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1 year ago

Which products would result from the double replacement reaction between mgcl2(aq) and na2co3(aq)?

1 answer:

8 0

Double displacement reactions are when the cations and anions of 2 compounds are exchanged. Cation of one reactant will form a new product with the anion of the other reactant and vice versa.
MgCl₂ --> Mg²⁺ + 2Cl⁻
Na₂CO₃ --> 2Na⁺ + CO₃²⁻
the cations and anions are exchanged therefore double displacement reaction is as follows;
MgCl₂(aq) + Na₂CO₃(aq) --> MgCO₃(s) + 2NaCl(aq)

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1 year ago

At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.400 M . N 2 ( g )

stiks02 [169]

Answer: The equilibrium concentration of NO after it is re-established is 0.55 M

Explanation:

For the given chemical equation:

$N_2(g)+O_(g)\rightleftharpoons 2NO(g)$

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=\frac{[NO]^2}{[N_2][O_2]}$ .....(1)

We are given:

$[NO]_{eq}=0.400M$

$[N_2]_{eq}=0.200M$

$[O_2]_{eq}=0.200M$

Putting values in expression 1, we get:

$K_{eq}=\frac{(0.400)^2}{0.200\times 0.200}\\\\K_{eq}=4$

Now, the concentration of NO is added and is made to 0.700 M

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

The equilibrium will shift in backward direction.

$N_2(g)+O_(g)\rightleftharpoons 2NO(g)$

Initial: 0.200 0.200 0.700

At eqllm: 0.200+x 0.200+x 0.700-2x

Putting values in expression 1, we get:

$4=\frac{(0.700-2x)^2}{(0.200+x)\times (0.200+x)}\\\\x=0.075$

So, equilibrium concentration of NO after it is re-established = (0.700 - 2x) = [0.700 - 2(0.075)] = 0.55 M

Hence, the equilibrium concentration of NO after it is re-established is 0.55 M

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2 years ago

if a sample of gas is intially at 1.8 atm,22.0 l, and 26.4 c, what will be the volume if the pressure is reduced by 0.8 atm and

Tems11 [23]

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2 years ago

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