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A 250 gram sample of water at the boiling point had 45.0 kj of heat added. how many grams of water were vaporized? heat of vaporization for water is 40.6 kj/mole.

Respuesta :

haneylia
You have 45 kJ of heat added and for every 40.6 kJ added, one mole is able to be vaporized. Find how many moles you can vaporize.

45 kJ / 40.6 kJ = 1.1 moles

Water is roughly 18 g per mole so 1.1 moles x 18 g per mol = 19.95 grams vaporized

Hope I helped!


dsdrajlin

When a 250 gram sample at the boiling point received 45.0 kJ of heat, 20.0 g of water were vaporized.

We have a 250 gram sample at the boiling point that received 45.0 kJ of heat (Q).

Since the heat of vaporization of water (ΔHvap) is 40.6 kJ/mol, we can calculate the moles (n) of water vaporized using the following expression.

[tex]45.0 kJ \times \frac{1mol}{40.6kJ} = 1.11 mol[/tex]

Then, we will convert 1.11 moles of water to grams using its molar mass (18.02 g/mol).

[tex]1.11 mol \times 18.02 g/mol = 20.0 g[/tex]

When a 250 gram sample at the boiling point received 45.0 kJ of heat, 20.0 g of water were vaporized.

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