A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a mass of 23.875 g for the crucible, lid, and sample. The mass of the empty crucible and lid was found earlier to be 22.652 g. He then heats the crucible to expel the water of hydration, keeping the crucible at red heat for 10 minutes with the lid slightly ajar. On colling, he finds the mass of crucible, lid, and contents to be 23.403 g. The sample was changed in the process to very light clue anhydrous CuSO4. How many moles of water are present per mole of CuSO4?
Answers
Answer:
There are present 5,5668 moles of water per mole of CuSO₄.
Explanation:
The mass of CuSO₄ anhydrous is:
23,403g - 22,652g = 0,751g.
mass of crucible+lid+CuSO₄ - mass of crucible+lid
As molar mass of CuSO₄ is 159,609g/mol. The moles are:
0,751g × = 4,7052x10⁻³ moles CuSO₄
Now, the mass of water present in the initial sample is:
23,875g - 0,751g - 22,652g = 0,472g.
mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid
As molar mass of H₂O is 18,02g/mol. The moles are:
0,472g × = 2,6193x10⁻² moles H₂O
The ratio of moles H₂O:CuSO₄ is:
2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668
That means that you have 5,5668 moles of water per mole of CuSO₄.
I hope it helps!
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in 5.164 g of Call2?
Answers
Answer:
1.06x10²² formula units
Explanation:
First we convert 5.164 g of CaI₂ into moles, using its molar mass:
- 5.164 g ÷ 293.887 g/mol = 0.0176 mol
Then we convert 0.0176 moles into formula units, using Avogadro's number, which relates the number of formula units present in 1 mol:
- 0.0176 mol * 6.023x10²³ FormulaUnits/mol = 1.06x10²² formula units
Answers
a) 1 HBr + 1 NaHCO₃ = 1 CO₂ + 1 H₂O + 1 NaBr
b) 2 HNO₃ + 1 K₂SO₃ = 1 H₂O + 2 KNO₃ + 1 SO₂
c) 2 HI + 1 Na₂S = 1 H₂S + 2 NaI
d) 1 (NH₄)₂SO₄ + 1 Ca(OH)₂ = 1 CaSO₄ + 2 H₂O + 2 NH₃
Answers
Answer:
1. Bronsted—Lowry acid
2. Bronsted—Lowry Base
3. Lower the pka
4. Lewis acids
Explanation:
Answers
Answer:
1,080 m
Explanation:
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Answer:
when gas condenses to liquid the quantity of energy converts.
Explanation:
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Answers
Answer:
6.1 kg
Explanation:
To obtain the total mass of the sample, we must first express each mass of the sample in the same unit of measurement.
Since the SI unit of mass is kilogram (kg), we shall express the total mass of the samples in kilogram (kg).
This is illustrated below:
Mass of the samples are:
M1 = 0.6160959 kg
M2 = 3.225 mg
M3 = 5480.7 g.
Conversion of 3.225 mg to kg
1 mg = 1×10¯⁶ kg
Therefore,
3.225 mg = 3.225 × 1×10¯⁶
3.225 mg = 3.225×10¯⁶ kg
Conversion of 5480.7 g to kg
1000 g = 1 kg
Therefore,
5480.7 g = 5480.7 /1000
5480.7 g = 5.4807 kg
Thus, we can obtain the total mass of the samples as follow:
M1 = 0.6160959 kg
M2 = 3.225×10¯⁶ kg
M3 = 5.4807 kg
Total mass =?
Total mass = M1 + M2 + M3
Total mass = 0.6160959 + 3.225×10¯⁶ + 5.4807
Total mass = 6.096799125 ≈ 6.1 kg
Therefore, the total mass of the samples is approximately 6.1 kg.