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Annette [7]
3 years ago
5

Determine the mass of 1 mol of gold (Au), of aluminum chloride (AlCl3), and of glucose (C6H12O6)

Chemistry
1 answer:
sleet_krkn [62]3 years ago
5 0

Answer: Mass of 1 mol of gold (Au) is 197g , of aluminum chloride(AlCl_3) is 133.5 g and of glucose (C_6H_{12}O_6) is 180 g

Explanation:  Mass of 1 mole of a substance is called its molar mass.

a) 1 mole of Au contains 6.023\times 10^{23}atoms and weigh 197g.

b) 1 mole of AlCl_3 will weigh=1\times {\text {mass of 1 mole of Aluminium}}+3\times {\text {mass of 1 mole of chlorine atom}}

Thus mass of 1 mole of AlCl_3 will be =1moles\times {27g/mol}+3moles\times {35.5g/mol}=133.5g

c) 1 mole of C_6H_{12}O_6 will weigh=6\times {\text {mass of 1 mole of carbon}}+12\times {\text {mass of 1 mole of hydrogen atom}}+6\times {\text {mass of 1 mole of oxygen}}

1 mole of C_6H_{12}O_6 will weigh=6\times 12g/mol+12\times 1g/mol+6\times 16g/mol=180g


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Answer:

1.930 * 10⁻⁹ mg of Mn⁺² are left unprecipitated.

Explanation:

The reaction that takes place is:

Mn⁺² + S⁻² ⇄ MnS(s)  

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If the pksp of MnS is 13.500, then the ksp is:

ksp=10^{-13.500}=3.1623*10^{-14}

From the problem we know that [S⁻²] = 0.0900 M

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3.1623*10⁻¹⁴= [Mn⁺²] * 0.0900 M

[Mn⁺²] = 3.514 * 10⁻¹³ M.

Now we can calculate the mass of Mn⁺², using the volume, concentration and atomic weight. Thus the mass of Mn⁺² left unprecipitated is:

3.514 * 10⁻¹³ M * 0.1 L * 54.94 g/mol = 1.930 * 10⁻¹² g = 1.930 * 10⁻⁹ mg.

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There are several different types of greenhouse gases. The major ones are carbon dioxide, water vapor, methane, and nitrous oxide. These gas molecules all are made of three or more atoms. The atoms are held together loosely enough that they vibrate when they absorb heat. Eventually, the vibrating molecules release the radiation, which will likely be absorbed by another greenhouse gas molecule. This process keeps heat near the Earth’s surface. Most of the gas in the atmosphere is nitrogen and oxygen, which cannot absorb heat and contribute to the greenhouse effect.

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