PLS HELP CHEM

Consider the insoluble compound silver bromide , AgBr . The silver ion also forms a complex with ammonia . Write a balanced net

Degger [83]

Answer:

- $AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

- $K=1.2x10^{-5}$

Explanation:

Hello,

In this case, by considering the dissolution of silver bromide:

$AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}$

And the formation of the complex:

$Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7$

We obtain the balanced net ionic equation by adding the aforementioned equations:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}$

So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:

$K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}$

Best regards.

4 0

3 years ago

1. When water is evaporated from a copper sulfate solution, copper sulfate remain.

djverab [1.8K]

Evaporation is used to separate a soluble solid from a liquid. For example, copper sulfate is soluble in water – its crystals dissolve in water to form copper sulfate solution. During evaporation, the water evaporates away leaving solid copper sulfate crystals behind.

6 0

3 years ago

Whether or not the process is observed in nature, which of the following could account for the transformation of gallium-67 to z

Sonja [21]

Answer:

Option a: positron emission.

Explanation:

In the transformation we have:

⁶⁷Ga → ⁶⁷Zn

The reaction is:

$^{67}_{Z}X \rightarrow ^{67}_{Z -1}Y$

For Ga to become Zn, the atom nucleus has to lose a proton, so in the given options, the reaction that involves the transformation of a proton is the option a, positron emission.

In a positron emission, a proton becomes into a neutron and a positron:

$^{A}_{Z}X \rightarrow ^{A}_{Z-1}Y + ^{0}_{+1}e$

Therefore, the correct answer is option a: positron emission.

I hope it helps you!

7 0

3 years ago

Ideal gas law, Please help me it's due soon

ICE Princess25 [194]

Answer:

is the equation of state of a hypothetical ideal gas

Explanation:

its formula is PV=nRT

5 0

2 years ago

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

3 0

3 years ago