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jekas [21]
3 years ago
11

6PbO + O2 → 2Pb304 a. How many grams of Pb304 are produced from 8.50 grams of lead(II) oxide?

Chemistry
1 answer:
AlladinOne [14]3 years ago
7 0

Answer:

8.70g of Pb3O4

Explanation:

6PbO + O2 → 2Pb3O4

Molar Mass of PbO = 207 + 16 = 223g/mol

The mass of the PbO from the balanced equation = 6 x 223 = 1338g

Molar Mass of Pb3O4 = (3x207) +(16x4) = 621 + 64 = 685g/mol

Mass of Pb3O4 from the balanced equation = 2 x 685 = 1370g

From the equation,

1338g oh PbO produced 1370g of Pb3O4.

Therefore, 8.50g of PbO will produce = (8.5 x 1370)/1338 = 8.70g of Pb3O4

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The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
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Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

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Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

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