Which of the following is not an indication of a chemical reaction?

Chemistry

1 answer:

eimsori [14]3 years ago

7 0

Answer: Option (3) is the correct answer.

Explanation:

A reaction due to which there occurs change in chemical composition of reactants is known as a chemical reaction.

For example, evolution of heat when two substances combine together shows a chemical reaction has taken place between the substances.

Whereas a reaction which does not bring any changes in chemical composition of the reactants is known as a physical reaction.

For example, boiling point, melting point, mass, volume etc are all physical properties.

Thus, we can conclude that out of the given options a substance melts when sitting in the Sun, is not an indication of a chemical reaction.

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What evidence do you see that a chemical reaction is taking place when you add hydrogen peroxide to potassium ioxide??

Yuri [45]

Answer:

Bubbling/Foaming

Heat is produced

Explanation:

A chemical reaction has occurred if

1. There is a change in colour

2. Formation of a precipitate

3. Formation of a gas

4. Change in temperature

5. Change of smell

In the reaction of hydrogen peroxide and potassium iodide, the solution quickly rises which shows the formation of gas. This means it is a chemical reaction.

The beaker/test tube/whatever you used should also become warm because it is an exothermic reaction. This means its a chemical reaction

3 0

3 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$