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4 years ago

Controls are defined as____

1 answer:

5 0

A scientific control is an experiment or observation designed to minimize the effects of variables other than the independent variable.

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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

sergeinik [125]

The question is incomplete, here is the complete question:

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

Answer: The percent yield of water in the reaction is 46.85 %.

Explanation:

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

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

For NaOH:

Given mass of NaOH = 77.0 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{77.0g}{40g/mol}=1.925mol$

For sulfuric acid:

Given mass of sulfuric acid = 72.6 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

So, 0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sulfuric acid produces 2 moles of water

So, 0.741 moles of sulfuric acid will produce = $\frac{2}{1}\times 0.741=1.482moles$ of water

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

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

$1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.68g$

To calculate the percentage yield of water, we use the equation:

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

Experimental yield of water = 12.5 g

Theoretical yield of water = 26.68 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{12.5g}{26.68g}\times 100\\\\\% \text{yield of water}=46.85\%$

Hence, the percent yield of water in the reaction is 46.85 %.

8 0

3 years ago

For each molecule below draw the structural formula of the molecule

torisob [31]

I think you forgot to post a picture

5 0

4 years ago

Explain why thinner containers are more effective for measuring the volume of liquids than shorter or wide containers?

Dmitry_Shevchenko [17]

Answer:THINNER contains more effective for measurin

Explanation:

4 0

3 years ago

Read 2 more answers

Illustrate the Lewis dot structure for AlH4^-

finlep [7]

Please find the attached image

3 0

3 years ago

At 700 K the equilibrium constant KC for the reaction between NO(g) and O2(g) forming NO2(g) is 8.7 × 106. The rate constant for

VMariaS [17]

Answer : The value of the rate constant for the forward reaction at 700 K is, $4.70\times 10^6$

Explanation :

The given chemical equilibrium reaction is:

$NO(g)+O_2(g)\rightleftharpoons NO_2(g)$

The expression for equilibrium constant is:

$K_c=\frac{[NO_2]}{[NO][O_2]}$

The expression for rate of forward and backward reaction is:

$R_f=K_f[NO][O_2]$

and,

$R_b=K_b[NO_2]$

As we know that at equilibrium rate of forward reaction is equal to rate of backward reaction.

$R_f=R_b$

$K_f[NO][O_2]=K_b[NO_2]$

$\frac{K_f}{K_b}=\frac{[NO_2]}{[NO][O_2]}$

$\frac{K_f}{K_b}=K_c$

Given:

$K_c=8.7\times 10^6$

$K_b=0.54M^{-1}s^{-1}$

Now put all the given values in the above expression we get:

$\frac{K_f}{K_b}=K_c$

$\frac{K_f}{0.54}=8.7\times 10^6$

$K_f=4.70\times 10^6$

Therefore, the value of the rate constant for the forward reaction at 700 K is, $4.70\times 10^6$

7 0

3 years ago