Chemistry, 30.07.2019 19:30 Itasouu

How many molecules of cacl2 are equivalent to 75.9 g cacl2

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Answer from: jadenp23

First, you need to find:
One mole of $CaCl_{2}$ 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 $CaCl_{2}$ , therefore, the atomic mass of $CaCl_{2}$ would be:

Atomic mass of $CaCl_{2}$ = Atomic mass of Ca + 2 * Atomic mass of Cl

Atomic mass of $CaCl_{2}$ = 40.078 + 2 * 35.45 = 110.978 g

Now,

110.978 g of $CaCl_{2}$ = 1 mole.
75.9 g of $CaCl_{2}$ = $\frac{75.9}{110.978} moles$ = 0.6839 moles.

Hence,
The total number of moles in 75.9g of $CaCl_{2}$ = 0.6839 moles

According to Avogadro's number,
1 mole = 1 * $6.022 * 10^{23}$ molecules
0.6839 moles = 0.6839 * $6.022 * 10^{23}$ molecules = $4.118*10^{23}$ molecules

Ans: Number of molecules in 75.9g of $CaCl_{2}$ = $4.118*10^{23}$ molecules

-i

Answer from: yangyang7

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.

Answer from: Quest

state the principle of separating two immiscible liquids by separating funnel. describe an activity with diagram to separate a mixture of water and kerosene oil. answer: immiscible layers separate out in layers depending on their densities in separating funne

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answer; i believe that the correct answer is (a) potassium(k)

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How many molecules of cacl2 are equivalent to 75.9 g cacl2...