Chemistry: A54-gram sample of an unknown material has a volume equal to 20 cm3 . based on its density, could the sample be aluminum? explain.

A54-gram sample of an unknown material has a volume

Ответ:

whereswoodruff

07.01.2022

D = m/V
d = 54/20
d = 2,7 g/cm³

Could the sample be aluminum?YES, this sample is aluminium - the density of chemical elements can be found in the periodic table.

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Ответ:

tiaholmes31

14.08.2022

Chemistry

δ N2(g) = 1.1825 g/L

Explanation:

δ ≡ m/vMw N2(g) = 28.0134 g/mol

ideal gas:

PV = RTn

∴ P = (837 torr)×( atm/760 torr) = 1.1013 atm

∴ T = 45.0 °C + 273.15 = 318.15 K

∴ R = 0.082 atm.L/K.mol

⇒ n/V = P/R.T

⇒ n/V = (1.1013 atm) / ((0.082 atm.L/K.mol)(318.15 k))

⇒ n/V = 0.0422 mol/L

⇒ δ N2(g) = (0.042 mol/L)×(28.0134 g/mol) = 1.1825 g/L

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A54-gram sample of an unknown material has a volume equal to 20 cm3
. based on its density,
could the sample be aluminum? explain.

Ответ:

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