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Varvara68 [4.7K]
3 years ago
10

What gas has formed when bubels form when water is boiled?

Chemistry
1 answer:
My name is Ann [436]3 years ago
4 0

The gas has formed when bubbles form when water is boiled are hydrogen and oxygen gas. They came from the reaction 2H2O ->  2H2 + O2 were O2 and N2 will separate after heat treatment.

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Air and water are both considered to be ____.
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Elements.

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Which of the following is not a type of thematic map? Help please
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Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating
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<u>Answer:</u> The percent yield of the reaction is 68.68%.

<u>Explanation:</u>

To calculate the mass of salicylamide, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g

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 salicylamide = 3.869 g

Molar mass of salicylamide = 137.14 g/mol

Putting values in equation 1, we get:

\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol

The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:

\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = \frac{1}{1}\times 0.0295=0.0295moles of iodo-salicylamide

Now, calculating the mass of iodo-salicylamide from equation 1, we get:

Molar mass of iodo-salicylamide = 263 g/mol

Moles of iodo-salicylamide = 0.0295 moles

Putting values in equation 1, we get:

0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g

To calculate the percentage yield of iodo-salicylamide, we use the equation:

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

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%

Hence, the percent yield of the reaction is 68.68%.

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