Answer:
Explanation:
Step 1. Calculate the pOH
pOH =-log[OH⁻]
pOH =-log(1.0 × 10⁻⁹)
pOH = 9.00
Step 2. Calculate the pH
pH + pOH = 14.00
pH + 9.00 = 14.00
Answer:
- <u>Yes, it is 14. g of compound X in 100 ml of solution.</u>
Explanation:
The relevant fact here is:
- the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.
That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.
Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.
With that, the solubility can be calculated from the followiing proportion:
- 84. g solute / 600 ml solution = y / 100 ml solution
⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.
The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.
<u>The answer is 14. g of solute per 100 ml of solution.</u>
According to Henderson–Hasselbalch Equation,
pH = pKa + log [Lactate] / [Lactic Acid]
As,
Ka of Lactic Acid = 1.38 × 10⁻⁴
pKa = -log Ka
pKa = -log 1.38 × 10⁻⁴
pKa = 3.86
So,
pH = 3.86 + log [0.10] / [0.13]
pH = 4.74 + log 0.769
pH = 4.74 - 0.11
pH = 4.63
Answer:
The correct answer will be "
".
Explanation:
We have:
Moles of solute (Na₂Co₃) = 
= 
= 
Now,
The volume of the solution will be:
⇒ 
∴ 
The formula to find the concentration will be:
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
102kg/hr * 1000g/hr * 1lb/453.6g * 24hr/day = 5404.0 lb/day
