The correct answer is option d, that is, atoms of the element.
As the atoms are neither destroyed nor created in a chemical reaction, the sum of the mass of the products in a reaction must be equivalent to the sum of the mass of the reactants.
The chemical reactions must be balanced, they must exhibit a similar number of atoms of each element on both the sides of the equation. As a consequence, the mass of the reactants must be equivalent to the mass of the products of the reaction.
Answer:
We can seprate oil and water by the process of seprating funnel
Answer : The expected coordination number of NaBr is, 6.
Explanation :
Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.
This is represented by,

When the radius ratio is greater than 0.155, then the compound will be stable.
Now we have to determine the radius ration for NaBr.
Given:
Radius of cation,
= 102 pm
Radius of cation,
= 196 pm

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.
The relation between radius ratio and coordination number are shown below.
Therefore, the expected coordination number of NaBr is, 6.
<span>128 g/mol
Using Graham's law of effusion we have the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = rate of effusion for gas 1
r2 = rate of effusion for gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Since the atomic weight of oxygen is 15.999, the molar mass for O2 = 2 * 15.999 = 31.998
Now let's subsitute the known values into Graham's equation and solve for m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
127.992 = m2
So the molar mass of the unknown gas is 127.992 g/mol.
Rounding to 3 significant figures gives 128 g/mol</span>