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3 years ago

Which of the following pairs of aqueous solutions will form a precipitate when mixed?

Chemistry

1 answer:

olya-2409 [2.1K]3 years ago

6 0

Answer:

The correct answer is A) Pb(C₂H₃O₂)₂ + Li₂SO₄

Explanation:

All salts of Na, K and ammonium are soluble.
All nitrates are soluble.
All chlorides are soluble, with the exception of AgCl, Hg₂Cl₂, PbCl₂ and CuCl.
All sulfates are soluble, with the exception of CaSO₄, SrSO₄, BaSO₄, PbSO₄, HgSO₄, Hg₂SO₄ and Ag₂SO₄.
All hydroxides are poorly soluble, with the exception of alkaline hydroxides, Sr(OH)₂, Ba(OH)₂ and NH4(OH)
All carbonates are poorly soluble, with the exception of alkaline carbonates and ammonium carbonate.
All sulfides are poorly soluble, with the exception of alkaline sulfides, alkaline earth sulfides, and ammonium sulfide.

The reaction A is:

Pb(C₂H₃O₂)₂ + Li₂SO₄ ⇒PbSO₄ + 2 LiC₂H₃O₂

Lead sulfate is a slightly soluble solid, therefore it precipitates.

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3 years ago

Silver (Ag) has two stable isotopes: 107Ag, 106.90 amu, and 109Ag, 108.90 amu. If the average atomic mass of silver is 107.87 am

inysia [295]

Answer:

The abundance of 107Ag is 51.5%.

The abundance of 109Ag is 48.5%.

Explanation:

The average atomic mass of silver can be expressed as:

107.87 = 106.90 * A1 + 108.90 * A2

Where A1 is the abundance of 107Ag and A2 of 109Ag.

Assuming those two isotopes are the only one stables, we can use the equation:

A1 + A2 = 1.0

So now we have a system of two equations with two unknowns, and what's left is algebra.

First we use the second equation to express A1 in terms of A2:

A1 = 1.0 - A2

We replace A1 in the first equation:

107.87 = 106.90 * A1 + 108.90 * A2

107.87 = 106.90 * (1.0-A2) + 108.90 * A2

107.87 = 106.90 - 106.90*A2 + 108.90*A2

107.87 = 106.90 + 2*A2

2*A2 = 0.97

A2 = 0.485

So the abundance of 109Ag is (0.485*100%) 48.5%.

We use the value of A2 to calculate A1 in the second equation:

A1 + A2 = 1.0

A1 + 0.485 = 1.0

A1 = 0.515

So the abundance of 107Ag is 51.5%.

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3 years ago

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german

Answer:

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Explanation:

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