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3 years ago

Describe the process by which ag+ ions are precipitated out of solution

Chemistry

2 answers:

sasho [114]3 years ago

8 0

Answer:

By adding a substance releasing a precipitating anion.

Explanation:

Hello,

In this case, silver precipitated out from substances such a silver nitrate which is largely soluble in water could be obtained via looking for another substance that when in solution, completely dissociates forming an anion in order to react with the silver cations released by the silver nitrate (just as an example because it could be any silver-releasing salt or compound) to form a poorly soluble silver substance. One of the widely used anions to precipitate silver out is the chloride one which yields a white precipitate of silver nitrate, this is attained via the drop-by-drop (qualitative) or the stiochiometric (quantitative) addition of the chloride-based salt which leads to the formation of the required precipitate.

Best regards.

MA_775_DIABLO [31]3 years ago

3 0

Describe the process by which Ag+ ions are precipitated out of solution. 4. In your testing, several precipitates are formed, and then dissolved as complexes.

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VMariaS [17]

Answer:- 245 kPa.

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FrozenT [24]

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At a certain temperature, 0.4611 mol of N 2 and 1.661 mol of H 2 are placed in a 4.50 L container. N 2 ( g ) + 3 H 2 ( g ) − ⇀ ↽

Monica [59]

Answer:

The equilibrium constant is 64.76

Explanation:

Step 1: Data given

Number of moles of N2 = 0.4611 moles

Number of moles of H2 = 1.661 moles

Volume = 4.50 L

At the equilibrium there is 0.1801 mol of N2

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3 (g)

Step 3: The initial moles

N2: 0.4611 mol

H2: 1.661 mol

NH3: 0 mol

Step 4: Calculate the moles reacted

N2: 0.4611 mol - 0.1801 mol = 0.281 mol

H2: 3 * 0.281 mol = 0.843 mol

NH3: 2 * 0.281 mol = 0.562 mol

Step 5: Calculate moles at the equilibrium

N2: 0.1801 mol

[N2] = 0.1801 mol / 4.50 L = 0.04002 mol/L

H2: 1.661 mol - 0.843 mol = 0.818 mol

[H2] = 0.818 mol / 4.50 L = 0.182 mol/L

NH3: 0.562 mol

[NH3] = 0.562 mol / 4.50 L = 0.125 mol/L

Step 6: Calculate Kc

Kc = [NH3]² / ([N2] [H2]³)

Kc = 0.125² /(0.04002*0.182³)

Kc = 64.76

The equilibrium constant is 64.76

3 0

3 years ago