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3 years ago

Please help!

Chemistry

1 answer:

sweet [91]3 years ago

4 0

Answer:

B) 1.30 ×10²⁴ molecules

Explanation:

According to Avogadro law,

Equal volume of all gases at same temperature and pressure have equal number of molecules.

For equal volume the amount of gas (moles) would be same.

V ∝ n

V = Kn

k = proportionality constant

V₁/n₁ = V₂/n₂

V₁ = initial volume

n₁ = initial moles

V₂ = final volume

n₂ = final moles

So if ammonia has 1.30 ×10²⁴ molecules, hydrogen will have same molecules at same conditions.

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Silver chloride, AgCl (Ksp = 1.8 x 10‒10), can be dissolved in solutions containing ammonia due to the formation of the soluble

Sergeeva-Olga [200]

Explanation:

The given reaction will be as follows.

$AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq)$ ............. (1)

$K_{sp}$ = $[Ag^{+}][Cl^{-}] = 1.8 \times 10^{-10}$

Reaction for the complex formation is as follows.

$Ag^{+}(aq) + 2NH_{3}(aq) \rightleftharpoons [Ag(NH_{3})_{2}]^{+}(aq)$ ........... (2)

$K_{f}$ = $\frac{[Ag(NH_{3})_{2}]}{[Ag^{+}][NH_{3}]^{2}} = 1.0 \times 10^{8}$

When we add both equations (1) and (2) then the resultant equation is as follows.

$AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)$ ............. (3)

Therefore, equilibrium constant will be as follows.

K = $K_{f} \times K_{sp}$

= $1.0 \times 10^{8} \times 1.8 \times 10^{-10}$

= $1.8 \times 10^{-2}$

Since, we need 0.010 mol of AgCl to be soluble in 1 liter of solution after after addition of $NH_{3}$ for complexation. This means we have to set

$[Ag^{+}]$ = $[Cl^{-}]$

= $\frac{0.010 mol}{1 L}$

= 0.010 M

For the net reaction, $AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)$

Initial : 0.010 x 0 0

Change : -0.010 -0.020 +0.010 +0.010

Equilibrium : 0 x - 0.020 0.010 0.010

Hence, the equilibrium constant expression for this is as follows.

K = $\frac{[Ag(NH_{3})^{+}_{2}][Cl^{-}]}{[NH_{3}]^{2}}$

$1.8 \times 10^{-2}$ = $\frac{0.010 \times 0.010}{(x - 0.020)^{2}}$

x = 0.0945 mol

or, x = 0.095 mol (approx)

Thus, we can conclude that the number of moles of $NH_{3}$ needed to be added is 0.095 mol.

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Answer:

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Answer:

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Explanation:

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