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3 years ago

How much total energy must be absorbed by a 150 g sample of ice at 0.0°C that it melts

Chemistry

1 answer:

zubka84 [21]3 years ago

7 0

Answer:

The answer to your question is Qt = 15750 cal

Explanation:

Data

mass = m = 150 g

Initial temperature = T1 = 0°C

Final temperature = T2 = 25°C

Process

1.- Calculate the Latent heat of fusion

Latent heat of fusion for water = l = 80 cal/g

Heat = Q

Formula

Q = ml

Substitution

Q = (150)(80)

Simplification and result

Q = 12000 cal

2.- Calculate the Specific heat

Specific heat of water = C = 1 cal/g°C

Formula

Q = mCΔT

Substitution

Q = (150)(1)(25 - 0)

Simplification and result

Q = 3750 cal

3.- Calculate the total heat

Qt = 12000 + 3750

Qt = 15750 cal

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A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate

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Answer: Empirical formula is $C_2H_5O$

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

Step 1 : Converting each of the given masses into their moles by dividing them by Molar masses.

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = $\frac{330g}{12g/mol}=27.5moles$

Moles of Hydrogen = $\frac{69.5g}{1g/mol}=69.5moles$

Moles of Oxygen = $\frac{220.2g}{16g/mol}=13.76moles$

Step 2: Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

$\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2$

$\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5$

$\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1$

Step 3: Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = $C_2H_5O$

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A mole represents a macroscopic quantity of matter that can be used in the laboratory.

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Answer:

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