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4 years ago

How many anions are in 0.500 g of MgBr2

Chemistry

1 answer:

Oksana_A [137]4 years ago

4 0

Given:

Mass of MgBr2 = 0.500 g

To determine:

Number of anions in 0.500 g MgBr2

Explanation:

Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol

Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles

Based on stoichiometry-

1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions

Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-

Now,

1 mole of Br- contains 6.023 * 10²³ anions

0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions

Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2

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Answer: The value of $K_c$ for given reaction is 0.465

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We are given:

Initial concentration of A = 0.650 M

Initial concentration of B = 1.35 M

Initial concentration of C = 0.300 M

Equilibrium concentration of A = 0.550 M

Equilibrium concentration of B = 0.400 M

For the given chemical equation:

$A+2B\rightarrow C$

Initial: 0.65 1.35 0.30

At eqllm: 0.65-x 1.35-2x 0.30+x

Evaluating the value of 'x'

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Equilibrium concentration of C = (0.30 + x) = (0.300 + 0.100) = 0.400 M

The expression of $K_c$ for above equation follows:

$K_c=\frac{[C]}{[A][B]^2}$

Putting values in above equation, we get:

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