LGrowth Calcium (Salze-77) Given be This LProduct of growth of Calcium, Lp(CaCO3) = 4.8·10-9 mol2/l2. How a lot Grams Calcium Resolves The Commission So what do you mean? in One of them Bathtub of 200 Of which:?

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Given be This LProduct of growth of Calcium, L_p(CaCO₃) = 4.8·10^-9 mol²/l². How a lot Grams Calcium Resolves The Commission So what do you mean? in One of them Bathtub of 200 Of which:?

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Lösung:

From Definitions following::

1. Es Applicable: 1·CaCO₃ ⇄ 1·Ca²⁺ + 1·CO2-3
2. This means that Applicable also: c(CaCO₃) = c(Ca²⁺) = c(CO2-3)
3. It 's following: from: L_p(CaCO₃) = c(Ca²⁺) · c(CO2-3)
       and Because z.B. (Ca²⁺⁾ = c(Ca²⁺)
       L_p(CaCO₃) = c(Ca²⁺) · c(Ca²⁺) = c²(Ca²⁺)     or
       L_p(CaCO₃) = c(CO2-3) · c(CO2-3) = c²(CO2-3)     or
       L_p(CaCO₃) = c(CaCO₃) · c(CaCO₃) = c²(CaCO₃)
4. → c(CaCO₃) = √(L_p) = = √ L_p = 6.93·10^-5 mol/l

Well, you know what? can by: c=n/V=m/M/V on 'm' Closed will be:

m = c(CaCO₃)·M(CaCO₃)·V = 6.93·10^-5 mol/l · 100.1 g/mol · 200 l = 1.39 Grams

On the The large Amount of water Seeing Resolves The Commission It 's just … Not much Calcium, In particular: about. 1.4 Grams.

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