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2 years ago

9

10.0 mL of a 0.100 mol L–1 solution of a metal ion M2+ is mixed with 10.0 mL of a 0.100 mol L–1 solution of a substance L. The f

ollowing equilibrium is established:
M2+(aq) + 2L(aq) Picture ML22+(aq)

At equilibrium the concentration of L is found to be 0.0100 mol L–1. What is the equilibrium concentration of ML22+, in mol L–1?

Someone Help Me

2 answers:

LenKa [72]2 years ago

8 0

<u>Answer:</u> The concentration of $ML_2^{2+}$ at equilibrium is 0.045 M.

<u>Explanation:</u>

We are given:

$[M^{2+}]_{initial}=0.100M$

$[L]_{initial}=0.100M$

For the given chemical equation:

$M^{2+}(aq.)+2L(aq.)\rightleftharpoons ML_2^{2+}(aq.)$

Initially: 0.100M 0.100M

At eqllm: 0.100 - x 0.100 - 2x x

We are also given:

$[L]_{eqllm}=0.0100M$

Equating the two values:

$0.0100=0.100-2x\\\\x=0.045M$

Hence, the concentration of $ML_2^{2+}$ at equilibrium is 0.045 M.

Mrac [35]2 years ago

7 0

The chemical reaction is
<span> M2+(aq) + 2L(aq) <==> ML22+(aq)
</span>Intial concentration 0.10 0.10
Change -x -2x +x
Equilibrium 0.10 - x 0.01 = 0.10 - 2x x
Solving for x
0.01 = 0.10 - 2x
x = 0.045
The equilibrium concentration of ML22+ is 0.045 mol L-1

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<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

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Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

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To formulate the formula of the compound, we need to follow some steps:

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<u>Step 2:</u> Calculating the mole ratio of the given elements.

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<u>Step 3:</u> Taking the mole ratio as their subscripts.

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