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Kruka [31]
3 years ago
8

Please help with the attached picture question 21, 22, 23

Chemistry
1 answer:
klasskru [66]3 years ago
5 0

Explanation:

21. The given molecule for cracking is tetradecane.

On cracking it forms one mole of decane (C10H22) and two moles of ethene gas.

The chemical equation is shown below:

C_1_4H_3_0->C_1_0H_2_2+2C_2H_4

22. The essential condition for the formation of an ester is the reaction of alcohol and acid in presence of concentrated sulfuric acid.

Thus among the given options, the first option is the correct one.

23. Isomers of butanol are shown below:

It is 2-butanol.

The position of -OH group changes to the second carbon.

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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
Mumz [18]

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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Need help with this.
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Answer:

Explanation:

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