Answer:
involves the splitting of an atom into two or more fragments
Explanation:
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:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u />
= 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag
