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Svetlanka [38]
3 years ago
8

A 10.0 g ice cube is placed into 250 g of water with an initial temperature of 20.0 C. If the water drops to a temperature of 16

.8 C, has a specific heat of 4.18 J/g*K, what is the enthalpy of fusion of the ice. Ignore the fact that the ice, once melted, has to be heated again.
Chemistry
1 answer:
Radda [10]3 years ago
3 0

the mass of ice taken = 10 g

the mass of water = 250 g

initial temperature of water = 20 C

the final temperature of water = 16. 8 C

specific heat of water = 4.18 J/g*K

the heat absorbed by ice to melt = heat loss by water

heat loss by water = mass X specific heat of water X change in temperature

heat loss by water = 250 X 4.18 X (20-16.8) = 3344 Joules

heat gained by ice = 3344 J

heat gained by ice = enthalpy of fusion X moles of ice

moles of ice = mass / molar mass = 10 / 18 = 0.56 moles

enthalpy of fusion = 3344 / 0.56 = 5971.43 J / mole

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Answer:

The answers to the question are

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Explanation:

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number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of  carbon to fore 0.3009 moles of  CO₂

The maximum mass of carbon dioxide that  can be formed = mass = moles × molar mass

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(b) The formula for  the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

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molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

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