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Oksanka [162]
3 years ago
6

_____ occurs when the product of the ion concentrations exceeds the Ksp.

Chemistry
2 answers:
arlik [135]3 years ago
4 0
Precipitation occurs when the product of the ion concentration exceeds the Ksp.
serious [3.7K]3 years ago
4 0

Answer : Precipitation occurs when the product of the ion concentrations exceeds the K__{sp}.

Explanation :

Ionic product : It is the product of the concentrations of the ions in solution raised to the same power by its stoichiometric coefficient in a solution of a salt. This takes place at any concentration. The ionic product is represented as, Q.

Solubility product constant : It is defined as the product of the concentration of the ions present in a solution raised to the power by its stoichiometric coefficient in a solution of a salt. This takes place at equilibrium only. The solubility product constant is represented as, K__{sp}.

There are three cases for the solubility :

  1. When Q this means that the solution is unsaturated solution and more solid will be dissolve.
  2. When Q=k_{sp} this means that the solution is saturated solution.
  3. When Q>k_{sp} this means that the solution is supersaturated solution and solid will be precipitate.

Hence, the precipitation occurs when the product of the ion concentrations exceeds the K__{sp}.

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How many liters of hydrogen gas are needed to react completely with 50.0 L of chlorine gas at STP ?
dolphi86 [110]

Answer:

50L of H2

Explanation:

First let us generate a balanced equation for the reaction

Cl2 + H2 —> 2HCl

From the equation above,

1L of Cl2 required 1L of H2 for complete reaction.

Therefore, 50L of Cl2 will also require 50L of H2 for complete reaction.

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The stock concentration of NaOH is 0.5M. 0.5M = 0.500mol/1.0L.
valentinak56 [21]

Answer:a) 0.1 mole. b) 4g. c) 2% d)  196 mL

Explanation: in 200mL , 0.1mole

mw NaOH = 40g/mol —> 4g in 0.1 mole

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density NaOH = 1g/mL so if 4g in 200 mL,  4mL , 196 mL water

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Who is the Father of Modern Periodic Table?
s344n2d4d5 [400]

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Dmitri Mendeleev

4 0
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Answer:

b

Explanation:

6 0
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NaClO3 &gt; NaCl + O2 <br> Balance
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Answer:

Balancing Strategies: To balance this reaction it is best to get the Oxygen atoms on the reactant side of the equation to an even number. Once this is done everything else falls into place. Put a "2" in front of the NaClO3. Change the coefficient in front of the O2.

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