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Jlenok [28]
3 years ago
10

The density of lead is 11.4g/cm3. What volume, in ft3, would be occupied by 10.0 g of lead?

Chemistry
1 answer:
yawa3891 [41]3 years ago
7 0

Answer:

3.10×10¯⁵ ft³.

Explanation:

The following data were obtained from the question:

Density (D) of lead = 11.4 g/cm³

Mass (m) of lead = 10 g

Volume (V) of lead =.?

Density (D) = mass (m) / Volume (V)

D = m/V

11.4 = 10 / V

Cross multiply

11.4 × V = 10

Divide both side by 11.4

V = 10 / 11.4

V = 0.877 cm³

Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:

1 cm³ = 3.531×10¯⁵ ft³

Therefore,

0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³

0.877 cm³ = 3.10×10¯⁵ ft³

Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.

Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.

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Answer:

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Explanation:

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Equilibrium constant of the reaction at 500 K =K_c=6.90\times 10^{-6}

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An expression of equilibrium constant is given as:

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[A] = (3-x) = 3 M - 0.002274 M =2.997726 M

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