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Drupady [299]
3 years ago
15

When the equation is correctly balanced, the coefficient of aluminum chloride will be ___.

Chemistry
1 answer:
ExtremeBDS [4]3 years ago
7 0

The answer for the following question is answered below.

<em>Therefore when the equation is balanced the co-efficient of aluminium chloride will be 2.</em>

Explanation:

The balanced equation for the following equation is:

AlCl_{3} + Ba(OH)_{2}  → Al(OH)_{3} + BaCl_{2}

To balance an equation:

  • Firstly we have to see whether the elements on the reactants and products side are same or not.
  • Then balance each element and see whether the equation  is balanced or not

<em>When 2 moles of aluminium chloride react with 3 moles of barium hydroxide on the reactants side then on the products side it yields 2 moles of aluminium hydroxide and 3 moles of barium chloride </em>

(i.e.)

2AlCl_{3} + 3Ba(OH)_{2}  → 2Al(OH)_{3} + 3BaCl_{2}

<em><u>Therefore when the equation is balanced the co-efficient of aluminium chloride will be 2.</u></em>

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Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI
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Answer :

AgI should precipitate first.

The concentration of Ag^+ when CuI just begins to precipitate is, 6.64\times 10^{-7}M

Percent of Ag^+ remains is, 0.0076 %

Explanation :

K_{sp} for CuI is 1\times 10^{-12}

K_{sp} for AgI is 8.3\times 10^{-17}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

CuI\rightleftharpoons Cu^++I^-

The expression for solubility constant for this reaction will be,

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

1\times 10^{-12}=0.0079\times [I^-]

[I^-]=1.25\times 10^{-10}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgI\rightleftharpoons Ag^++I^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][I^-]

8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M

[Ag^+]=6.64\times 10^{-7}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{6.64\times 10^{-7}}{0.0087}\times 100

Percent of Ag^+ remains = 0.0076 %

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Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

\alpha = \sqrt{4.9629*10^{-5}}

\alpha = 7.044*10^{-3

percentage% (∝) = 7.044*10^{-3}*100

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For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

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\alpha = 0.0130\\

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= 1.30%

For Part C; where Concentration of C= 1.92*10^{-2} M

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\alpha = \sqrt{6.802*10^{-4}}

\alpha =0.02608

percentage% (∝) = 0.02608  × 100%

= 2.60%

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