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seraphim [82]
3 years ago
15

Briefly explain whether the following procedural errors would result in either an incorrectly high or incorrectly low calculated

percent Cu recovery for this experiment.
A) This solution was not basic when it was heated in part 3. ( in part 3 i convertedCu(OH)2 to CuO).

B) A slightly blue solution was decanted from Cu in part V. (in part 5 i reduced Cu(H20)6 ions with zink)

C) In part 5 the water in the beaker boiled away, exposing the evaporating dish to excess heat (same as above).
Chemistry
1 answer:
seropon [69]3 years ago
7 0
<span>A) This solution was not basic when it was heated in part 3. ( in part 3 i convertedCu(OH)2 to CuO).
Incorrectly low, because not all copper compounds will precipitate out

B) A slightly blue solution was decanted from Cu in part V. (in part 5 i reduced Cu(H20)6 ions with zink)
Incorrectly low, because some copper were thrown away

C) In part 5 the water in the beaker boiled away, exposing the evaporating dish to excess heat (same as above).
incorrectly high, because other compounds might be present as well </span>
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The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

<u>Step 1:</u> Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

<u>Step 2:</u> Calculate moles of Ca^2+

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<u>Step 3: </u>Calculate moles of SO4^2-

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Moles SO4^2- = 0.05 moles

<u>Step 4: </u>Calculate total volume

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<u>Step 5: </u> Calculate Q

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Qsp = (0.050)(0.050 )=0.0025 >> Ksp

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<u> Step 6:</u> Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

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<u> Step 7:</u> Calculate total CaSO4 dissolved

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<u>Step 8:</u> Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

<u>Step 9:</u> Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

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