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Brums [2.3K]
3 years ago
12

1.0 mole of KCl(aq) is added to which solution to produce a precipitate?

Chemistry
1 answer:
taurus [48]3 years ago
7 0

Answer:

Ag+

Explanation:

If you imagine as if the problem were double replacement, you would pair the Cl with one of the following ions provided in the choices. As seen on Table F, Ag+ paired with Cl- produces an insoluble compound, hence the precipitate. All the other ions shown in the multiple choice section, when paired with Cl- will produce a soluble compound, as a result NOT a precipitate.

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A student is titrating 50 mL of 0.32 M NH3 with 0.5 M HCl. How much hydrochloric acid must be added to react completely with the
Sveta_85 [38]

Answer:

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

Explanation:

Using the formula

Ca Va = Cb Vb

Cb = 0.32 M

Vb = 50 mL = 50/1000 = 0.050L

Ca = 0.5 M

Va =?

Substituting for Va in the equation, we obtain:

Va = Cb Vb / Ca

Va = 0.32 * 0.05 / 0.5

Va = 0.016 / 0.5

Va = 0.032 L

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

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3 years ago
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3 years ago
How many grams of barium sulfate, BaSO 4 , be precipitated when 2.25 moles of sodium sulfate , Na 2 SO 3 , reacts with an excess
gulaghasi [49]

Answer:

525.1 g of BaSO₄ are produced.

Explanation:

The reaction of precipitation is:

Na₂SO₄ (aq) + BaCl₂ (aq) →  BaSO₄ (s) ↓  +  2NaCl (aq)

Ratio is 1:1. So 1 mol of sodium sulfate can make precipitate 1 mol of barium sulfate.

The excersise determines that the excess is the  BaCl₂.

After the reaction goes complete and, at 100 % yield reaction, 2.25 moles of BaSO₄ are produced.

We convert the moles to mass: 2.25 mol . 233.38 g/mol = 525.1 g

The precipitation's equilibrium is:

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7 0
2 years ago
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