/mol) / 1 mol Energy ≈ (1.19545 kJ) / 1 mol ≈ 1.20 kJ

Therefore, the energy needed to melt 4.50 grams of iron is approximately 1.20 kJ.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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