The water molecules are not completely removed so additional heating is required.
Explanation:
We have the copper (II) sulfate pentehydrate with the chemical formula CuSO₄ · 5H₂O.
molar mass of CuSO₄ · 5H₂O = 159.6 + 5 × 18 = 249.6 g/mole
Knowing this, we devise the following reasoning:
if in 249.6 g of CuSO₄ · 5H₂O there are 90 g of H₂O
then in 8 g of CuSO₄ · 5H₂O there are Y g of H₂O
Y = (8 × 90) / 249.6 = 2.88 g of water
mass of dried CuSO₄ = mass of CuSO₄ · 5H₂O - mass of H₂O
mass of dried CuSO₄ = 8 - 2.88 = 5.12 g
5.12 g is less that the weighted mass of 6.50 g. We deduce from this that the sample needs additional heating in order to remove all the water (H₂O) molecules.
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Answer:
The no. of protons = atomic number
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It will be different according to how each of them is easily warmed up when using thermal energy.
Water heats at 100 celsius and all the other solutions have slightly different boiling points
The sun heats the land
hot air rises
cold air from the ocean/body of water rushes in under the hot air
wind
The molar heat of fusion for iron with a mass of 200.0g releases 9,840 cal when it freezes at its freezing point is 2,747.7 cal/mol.
<h3>How to calculate molar heat of fusion?</h3>
The heat of fusion of a substance can be calculated by using the following formula:
Q = m∆H
Where:
- Q = quantity of heat
- m = mass
- ∆H = change in temperature of fusion
However, the quantity of heat has been given as 9840calories. The molar heat of fusion of iron can be calculated by dividing the heat of fusion by the number of moles of iron.
Moles of iron = mass ÷ molar mass
moles = 200g ÷ 55.8g/mol
moles = 3.58moles
molar heat of fusion = 9840 cal ÷ 3.58mol
molar heat of fusion = 2748.6 cal/mol
Therefore, the molar heat of fusion for iron with a mass of 200.0g releases 9,840 cal when it freezes at its freezing point is 2,747.7 cal/mol.
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