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alexandr1967 [171]
1 year ago
13

A substance occupies one half of an open container. The atoms of the substance are closely packed but are still able to slide pa

st each other.
What is most likely the phase of the substance?

O: Gas
O: liquid
O: Solid and gas
O: liquid and solid

Pls help ;-;
Chemistry
1 answer:
SCORPION-xisa [38]1 year ago
8 0

For a substance to occupy one half of an open container with the atoms being able to slide past each other, the most likely phase of the substance would be liquid.

Atoms of gases would not occupy just one-half of an open container because atoms of gases will diffuse and spread out from an open container.

Atoms of solids are fixed about a particular position and will most likely not be able to slide past each other in an open container.

Only the atoms of liquid take the shapes of their containers and do not have the capacity to diffuse out ordinarily. They are also able to flow despite the closeness of the atoms and would occasionally slide past each other.

More on states of matter can be found here: brainly.com/question/9402776

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
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Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

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Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

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

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

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

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

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2 years ago
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Explanation:

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The first step is to balance the equation:


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Check the balance


element          left side        right side

C                    3                  3  
H                    8                  4*2 = 8
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Then you have the molar ratios:


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Now you have 40 moles of O2 so you make the proportion:

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