Answer:
1.44 x 10²⁵ ions of Na⁺
Explanation:
Given parameters:
Mass of NaCl = 1.4kg = 1400g
Unknown:
Number of ions of sodium = ?
Solution:
The compound NaCl in ionic form can be written as;
NaCl → Na⁺ + Cl⁻
In 1 mole of NaCl we have 1 mole of sodium ions
Now, let us find the number of moles in NaCl;
Number of moles = 
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Number of moles = 
= 23.93mol
So;
Since 1 mole of NaCl gives 1 mole of Na⁺
In 23.93 mole of NaCl will give 23.93 mole of Na⁺
1 mole of a substance = 6.02 x 10²³ ions of a substance
23.93 mole of a substance = 6.02 x 10²³ x 23.93
= 1.44 x 10²⁵ ions of Na⁺
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
oxidation-reduction or redox reaction
Explanation:
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