Answer is suspension.
Lets define all options.
<h3>Suspension:</h3>
In suspension the solute does not dissolve in liquid. When placed on table for some time, it will settle down at the bottom of the beaker. We can separate particles of solute easily from solvent through filtration.
<h3>Colloid:</h3>
In colloid particles of solute does not dissolve in liquid neither it is settle down. It floats through the solvent. It cannot be separated by filtration.
<h3>Solution:</h3>
In solution the particles of solute dissolve in to the solvent. We cannot identify them as separate. We cannot separate them by filtration. Salt and water solution is an example of it. Evaporation is the technique that is required to separate them.
<h3>Compound:</h3>
In compound, the two elements combine to form a new thing. Resultant/ compound have new or different properties other than its ingredients.
Now, the question was which of the following allow to settle out when sit on a table, so the answer is suspension. Suspension allows the particles to settle out when sit on a tables for some time.
<span>C) 4.5 billion years old</span>
Answer:
Part A:
to two significant figures
Part B:
to two significant figures
Part C:
to two significant figures
Explanation:
Given that :
mass of the hydrogen = 0.30 g
the molar mass of hydrogen gas molecule = 2 g/mol
we all know that:
number of moles = mass/molar mass
number of moles = 0.30 g /2 g/mol
number of moles = 0.15 mol
For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = 
For Part A:




to two significant figures
Part B. For hot temperature, 




to two significant figures
Part C. For an extremely hot temperature, 




to two significant figures
Answers
The car's forward motion is opposed by the friction between the road and the tires and by the resistance of the air.
Answer:
1. B has no acceleration because the straight line shows that it's a constant speed not speeding up or down.
2. A because you can see the decline in speed as time goes on