Chemistry, 06.07.2019 21:50 chrisraptorofficial

Q3.6: calculate the number of po43- ions present in 23.7 g of ca3(po4)2.

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Moles of ca3(po4)2 = 23.7 / 310.17 = 0.076

moles of (PO4)3- = 0.076 x 2 = 0.152

now, no. of ions = 0.152 x 6.022 x 10^{23}

= 9.2 x 10^{22}

Answer from: AeelynRamos

23.7 g of Ca₃(PO₄)₎₂ contains 9.20 x 10^22 ions of PO₄³⁻

23.7 g of Ca₃(PO₄)₎₂ contains 0.1528 moles of PO₄³⁻ which is equivalent to 9.20 x 10^22 ions of PO₄³⁻

Further Explanation:An AtomAn atom is the smallest particle of an element that can take part in a chemical reaction. For example, sodium atom, Na. A MoleculeA molecule is a substance that is made up of one or more atoms. The atoms in a molecule may be similar or different making a compound. A compound A compound is a substances that contains two or more different atoms that are bonded together. When the atoms are similar the substance is known as a molecule, therefore not all molecules are compounds.Ca₃(PO₄)₎₂ is an example of a compound that contains three calcium ions and two Phosphate ions.

Therefore;

Ca₃(PO₄)₎₂ ionizes to give out 3 Ca²⁺ ions and 2 PO₄³⁻ ions

Ca₃(PO₄)₎₂ → 3Ca²⁺ + 2PO₄³⁻

Required to determine the number of ions of PO₄³⁻ ions

Step 1: Moles of Ca₃(PO₄)₎₂

Number of moles = Mass /Molar mass

Molar mass of Ca₃(PO₄)₎₂ = 310.1767 g/mol

Thus;

Moles = 23.7 g/310.1767 g/mol

= 0.0764 moles

Step 3: Moles of PO₄³⁻ ions

1 mole of Ca₃(PO₄)₎₂ ionizes to give 2 moles PO₄³⁻ ions

Therefore;

0.0764 moles will produce;

= 0.0764 moles x 2

= 0.0152 moles PO₄³⁻ ions

Step 3: Number of PO₄³⁻ ions

1 mole of PO₄³⁻ ions contains 6.022 x 10^23 ions

Therefore;

0.01528 moles PO₄³⁻ ions contains;

= 0.01528 x 6.022 x 10^23 ions

= 9.20 x 10^22 ions

Keywords: Atoms, ions, Moles, molar mass

Learn more about; Atom: Molecules Compounds Avogadro’s constant: Molecular mass:

Level: High school

Subject: Chemistry

Topic: Moles

Sub-topic: Avogadro's constant

Answer from: Quest

for points guys

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Q3.6: calculate the number of po43- ions present in 23.7 g of ca3(po4)2....

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