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olya-2409 [2.1K]
3 years ago
11

Be sure to answer all parts. What can be added to the equilibrium AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl−(aq) (a) that will shi

ft it to the left? (Select all that apply.) AgCl NH3 Ag(NH3)2+ Cl− (b) that will shift it to the right? (Select all that apply.) AgCl NH3 Ag(NH3)2+ Cl− (c) that will not shift it in either direction? (Select all that apply.) AgCl NH3 Ag(NH3)2+ Cl−
Chemistry
1 answer:
lora16 [44]3 years ago
6 0

Answer:

a) Ag(NH₃)₂⁺,  Cl⁻.

b) NH₃.

c) AgCl.

Explanation:

Based on LeChatelier's law, a system in chemistry can change responding to a disturbance of concentration, temperature, etc. in order to restore a new state.

In the reaction:

AgCl(s) + 2NH₃(aq) ⇌ Ag(NH₃)₂⁺(aq) + Cl⁻(aq)

When reactants are added, the system will produce more products restoring the equilibrium and vice versa. A reactant in solid state doesn't take part in the equilibrium, thus:

a) Ag(NH₃)₂⁺,  Cl⁻. The addition of products will shift the equilibrium to the left

b) NH₃. The addition of reactant will shift the equilibrium to the right.

c) As AgCl is in solid phase, will not shift the equilibrium in either direction.

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If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.

We solve this problem by first writing the equation
N2 + 3H2 = 2NH3
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3 years ago
Write the rates for the following reactions in terms of the disappearance of reactants and appearance of products: (a) 302 .....
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Explanation:

\mathbf{From \  the  \ information \  given:} \\ \\ \mathbf{The \  rates \  of  \ the \ f ollowing \  reactions \  can \  be \  expressed  \ as \  follows:}

(a)

\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}

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(c)

\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} =  \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt}   }

(d)

\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\  \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}

(e)

\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\  \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}

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True

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Mass plus atomic and number

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