They're based on hydrogen.
The answer you need is Volume.
Every mole of MgCl2 reacts with 2 moles of KOH, therefore the 4 moles of KOH will only react with 2 moles of MgCl2, making it the limiting reagent and therefore KOH determines how much Mg(OH)2 is produced.
3.0 × 10¹¹ RBC's (or) 3E11 RBC's
Solution:
Step 1: Convert mm³ into L;
As,
1 mm³ = 1.0 × 10⁻⁶ Liters
So,
0.1 mm³ = X Liters
Solving for X,
X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³
X = 1.0 × 10⁻⁷ Liters
Step 2: Calculate No. of RBC's in 5 Liter Blood:
As given
1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's
So,
5.0 Liters of Blood will contain = X RBC's
Solving for X,
X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters
X = 3.0 × 10¹¹ RBC's
Or,
X = 3E11 RBC's
Answer:
11.9 is the pOH of a 0.150 M solution of potassium nitrite.
Explanation:
Solution : Given,
Concentration (c) = 0.150 M
Acid dissociation constant = 
The equilibrium reaction for dissociation of
(weak acid) is,

initially conc. c 0 0
At eqm.

First we have to calculate the concentration of value of dissociation constant
.
Formula used :

Now put all the given values in this formula ,we get the value of dissociation constant
.



By solving the terms, we get

No we have to calculate the concentration of hydronium ion or hydrogen ion.
![[H^+]=c\alpha=0.150\times 0.0533=0.007995 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%3D0.150%5Ctimes%200.0533%3D0.007995%20M)
Now we have to calculate the pH.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


pH + pOH = 14
pOH =14 -2.1 = 11.9
Therefore, the pOH of the solution is 11.9