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vovangra [49]
3 years ago
12

How much heat energy is needed to heat 250 g of water from 200C to its boiling point and then completely vaporize it?

Chemistry
1 answer:
alexandr1967 [171]3 years ago
8 0

Answer: 40.66kJ/mol.

Explanation:Assuming that pressure is equal to Explanation:Assuming that pressure is equal to 1 atm Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at

Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at 100

Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at 100∘

Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at 100∘C

Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at 100∘C to vapor at

Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at 100∘C to vapor at 100

Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at 100∘C to vapor at 100∘

Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at 100∘C to vapor at 100∘C

Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at 100∘C to vapor at 100∘C .

Explanation:Assuming that pressure is equal to 1 atm , boiling water at its boiling point implies providing it with enough heat to turn it from liquid at 100∘C to vapor at 100∘C .The amount of heat needed to allow one mole of water to undergo this phase change is called the enthalpy change of vaporization,

nge of vaporization, Δ

nge of vaporization, ΔH

nge of vaporization, ΔHvap

nge of vaporization, ΔHvap .

nge of vaporization, ΔHvap .For water at

nge of vaporization, ΔHvap .For water at 100

nge of vaporization, ΔHvap .For water at 100∘

nge of vaporization, ΔHvap .For water at 100∘C

nge of vaporization, ΔHvap .For water at 100∘C , the enthalpy change of vaporization is equal to

nge of vaporization, ΔHvap .For water at 100∘C , the enthalpy change of vaporization is equal toΔ

nge of vaporization, ΔHvap .For water at 100∘C , the enthalpy change of vaporization is equal toΔH

nge of vaporization, ΔHvap .For water at 100∘C , the enthalpy change of vaporization is equal toΔHvap

nge of vaporization, ΔHvap .For water at 100∘C , the enthalpy change of vaporization is equal toΔHvap=

nge of vaporization, ΔHvap .For water at 100∘C , the enthalpy change of vaporization is equal toΔHvap=40.66 kJ/mol

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The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.
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Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

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