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Lerok [7]
3 years ago
10

When 45 g of an alloy, at 25oC, are dropped into 100.0 g of water, the alloy absorbs 956 J of heat. If the final temperature of

the alloy is 37oC, what is its specific heat?
Chemistry
1 answer:
taurus [48]3 years ago
3 0

Answer:

The answer to the question is

The specific heat capacity of the alloy = 1.77 J/(g·°C)

Explanation:

To solve this, we list out the given variables thus

Mass of alloy = 45 g

Initial temperature of the alloy = 25 °C

Final temperature of the alloy = 37 °C

Heat absorbed by the alloy = 956 J

Thus we have

ΔH = m·c·(T₂ - T₁) where  ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C  and m = mass of the alloy = 45 g

∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c  or

c = 956 J/(540 g·°C) = 1.77 J/(g·°C)

The specific heat capacity of the alloy is 1.77 J/(g·°C)

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Answer:

Molecular formula = C20H30

Explanation:

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Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01

Also from the question, moles of H2O = 0.135/18= 0.0075mole

Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H

To get the empirical formula, divide by smallest number of mole

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Answer:

See explanation and image attached

Explanation:

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