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Veronika [31]
4 years ago
10

The Solubility Product Constant for calcium bromide is > 1. If calcium bromide is dissolved in water you can say that the equ

ilibrium concentrations of calcium and bromide ions are: ... ...A. High ...B. Moderate ...C. Low The solubility of calcium bromide in water is: ... ...A. High ...B. Moderate ...C. Low
Chemistry
1 answer:
DiKsa [7]4 years ago
7 0

Answer: (1). If calcium bromide is dissolved in water you can say that the equilibrium concentrations of calcium and bromide ions are high.

(2). The solubility of calcium bromide in water is high.

Explanation:

We know that for a reaction its solubility product is equal to the product of concentration of ions present.

For example, CaBr_{2} \rightarrow Ca^{2+} + 2Br^{-}

      K_{sp} = [Ca^{2+}][Br^{-}]^{2}

As it is given that for this reaction K_{sp} > 1. Hence, the product of concentration of both these ions will be greater than 1 which means that products are favored.

Therefore, equilibrium concentrations of calcium and bromide ions are high.

Also, more is the number of these ions formed more will be the solubility of the compound (CaBr_{2}).

As a result, the solubility of calcium bromide in water is high.

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3 years ago
A student adds 0. 0030 mol of hcl to 100 ml of a 0. 10 m solution of a r2nh, a weak base. The ph of the solution is found to be
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Since an acidic salt solution is produced when a strong acid neutralizes a weak base, the pH of the salt solution formed when HCl is added to R2NH will be less than 7.

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A neutralization reaction is the react ion between an acid and a base to form salt and water only.

Neutralization reactions can either produce a neutral solution, an acidic solution or an alkaline solution at equivalence point.

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Read 2 more answers
What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as
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CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

<span>CaCl2 ---> Ca2+  +  2 Cl-</span>

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

 

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

 

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

<span>Cl- = 0.4 moles             (ANSWER)</span>

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3 years ago
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