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svetlana [45]
2 years ago
4

PLEASE HELP, OVERDUE ASSIGNMENT.

Chemistry
1 answer:
Keith_Richards [23]2 years ago
4 0

3rd statement about this acidic reaction is not true

In acidic reactions,  Litmus paper is available in red and blue varieties and is typically more trustworthy. To determine whether a solution is normal or acidic, use litmus paper. The acid or basic strength is not precisely outlined in Litmus's paper. Thus, Base causes red litmus to turn blue.

The pH scale has numbers 0 through 14. Seven is the neutral pH. Over 7 is a base, and anything below 7 is acidic.

H+ ions are present in all acidic reactions. As a result, whether mineral or organic, all acid solutions transmit electricity. They allow an electric current to flow through them.

To learn more about acidic reaction please visit -
brainly.com/question/3069273
#SPJ4

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Explanation:

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At a certain temperature, 0.800 mol SO 3 is placed in a 1.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil
Elanso [62]

Answer : The value of equilibrium constant (K) is, 0.004

Explanation :

First we have to calculate the concentration of SO_3\text{ and }O_2

\text{Concentration of }SO_3=\frac{\text{Moles of }SO_3}{\text{Volume of solution}}=\frac{0.800mol}{1.50L}=1.2M

and,

\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{0.150mol}{1.50L}=0.1M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                       2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

Initial conc.       1.2             0               0

At eqm.          (1.2-2x)        2x               x

As we are given:

Concentration of O_2 at equilibrium = x = 0.1 M

The expression for equilibrium constant is:

K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}

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

K_c=\frac{(2x)^2\times (x)}{(1.2-2x)^2}

K_c=\frac{(2\times 0.1)^2\times (0.1)}{(1.2-2\times 0.1)^2}

K_c=0.004

Thus, the value of equilibrium constant (K) is, 0.004

3 0
4 years ago
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