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Katena32 [7]
3 years ago
6

If a liquid substance is transferred to a different container, what can be predicted about the volume of the liquid in the new c

ontainer?
Chemistry
1 answer:
taurus [48]3 years ago
8 0

The same

Explanation:

If a liquid substance is transferred to a different container, the volume of the liquid in the new container will remain the same.

The volumes of liquids are fixed and does not change. Wherever they are contained, just like solids, they maintain their constant space.

  • Volume is the amount of space occupied by a body.
  • Gases do not have fixed volume as they fill their containers and they take up the shape.
  • Solids and liquids have a fixed volume.
  • They do not change their volume.

learn more:

State of matter brainly.com/question/10972073

#learnwithBrainly

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Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p
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Answer : The equilibrium constant K_c for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of Br_2.

\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}

\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M

Now we have to calculate the dissociated concentration of Br_2.

The balanced equilibrium reaction is,

                              Br_2(g)\rightleftharpoons 2Br(aq)

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of Br_2 = \alpha = 1.2 %

So, the dissociate concentration of Br_2 = C\alpha=1.731M\times \frac{1.2}{100}=0.2077M

The value of x = 0.2077 M

Now we have to calculate the concentration of Br_2\text{ and }Br at equilibrium.

Concentration of Br_2 = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of Br = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

K_c=\frac{[Br]^2}{[Br_2]}

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

K_c=\frac{(0.4154)^2}{1.5233}=0.1133

Therefore, the equilibrium constant K_c for the reaction is, 0.1133

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