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3 years ago

Find the relative equations from the table shown: (40 pts.)

Chemistry

1 answer:

Artemon [7]3 years ago

6 0

Answer:

orbitals s p d f

l(orbital number) 0 1 2 3

energetic region n l Er= n + l

7f 7 3 7+3 = 10

5d 5 2 5+2 = 7

3s 3 0 3 + 0 = 3

6p 6 1 6 + 1 = 7

4f 4 3 4 + 3 = 7

6d 6 2 6 + 2 = 8

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The electrolysis of water forms H2 and O2. 2H2O 2H2 + O2 What is the percent yield of O2 if 10.2 g of O2 is produced from the de

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Answer: The percent yield of oxygen gas is 67.53 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of water = 17.0 g

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Putting values in equation 1, we get:

$\text{Moles of water}=\frac{17.0g}{18g/mol}=0.944mol$

For the given chemical equation:

$2H_2O\rightarrow 2H_2+O_2$

By Stoichiometry of the reaction:

2 moles of water produces 1 mole of oxygen gas

So, 0.944 moles of water will produce = $\frac{1}{2}\times 0.944=0.472moles$ of oxygen gas

Now, calculating the mass of oxygen gas from equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 0.472 moles

Putting values in equation 1, we get:

$0.472mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.472mol\times 32g/mol)=15.104g$

To calculate the percentage yield of oxygen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of oxygen gas = 10.2 g

Theoretical yield of oxygen gas = 15.104 g

Putting values in above equation, we get:

$\%\text{ yield of oxygen gas}=\frac{10.2g}{15.104g}\times 100\\\\\% \text{yield of oxygen gas}=67.53\%$

Hence, the percent yield of oxygen gas is 67.53 %.

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4 years ago

Read 2 more answers

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