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4 years ago

I will give brainliest!! How long is the pencil? Make sure to use the appropriate amount of significant figures.

Chemistry

1 answer:

8090 [49]4 years ago

5 0

Answer: Option (c) is the correct answer.

Explanation:

The bottom of pencil is placed at the starting point of scale. Whereas the tip of pencil depicts the end point of its length.

The bottom of pencil is at 0 mm and tip of pencil is at 18.73 mm. The appropriate amount of significant figures is 18.73 mm.

Therefore, we can conclude that out of the given options, pencil is 18.73 mm long.

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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

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4 years ago

Please help will mark Brainly !! PLEASE HELP !

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