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Sphinxa [80]
4 years ago
14

What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c

. PbSO4 d. PbSO2 e. PbS2O3
Chemistry
2 answers:
brilliants [131]4 years ago
8 0

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and  (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

GalinKa [24]4 years ago
6 0

Answer:

The empirical formula is PbSO4 (option C)

Explanation:

Step 1: Data given

Suppose the mass of a compound is 100.0 grams

The compound contains:

68.3 % lead = 68.3 grams

10.6 % sulfur = 10.6 grams

Rest = oxygen

Molar mass Pb = 207.2 g/mol

Molar mass S = 32.065 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate mass O

MAss O = 100 - 68.3 - 10.6 = 21.1 grams

Step 3: Calculate moles

Moles = mass / molar mass

Moles Pb = 68.3 grams / 207.2 g/mol

Moles Pb = 0.3296 moles

Moles S = 10.6 grams / 32.065 g/mol

Moles S = 0.3306 moles

Moles O = 21.1 grams / 16.0 g/mol

Moles O = 1.319 moles

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

Pb: 0.3296/0.3296 = 1

S: 0.3306/0.3296 = 1

O: 1.319/0.3296 = 4

The empirical formula is PbSO4 (option C)

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4 0
3 years ago
For test 6 (150 hevy 50 light particles), what is the pressure contribution from the heavy particles?
lawyer [7]

Answer:

The pressure contribution from the heavy particles is 17.5 atm

Explanation:

According to Dalton's law of partial pressures, if there is a mixture of gases which do not react chemically together, then the total pressure exerted by the mixture is the sum of the partial pressures of the individual gases that make up the mixture.

In the simulation:

the pressure of the 50 light particles alone was determined to be 5.9 atm, the pressure of the 150 heavy particles alone was measured to be 17.5 atm,

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45.7 grams of calcium chloride reacts with an excess of aluminum oxide. How many grams of aluminum chloride will be produced
damaskus [11]
Molar mass (CaCl2) = 40.1 +2*35.5 = 111.1 g/mol
Molar mass (AlCl3) = 27.0 +3*35.5= 133.5 g/ mol

                                               
3CaCl2+Al2O3 -------->3CaO +2AlCl3
mole from reaction              3 mol                                              2 mol
mass from reaction         3mol* 111.1g/mol                             2 mol*133.5g/mol
                                               333.3 g                                            267.0 g
mass from problem              45.7 g                                               x g

Proportion:
  333.3 g  CaCl2  -------   267.0 g AlCl3
  45.7 g   CaCl2   --------   x g    AlCl3

x=45.7*267.0/333.3= 36.6 g AlCl3
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3 years ago
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