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Gemiola [76]
3 years ago
14

Please help, with step by step work

Chemistry
1 answer:
natali 33 [55]3 years ago
5 0

\qquad ☀️\pink{\bf{ {Answer  = \: \:   85.57g }}}

Molar mass of \bf Cu_2O

\qquad \twoheadrightarrow\sf 63.546 \times 2 +16

\qquad \pink{\twoheadrightarrow\bf 143.092 g}

<u>As we know</u>–

1 mol =\bf 6.02×10^{23} formula units

1 mol\bf Cu_2O = 143.092 g = \bf 6.02×10^{23}formula units

Henceforth –

\bf 3.60×10^{23} formula units \bf Cu_2O–

\qquad \sf :\implies \dfrac{143.092 \times3.60×10^{23  }}{6.02×10^{23}}

\qquad \sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23  }}}{6.02×\cancel{10^{23}}}

\qquad \pink{:\implies\bf 85.57 g}

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Answer:

Initial volume of the container (V1) = 1.27 L (Approx)

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Number of mol (n2) = (5.67 +2.95) x 10⁻² = 8.62 x 10⁻²

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Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

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K_{sp}=[Cu^+][Cl^-]

Putting values in above equation, we get:

1.0\times 10^{-6}=[Cu^+]\times 0.021

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For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.

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