Answer:
6.25 μg/mL
Explanation:
When a dilution is made, the mass of the solute is conserved (Lavoiser's law), so the mass pipetted will be the mass in the assay. The mass is the concentration (C) multiplied by the volume (V). If the pipet solution is called 1, and the assay 2:
m1 = m2
C1*V1 = C2*V2
C1 = 250 μg/mL
V1 = 25 μL
V2 = 975 μL + 25 μL = 1000 μL (is the final volume of the assay after the addition of LDH)
250*25 = C2*1000
C2 = 6.25 μg/mL
Mole ratio:
MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl
2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)
moles KOH = 4 x 2 / 1
= 8 moles of KOH
molar mass KOH = 56 g/mol
mass of KOH = n x mm
mass of KOH = 8 x 56
= 448 g of KOH
hope this helps!
Answer:
- <u>Alkaline or basic solution </u>(alkaline and basic means the same)
Explanation:
According to the <em>pH</em>, solutions may be classified as neutral, acidic, or alkaline (basic).
This table shows such classification:
pH classification
7 neutral
> 7 alkaline or basic
< 7 acidic
Thus, since the pH of the solution is 8.3, which is greater than 7, the solution is classified as basic (alkaline).
Additionally, you must learn that pH is a logarithmic scale for the concentration of hydronium ions in the solution.
You can calculate the concentration of hydronium ions using antilogarithm properties:
![pH=-log[H_3O^+]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-8.3}=0.00000000501](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-pH%7D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-8.3%7D%3D0.00000000501)
NaOH solutions are alkaline solutions, bases, according to Arrhenius model, because they contain OH⁻ ions and release them when ionize in water.