Answer:
2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.
Explanation:
<em>∵ pH = - log[H₃O⁺]</em>
∴ 4.6 = - log[H₃O⁺].
∴ log[H₃O⁺] = - 4.6.
∴ [H₃O⁺] = 2.51 x 10⁻⁵.
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
[H₃O⁺] = 2.51 x 10⁻⁵ M.
∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.51 x 10⁻⁵ M) = 3.98 × 10⁻¹⁰ M ≅ 4.0 × 10⁻¹⁰ M.
<em>So, the right choice is: 2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.</em>
Answer: The final volume of this solution is 0.204 L.
Explanation:
Given: Molarity of solution = 2.2 M
Moles of solute = 0.45 mol
Molarity is the number of moles of solute present divided by volume in liters.
Substitute the values into above formula as follows.
Thus, we can conclude that the final volume of this solution is 0.204 L.
Answer:
b) 2.0 mol
Explanation:
Given data:
Number of moles of Ca needed = ?
Number of moles of water present = 4.0 mol
Solution:
Chemical equation:
Ca + 2H₂O → Ca(OH)₂ + H₂
now we will compare the moles of Ca and H₂O .
H₂O : Ca
2 : 1
4.0 : 1/2×4.0 = 2.0 mol
Thus, 2 moles of Ca are needed.
The question ask for the percentage of the abundance of galium-69 where there is two isotopes of galium: the 69Ga and the 71Ga. The average atomic mass of gallium is 69.723 amu. So the formula would be <span>69.723amu=(%x)∗(68.926amu)+(1−%x)∗(70.025amu) and the answer to this is 1.58%</span>
