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:
Solving by the method of exponential growth.
bacteria = 2
after one hr = 2² = 4
after 2nd hr = 2³ = 8
after 3rd hr = 2⁴ = 16
after 4th hr = 2⁵ = 32
Answer:
Explanation:
Hello there!
In this case, given the T-V variation, we understand it is possible to apply the Charles' law as shown below:
Thus, since we are interested in the initial temperature, we can solve for T1, plug in the volumes and use T2 in kelvins:
Best regards!
Assuming that the solution is simply an aqueous solution
so that it is purely made of NaClO4 (the solute) and water (the solvent), then
I believe the dissolved species would only be the ions of NaClO4, these are:
Na+
ClO4 -
The Empirical Formula is
<span>a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.</span>
