Density (p) is defined as the mass (m) per unit volume (v) or:
p = m/v
Using this relationship, the volume is:
v = m/p
Using the given values of mass of 80 grams and density of 8 g/cm3, the sample volume is:
v = 80 grams/8 grams/cm3
v = 10 cm3
The final answer is 10 cm3.
The reason for this is that <span>these two ketones are so small that they have only one possible ketone. So the number is usually omitted. Normally the ketone group needs a number but these two are exceptions</span>
Answer: The nuclide symbol of X is 
Explanation:
The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.
The general equation representing alpha decay process is:

For the given equation :

As the atomic number and mass number must be equal on both sides of the nuclear equation:

Thus the nuclide symbol of X is 
There are 27.67834 g/mol in 3.73 moles of lioh
Answer:
The atomic mass of X is 204.5 amu.
Explanation:
We know that metals react with oxygen forming a metallic oxide, according to the following equation:
4 X + O₂ ⇄ 2 X₂O
The oxide is formed just by <em>just</em> two elements, the metal and oxygen. The total mass of the oxide is 1.4158g and the mass of the metal in the oxide must be 1.3625g because of the Law of conservation of mass. Then, we can substract the mass of the metal to obtain the mass of oxygen.
mass O + mass X = mass X₂O
mass O = mass X₂O - mass X = 1.4158g - 1.3625g = 0.0533g
So, for every 0.0533g of oxygen there are 1.3625 g of the metal X. In the formula X₂O there is 1 mol of atoms of oxygen, which has a molar mass of 16 g/mol. We can use this data to find out the mass of the metal in the oxide.

Given in the formula there are 2 moles of atoms of X, the molar mass should be half of 409g, i.e., 204.5g/mol. If a mol of X has a mass of 204.5 g, an atom of X has a mass of 204.5 amu, according to its definition.