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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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A 0.2 g sample of pyrolusite is analyzed for manganese content as follows. Add 50.0 mL of 0.1 M solution of ferrous ammonium sul
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Answer:

66.7%

Explanation:

The reaction for the titration of the excess ferrous ion is:

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We calculate the moles of Fe⁺² from the used moles of KMnO₄:

  • 0.02 M * 15.0 mL = 0.30 mmol KMnO₄
  • 0.3 mmol KMnO₄ * \frac{5mmolFe^{+2}}{1mmolKMnO_4} = 1.5 mmol Fe⁺²

Then we substract those 0.30 mmol from the original amount used:

  • 0.1 M * 50.0 mL = 5.0 mmol Fe⁺²
  • 5.0 - 1.5 = 3.5 mmol Fe⁺²

The reaction between ferrous ammonium sulfate and MnO₂ is:

  • 2Fe⁺² + MnO₂ + 4H⁺ → 2Fe³⁺ + Mn²⁺ + 2H₂O

So we convert those 3.5 mmol Fe⁺² that were used in this reaction to MnO₂ moles:

  • 3.5 mmol Fe⁺²  * \frac{1mmolMnO_2}{2mmolFe^{+2}}= 1.75 mmol MnO₂

Then we convert MnO₂ to Mn₃O₄, using the reaction:

  • 3MnO₂ → Mn₃O₄ + O₂
  • 1.75 mmol MnO₂ * \frac{1mmolMn_3O_4}{3mmolMnO_2} = 0.583 mmol Mn₃O₄

Finally we convert Mn₃O₄ moles to grams:

  • 0.583 mmol Mn₃O₄ * 228.82 mg/mmol = 133.40 mg Mn₃O₄

And calculate the percent

  • 0.2 g = 200 mg
  • 133.40 / 200 * 100% = 66.7%
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