The amount of KNO₃ precipitated out of solution when you cooled the solution from 100°C to -22°C is obtained from the solubility curve
<h3>What is a solubility curve?</h3>
A solubility curve is a curve of the solubility of a solute against temperature.
The solubility curve shows that the solubility of different solute at different temperatures.
The solubility curve of KNO₃ is as shown. Solubility at 100 °C and -22 °C is not shown in the curve.
However, the amount of KNO₃ precipitated out of solution when cooled the solution from 100 °C to -22 °C can be determined by subtracting the amount of solute dissolve at -22 °C from that dissolved at 100 °C.
In conclusion, the solubility curve is used to determined the amount of solute dissolved in a given volume of solvent at different temperatures.
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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
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Answer:
Explanation:
Mineral
The naturally occurring mineral anglesite, PbSO4, occurs as an oxidation product of primary lead sulfide ore, galena.
Basic and hydrogen lead sulfates
A number of lead basic sulfates are known: PbSO4·PbO; PbSO4·2PbO; PbSO4·3PbO; PbSO4·4PbO. They are used in manufacturing of active paste for lead acid batteries. A related mineral is leadhillite, 2PbCO3·PbSO4·Pb(OH)2.
At high concentration of sulfuric acid (>80%), lead hydrogensulfate, Pb(HSO4)2, forms.[4]
Chemical properties
Lead(II) sulfate can be dissolved in concentrated HNO3, HCl, H2SO4 producing acidic salts or complex compounds, and in concentrated alkali giving soluble tetrahydroxidoplumbate(II) [Pb(OH)4]2− complexes.
PbSO4(s) + H2SO4(l) ⇌ Pb(HSO4)2(aq)
PbSO4(s) + 4NaOH(aq) → Na2[Pb(OH)4](aq) + Na2SO4(aq)
Lead(II) sulfate decomposes when heated above 1000 °C:
PbSO4(s) → PbO(s) + SO3(g)
Pretty sure the answer is true
Answer:
<h2>The answer is 1.5 L</h2>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
Since we are finding the new volume
From the question we have
We have the final answer as
<h3>1.5 L</h3>
Hope this helps you