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Andre45 [30]
3 years ago
13

What is Avogadro's number, and why is it useful? (3 points)

Chemistry
1 answer:
SashulF [63]3 years ago
4 0

Answer:

6.022 x 10²³; it is a conversion factor between moles and number of particles

Explanation:

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole of hydrogen  = 6.022 × 10²³ atoms of hydrogen

238 g of uranium = 1 mole of uranium = 6.022 × 10²³ atoms of uranium

By taking ions:

62 g of NO⁻₃ =  1 mole of  NO⁻₃  = 6.022 × 10²³ ions of  NO⁻₃

96 g of SO₄²⁻ = 1 mole of SO₄²⁻ =  6.022 × 10²³ ions of  SO₄²⁻

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Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s
Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

                        N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where N is amount of radioactive nuclide after t time, N_{0} is initial amount of radioactive nuclide and t_{\frac{1}{2}} is half life of radioactive nuclide

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1.5mol C3H8 from C3H8+5O2-->3CO2+4H2O .how many grams of carbon dioxide are produced
koban [17]

Answer:

\large \boxed{\text{200 g CO}_{{2}}}

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ:                                 44.01

            C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol:    1.5

1. Calculate the moles of CO₂

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\rm  \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}

2. Calculate the mass of CO₂.

\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2}  \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}

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