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irina1246 [14]
3 years ago
9

What is the mass of 1.6 mol of aluminum atom?

Chemistry
1 answer:
PIT_PIT [208]3 years ago
8 0
Answer = 43.2

Because to find the mass you multiply the amount of moles ( 1.6) by the mass of the element ( mass of aluminium is 27) so
1.6 x 27 = 43.2g

Hope this helps
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Which has the highest density, water at 0c, water at 4c, water at 6c, water at 8c
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Answer:

water at 0C

Explanation:

The colder the water is, the denser it is, so the water here with the lowest temperature, is 0C

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2 years ago
The C=C bond in 2-cyclohexenone produces an unusually strong signal. Explain using resonance structuresBased on the above resona
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Answer:

When two single single bonds separated by a double bond (e.g C=C-C=C or C=C-C=O in the case of 2-cyclohexenone), the effect of resonance among those there bonds will be observed.

Explanation:

Since the Oxygen atom has higher electronegativity, it will cause the electrons in the resonance bonds 'flow' toward the Oxygen atom, so that the C=C will 'lose' some electron. The signal read for that bond will be different from other alkene structure.

Attachment is the resonance structure of 2-cyclohexene.

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A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h
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Answer:

5.758  is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d =998.2 kg/m^3

1 kg = 1000 g

1 m^3=10^6 cm^3

998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3

Mass of water ,m'= m - M = 45.65 g -  27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

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V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' =  9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3

Density of metal = d'

Volume of metal = v' =\frac{M'}{d'}

Difference in volume will give volume of metal ingot.

v' = v - V

v'=19.73 cm^3-18.08 cm^3=

v'=1.65 cm^3

Since volume cannot be in negative .

Density of the metal =d'

=d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3

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