Answer:
HgSO₄
Explanation:
% => g => moles => ratio => reduce => empirical ratio
%Hg = 67.6% => 67.6g/201g/mol = 0.34mol
%S = 10.8% => 10.8g/32g/mol = 0.34mol
%O = 21.6% => 21.6g/16g/mol = 1.35mol
Hg:S:O => 0.34:0.34:1.35
Reduce to whole number ratio by dividing by the smaller mole value...
Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4
∴ Empirical Formula is HgSO₄
The empirical formula : CH₃
<h3>Further explanation</h3>
Given
2.5 g sample
2.002 g Carbon
Required
The empirical formula
Solution
Mass of Hydrogen :
= 2.5 - 2.002
= 0.498
Mol ratio C : H :
C : 2.002/12 = 0.167
H : 0.498/1 = 0.498
Divide by 0.167 :
C : H = 1 : 3
If you add them up: 2+2+6 = 10, you will know it's Neon, which is written as Ne
Electrons are valence and free moving so they take place in charge transfer
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:
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