Which of the following describes a gas?

Chemistry

2 answers:

zimovet [89]3 years ago

6 0

A is the answer. A gas never has a fixed volume or shape

Setler79 [48]3 years ago

4 0

Answer:

A. Does not have a fixed volume or a fixed shape.

Explanation:

A liquid is matter that has a fixed volume but not a fixed shape. A gas is matter that has neither a fixed volume nor a fixed shape. Like a gas, plasma lacks a fixed volume and shape.

Liquids have a fixed volume but no fixed shape. This is because the particles in a liquid are further apart than the particles in a solid, allowing them to easily slide past each other. Solids have a fixed shape because their molecules are so tightly packed they cannot do much more than vibrate against each other.

You might ask, What is a definite shape?

Any matter that is a solid has a definite shape and a definite volume. The molecules in a solid are in fixed positions and are close together. Although the molecules can still vibrate, they cannot move from one part of the solid to another part. As a result, a solid does not easily change its shape or its volume.

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guajiro [1.7K]

It is two or more objects and different things that can be removed from each other.

7 0

3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.