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Ludmilka [50]
3 years ago
7

Do the following problem using significant figures. 4.5 cm x 2000.0 cm x 4 cm = I cm^3

Chemistry
1 answer:
Anon25 [30]3 years ago
3 0

Given :

Measurements 4.5\ cm\times 2000.0\ cm\times 4\ cm .

To Find :

The multiplication using significant figures .

Solution :

We know , the least number of significant figures in any number of the problem determines the number  of significant figures in the answer.

So , significant figure( S.F) in 4.5 cm is 2 .

S.F in 2000.0 cm is 5 .

S.F in 4 cm is 1 .

So , minimum S.F is 1 .

Therefore , S.F of their product is 1 .

Now , multiplying them we will get :

M=4.5\times 2000.0\times 4 \ cm^3\\\\M=36000\ cm^3

Now , converting 36000 in 1 significant digit .

First we will write it in exponential 2 S.F form.

3.6\times 10^4 \ cm^3

Now , to convert 3.6 into 1 S.F we should round off this .

So , round off 3.6 we will get 4 .

Therefore , The multiplication using significant figures is 4\times 10^4\ cm^3 .

Hence , this is the required solution .

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At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:
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<u>Answer:</u> The equilibrium concentration of ICl is 0.27 M

<u>Explanation:</u>

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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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