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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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An increase in the atomic number________the atomic radius moving from left to right across a period.
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4 0
3 years ago
An iron nail rusts when exposed to oxygen. According to the following reaction, how many moles of iron(III) oxide will be formed
Zigmanuir [339]

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0.453 moles

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From the equation,  mass of O2 involved = 16 x 2 x 3 = 96g

                                 mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2

                                                                            = 100g

                Therefore 96g of O2 produced 100g of Fe2O3

                                  32.2g of O2 Will produce   100x32.2/96

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Converting it to mole using   number of mole = mass/molar mass

but molar mass of Fe2O3 = 26 + (16 X 3)

                                           = 74g/mole

Therefore number of mole of 33.54g of Fe2O3 = 33.54/74

                                                                           = 0.453 moles

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