How many molecules of cacl2 are equivalent to 75.9 g cacl2

First, you need to find:
One mole of is equivalent to how many grams?

Well, for this you have to look up the periodic table. According to the periodic table:
The atomic mass of Calcium Ca = 40.078 g (See in group 2)
The atomic mass of Chlorine Cl = 35.45 g (See in group 17)

As there are two atoms of Chlorine present in , therefore, the atomic mass of  would be:

Atomic mass of   = Atomic mass of Ca + 2 * Atomic mass of Cl

Atomic mass of = 40.078 + 2 * 35.45 = 110.978 g

Now,

110.978 g of = 1 mole.
75.9 g of  =  = 0.6839 moles.

Hence,
The total number of moles in 75.9g of = 0.6839 moles

According to Avogadro's number,
1 mole = 1 *  molecules
0.6839 moles = 0.6839 * molecules =  molecules

Ans: Number of molecules in 75.9g of   =   molecules

-i

4.11 * 10^ 23 molecules

Explanation:

1) Data:

Compound: CaCl2

mass, m = 75.9 g

unknown, N = ?

2) Formulas:

N = n * Avogadro's number = n * 6.022 * 10^23 molecules / mol

n = mass in grams / molar mass

molar mass = sum of the masses of the atoms in the molecular formula

3) Solution

molar mass of CaCl2:

      mass of Ca = 1 * atomic mass of Ca = 40.1 g/mol
      mass of Cl2 = 2 * atomic mass of Cl2 = 2 * 35.5 g/mol = 71.0 g/mol
      molar mass = 40.1 g/mol + 71.0 g/mol = 111.1 g/mol

n = m / molar mass = 75.9 g / 111.1 g/mol = 0.683 mol

N = n * 6.022 * 10^23 = 0.683 mol * 6.022 * 10^23 molecules / mol = 4.11 * 10^23 molecules, which is the answer.

No. an alpha particle is a pair of protons and a pair of neutrons, i.e. a helium