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creativ13 [48]
3 years ago
12

Which process is an example of a chemical change?

Chemistry
2 answers:
Ivanshal [37]3 years ago
8 0
A metal rail rusting in damp weather.
jonny [76]3 years ago
3 0

Answer:

The correct answer is :<u><em>a metal railing rusting in damp weather</em></u> is an example of a chemical change.

Explanation:

Physical property is defined as the property of a substance which becomes evident during physical changes. Example: Melting point

A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.

Example: Melting of ice

Chemical property is defined as the property of a substance which becomes evident during chemical changes. Example: Reactivity with other substances

A chemical change is defined as a change in which a change in chemical composition takes place. A new substance is formed in these reactions.

Example: Corrosion of iron called as rusting.  

Rust is ferric oxide Fe_2O_3.

4Fe+3O_2\righatarrow 2Fe_2O_3

A metal railing rusting in damp weather  is an example of chemical change because composition of an iron rail is changing.

4Fe+3O_2\rightarrow 2Fe_2O_3

Where as other examples given are the examples of physical change.

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<u>Answer:</u> The equilibrium concentration of ICl is 0.27 M

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

Initial concentration of iodine gas = \frac{0.45}{2}=0.225M

Initial concentration of chlorine gas = \frac{0.45}{2}=0.225M

For the given chemical equation:

2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

                      I_2(g)+Cl_2(g)\rightarrow 2ICl(g);K_c=\frac{1}{0.11}=9.091

<u>Initial:</u>         0.225      0.225

<u>At eqllm:</u>   0.225-x    0.225-x     2x

The expression of K_c for above equation follows:

K_c=\frac{[ICl]^2}{[Cl_2][I_2]}

Putting values in above equation, we get:

9.091=\frac{(2x)^2}{(0.225-x)\times (0.225-x)}\\\\x=0.135,0.668

Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M

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3 years ago
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