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Kamila [148]
3 years ago
10

Solve by a 1 variable equation -3(x-3)=x+9-4x

Chemistry
1 answer:
Otrada [13]3 years ago
5 0
-3(x - 3) = x + 9 - 4x

Firstly, before we simplify the whole thing altogether, let's simplify each respective side. Distribute accordingly on the left side (the number outside the parentheses is distributed to the numbers inside the parentheses), and we can add the numbers with variables together on the right side.

-3x + 9 = -3x + 9

Now, let's add 3x to both sides, to simplify both sides now. This leaves us with: 1, I believe, since -3 ÷ -3 is 1.

x = x, and since x also equals to 1, our answer is 1. And once you check, substituting 1 for x, you will find that 1 is the correct answer.
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Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the
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Answer : The equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

Solution : Given,

Equilibrium constant, K_c=1.8\times 10^{-5}

Initial concentration of HC_2H_3O_2 = 0.260 m

Let, the 'x' mol/L of H_3O^+ are formed and at same time 'x' mol/L of HC_2H_3O_2 are also formed.

The equilibrium reaction is,

                  HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)

Initially                0.260 m                       0                 0

At equilibrium    (0.260 - x)                     x                 x

The expression for equilibrium constant for a given reaction is,

K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

Now put all the given values in this expression, we get

1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}

By rearranging the terms, we get the value of 'x'.

x=2.154\times 10^{-3}m

Therefore, the equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

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3 years ago
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