Answer:
A 50-mL volumetric cylinder with 0.1-mL accuracy scale should be used for this purpose since three significant figures of accuracy are required.
Explanation:
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A 50-mL volumetric cylinder with 0.1-mL accuracy scale should be used for this purpose since three significant figures of accuracy are required.
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Given the solubility of strontium arsenate is 0.0480 g/l . we have to convert it into mol/L by dividing it over molar mass (540.7 g/mol)
Molar solubility = 0.0480 / 540.7 = 8.9 x 10⁻⁵ mol/L
Dissociation equation:
Sr₃(AsO₄)₂(s) → 3 Sr²⁺(aq) + 2 AsO₄³⁻(aq)
3 s 2 s
Ksp = [Sr²⁺]³ [AsO₄³⁻]²
= (3s)³ (2s)²
= 108 s⁵
Ksp = 108 (8.9 x 10⁻⁵) = 5.95 x 10⁻¹⁹
Answer is: <span>the molarity of this glucose solution is 0.278 M.
m</span>(C₆H₁₂O₆<span>) = 5.10 g.
n</span>(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆<span>) .
</span>n(C₆H₁₂O₆) = 5.10 g ÷ 180.156 g/mol.
n(C₆H₁₂O₆<span>) = 0.028 mol.
</span>V(solution) = 100.5 mL ÷ 1000 mL/L.
V(solution) = 0.1005 L.
c(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ V(solution).
c(C₆H₁₂O₆) = 0.028 mol ÷ 0.1005 L.
c(C₆H₁₂O₆<span>) = 0.278 mol/L.</span>
Explanation:
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