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Finger [1]
3 years ago
8

How many moles of Ca(NO3)2 must be added to 1.0 L of a 0.255 M HF solution to begin precipitation of CaF2(s)

Chemistry
1 answer:
yulyashka [42]3 years ago
7 0

Answer:

6.1×10^(-10) mol

Explanation:

CaF2(s) -----------> Ca2+(aq) +2F-(aq)

          Ksp = [Ca2+][F-]^2

  Given data

   [F-] = 0.255 M

∴ The concentration of Ca2+required to start the precipitation

                 [Ca2+] = Ksp / [F-]2

                            = (4.0 X 10^-11) /(0.255)^2

                            = 6.1×10^(-10) M

Here the volume is 1.0 L  

∴ The number of molesof  Ca(NO3)2 must be added to 1.0 L of a1.0 L of a 0.255 M HF solution to begin precipitation ofCaF2 (s)=   6.1×10^(-10) mol

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6c.Calculate the maximum volume, in dm3, of chlorine gas at Stp that can be obtained from 23.4 tonnes of molten sodium chloride.
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We'll begin by converting 23.4 tonnes to grams (g). This can be obtained as follow:

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Next, we shall determine the number of mole of chlorine, Cl₂ produced from the reaction. This can be obtained as follow:

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From the balanced equation above,

2 moles of NaCl reacted to produce 1 mole of Cl₂.

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2×10⁵ moles of Cl₂ at stp = 4.48×10⁶ dm³

Thus, the volume of chlorine obtained from the reaction is 4.48×10⁶ dm³

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