Answer:
49.86 × 10²³ atoms of Al
Explanation:
Given data:
Number of moles of Al = 8.28 mol
Number of atoms = ?
Solution:
The given problem will solve by using Avogadro number.
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 = 6.022 × 10²³ atoms of hydrogen
For 8.28 moles of Al:
1 mole = 6.022 × 10²³ atoms of Al
8.28 mol×6.022 × 10²³ atoms / 1mol
49.86 × 10²³ atoms of Al
Answer:
the concentration of bicarbonate is <em>[HCO₃⁻] = 0,03996 M </em>and carbonate is <em>[CO₃²⁻] = 3,56x10⁻⁵ M.</em>
Explanation:
Carbonate-bicarbonate is:
HCO₃⁻ ⇄ CO₃²⁻ + H⁺ With pka = 10,25
Using Henderson-Hasselbalach formula:
pH = pka + log₁₀![\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B3%7D%5E%7B2-%7D%5D%7D%7B%5BHCO_%7B3%7D%5E-%5D%7D)
7,2 = 10,25 + log₁₀![\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B3%7D%5E%7B2-%7D%5D%7D%7B%5BHCO_%7B3%7D%5E-%5D%7D)
8,91x10⁻⁴ =
<em>(1)</em>
Also:
0,040 M = [CO₃²⁻] + [HCO₃⁻] <em>(2)</em>
Replacing (2) in 1:
<em>[HCO₃⁻] = 0,03996 M</em>
Thus:
<em>[CO₃²⁻] = 3,56x10⁻⁵ M</em>
I hope it helps.
A nuclear reactionIt’s like another particle with the release of energy