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dalvyx [7]
3 years ago
9

How many total ions are present in 347g of cacl2?

Chemistry
1 answer:
vladimir1956 [14]3 years ago
8 0

In 1 mole of CaCl_{2}, there are 3 moles of ions, 1 mole of Ca^{2+} and 2 mole of Cl^{-1}.

CaCl_{2}\rightarrow Ca^{2+}+2Cl^{-}

Molar mass of CaCl_{2} is 110.98 g/mol. Calculating number of moles from given mass as follows:

n=\frac{m}{M}=\frac{347 g}{110.98 g/mol}=3.12 mol

Thus, number of moles of ions will be 3\times 3.12 mol=9.38 mol.

Since, 1 mole of any substance has 6.023\times 10^{23} units of that substance where 6.023\times 10^{23}  is Avogadro's number.

Thus, 9.38 mol of ions will have 9.38\times 6.023\times 10^{23}=5.65\times 10^{24} number of ions.

Therefore, total number of ions in 347 g of CaCl_{2} is  5.65\times 10^{24}.


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<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × 10^{23} x 0.524 mole

3.155601758 atoms =3.155 atoms

B. The number of SO_2 molecules in 9.87 moles of SO_2:

6.02214076 × 10^{23} x 9.87

59.4385293 molecules= 59.43 molecules

C. The moles of Fe in 1.40 x 10^{22} atoms of Fe:

1.40 x 10^{22} ÷ 6.02214076 × 10^{23}

0.2324754694 x 10^{-1} moles.

0.23 x 10^{-1} moles.

D. The moles of C_2H_6O in 2.30x10^{24} molecules of C_2H_6O:

2.30x10^{24} ÷ 6.02214076 × 10^{23}

3.819239854 moles=3.81 moles

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

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