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3 years ago

The fluorocarbon compound C2Cl3F3 has a normal boiling pointof

Chemistry

1 answer:

Dennis_Churaev [7]3 years ago

4 0

Answer:

5.21 kJ

Explanation:

The transformation happens in steps. First, the liquid solution gains heat (Q1) and has its temperature elevated from 5.00°C to 47.6°C. Then, it vaporizes at a constant temperature, with the heat of vaporization (Q2). After that, the substance gains heat (Q3) and the temperature increases to 82.00°C.

The heat of the transformation with the change in temperature can be calculated by:

Q = m*c*ΔT, where m is the mass, c is the specific heat and ΔT the temperature variation.

The heat of the phase change is the heat of vaporization multiplied by the number of moles of the substance.

The total heat required is the sum of the heats of the steps. The molar mass of C2Cl3F3 is 187.4 g/mol. The number of moles is the mass divided by the molar mass:

n = 25/187.4 = 0.1334 mol

So,

Q1 = 25.0 g * 0.91 J/g.K * (47.6 - 5.00)K = 969.15 J

Q2 = 0.1334 mol * 27.49 kJ/mol = 3.6672 kJ = 3667.2 J

Q3 = 25.0 g * 0.67 J/g.K * (82.00 - 47.6)K = 576.2 J

Q = 969.15 + 3667.2 + 576.2

Q = 5212.55 J

Q = 5.21 kJ

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9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r

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Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

$\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O$

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

$\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\ \\ Percent\text{ }yield=0.9484\times100\% \\ \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}$

Therefore, the percent yield for the reaction is 95%.

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