Answer:
5.17.
Explanation:
<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
</em>
[OH⁻] = 1.5 x 10⁻⁹ M.
∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(1.5 x 10⁻⁹ M) = 6.66 × 10⁻⁶ M.
∵ pH = - log[H₃O⁺]
<em>∴ pH = - log(6.66 × 10⁻⁶ M) = 5.17.
</em>
<span>A) Frequency due to motion
As you are moving toward the object that is making the sound the frequency is higher. When you begin to move away the frequency lowers.
</span>
<u>Answer:</u> The solubility of oxygen at 682 torr is 
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:

Or,

where,
are the initial concentration and partial pressure of oxygen gas
are the final concentration and partial pressure of oxygen gas
We are given:
Conversion factor used: 1 atm = 760 torr

Putting values in above equation, we get:

Hence, the solubility of oxygen gas at 628 torr is 
Um, I think it’s: k is potassium and F is fluorine so potassium Fluoride
I would say 2 because co2 goes out and o goes in