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Vesna [10]
3 years ago
13

A solution contains 0.0150 m pb2 (aq) and 0.0150 m sr2 (aq). if we add so42–(aq), what will be the concentration of pb2 (aq) whe

n srso4(s) begins to precipitate
Chemistry
1 answer:
polet [3.4K]3 years ago
3 0
Ksp of SrSO4 = 3.2 x 10-7, so it will precipitate when Ksp > 3.2 x 10-7  
Same way, calculating for 0.0150 M Sr2 SrSO4 precipitates,
 when [ SO4 2-] > (3.2 x 10-7 / 0.0150) = 2.133 x 10^-05.
 Thus SrSO4 precipitates when SO4 2- > 2.133 x 10^-05 
 Now the Pb2+ at this point = 2.133 x 10^-05 x 0.0150 =3.2 x 10^-07 
 Hence Ksp for PbSO4 = 1.6 x 10-8 
 the solubility product for PbSO4 is more and the PbSO4 precipitates as
follows [Pb2+] = 1.6 x 10-8 / 2.133 x 10^-05 = 7.50 x 10^-04

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Answer:

d = 0.93 g/cm³

Explanation:

Given data:

Mass of object = 28 g

Volume of object = 3cm×2cm×5cm

density of object = ?

Solution:

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Volume of object = 30 cm³

Density of object:

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3 0
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A first-order reaction is 45% complete at the end of 43 minutes. What is the length of the half-life of this reaction
shtirl [24]

The half-life of the reaction is 50 minutes

Data;

  • Time = 43 minutes
  • Type of reaction = first order
  • Amount of Completion = 45%

<h3>Reaction Constant</h3>

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The reactant left = (1 - 0.45) X_0 = 0.55 X_0 = X

For a first order reaction

\ln(\frac{x}{x_o}) = -kt\\ k = \frac{1}{t}\ln (\frac{x_o}{x}) \\ k = \frac{1}{43}\ln (\frac{x_o}{0.55_o})\\ k = 0.013903 min^-^1

<h3>Half Life </h3>

The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.

x = \frac{x_o}{2} \\t = t_\frac{1}{2} \\t_\frac{1}{2} = \frac{1}{k}\ln(\frac{x_o}{x_o/2})\\

Substitute the values

t_\frac{1}{2} = \frac{1}{k}\ln(2)=\frac{0.6931}{0.013903}\\t_\frac{1}{2}= 49.85 min = 50 min

The half-life of the reaction is 50 minutes

Learn more on half-life of a first order reaction here;

brainly.com/question/14936355

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2 years ago
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topjm [15]

Explanation:

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