Given:
Diprotic weak acid H2A:
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9.
Concentration = 0.0650 m
Balanced chemical equation:
H2A ===> 2H+ + A2-
0.0650 0 0
-x 2x x
------------------------------
0.065 - x 2x x
ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)
solve for x and determine the concentration at equilibrium.
The most likely bond between element X and Iodine would be an ionic, or electrovalent, bond. Iodine has seven electrons in its outer shell, also known as the valence shell. To become perfectly stable, it needs only a single electron from another element. Hence no sharing of electron takes place (usually), which is the condition required for it to be covalent bonding. Hence it's most likely an ionic bonding/
Answer:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in

<u>= 4.75 × 10²³ molecules of Propane</u>.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =



<u>= 34.72 g</u>