Answer:
1.5 × 10⁻¹¹ M
Explanation:
Step 1: Given data
- Concentration of OH⁻ ([OH⁻]): 6.6 × 10⁻⁴ M
- Concentration of H⁺ ([H⁺]): ?
Step 2: Consider the self-ionization of water
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
Step 3: Calculate the molar concentration of H⁺
We will use the equilibrium constant for the self-ionization of water (Kw).
Kw = 1.0 × 10⁻¹⁴ = [H⁺] × [OH⁻]
[H⁺] = 1.0 × 10⁻¹⁴ / [OH⁻]
[H⁺] = 1.0 × 10⁻¹⁴ / 6.6 × 10⁻⁴
[H⁺] = 1.5 × 10⁻¹¹ M
Answer:
[OH-] = 6.17 *10^-10
Explanation:
Step 1: Data given
pOH = 9.21
Step 2: Calculate [OH-]
pOH = -log [OH-] = 9.21
[OH-] = 10^-9.21
[OH-] = 6.17 *10^-10
Step 3: Check if it's correct
pOH + pH = 14
[H+]*[OH-] = 10^-14
pH = 14 - 9.21 = 4.79
[H+] = 10^-4.79
[H+] = 1.62 *10^-5
6.17 * 10^-10 * 1.62 * 10^-5 = 1* 10^-14
Avogadro's Law is the equation of 6.022 x 10^23 so I hope this helps with the question you are trying to ask
Answer: It's the anode broski (B)
Explanation: I'm taking the Chem summer course too broski, this was the correct answer. Cheers broski
b is fastest, having the lowest activation energy (35 kJ) and is an exothermic reaction, releasing energy in the form of heat
c is slowest, having the highest activation energy (55 kJ) and is an endothermic reaction, taking in energy from its surroundings
