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4 years ago

Choose the following statements that show evidence for a chemical reaction.

Chemistry

1 answer:

Leno4ka [110]4 years ago

8 0

Explanation:

A chemical reaction is defined as the reaction in which bonds between the reactants either break or form which leads to the formation of a new substance.

For example, $2Li + Cl_{2} \rightarrow 2LiCl$

So, when we drop a sodium metal into water then it produces a frizzing sound which shows the metal is reacting with water.

We know that when two aqueous solutions chemically react with each other then it may lead to the formation of an insoluble substance which is known as precipitate.

This means that formation of a precipitate is also a chemical reaction.

Thus, we can conclude that following are the statements which show evidence for a chemical reaction.

Dropping sodium metal into water produces fizzing.

Mixing two aqueous solutions produces a precipitate.

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CoPO4 (s) + H3O+ (aq) -----> HPO42- (aq) + Co3+ (aq) + H2O *(l)

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The sulfuric acid–catalyzed reaction of isopentyl alcohol with acetic acid to form isopentyl acetate is performed. This reaction

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Answer:

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Explanation:

Distillation is the process of separating a mixture of substances based on differences in boiling points. During distillation, the compound with the lowest/least boiling point is distilled and collected first and then the one with the next least boiling point and it goes on like that.

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Acetone has a density of 0.7857 g/cm^3. What is the volume in milliliters of 7.70g of acetone

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3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

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Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$