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MAVERICK [17]
4 years ago
6

To determine whether a shiny gold-colored rock is actually gold, a chemistry student decides to measure its heat capacity. She f

irst weighs the rock and finds that it has a mass of 4.7 g. She then finds that, upon absorption of 52.7 J of heat, the rock undergoes a rise in temperature from 25 ∘C to 57 ∘C. Find the specific heat capacity of the substance comprising the rock.
Chemistry
1 answer:
zhenek [66]4 years ago
6 0

Answer:

The answer to your question is: Cp = 350.4 J/ kg°C

Explanation:

Data

mass = m = 4.7 g   convert grams to kilograms  0.0047 kg

heat= Q = 52.7 J

Temperature = 25 - 57 °C

Formula

Q = mCpΔT    solve for Cp

Cp = Q / mΔT

Process

Cp = 52.7 / [(0.0047)(57 - 25)]            Substitution

Cp = 52.7 / 0.1504                                

Cp = 350.4 J/ kg°C

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<u>Answer:</u> The empirical formula for the given compound is C_3H_6O

<u>Explanation:</u>

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where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

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We know that:

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Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, \frac{12}{44}\times 0.00632=0.00172g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, \frac{2}{18}\times 0.00258=0.000286g of hydrogen will be contained.

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To formulate the empirical formula, we need to follow some steps:

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Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 4.83\times 10^{-5}mol

For Carbon = \frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3

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For Oxygen  = \frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is C_3H_{6}O_1=C_3H_6O

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