The water cycle regardless if it is in a lake, our bodies, food, or underground.
When PH + POH = 14
∴ POH = 14 -7 = 7
when POH = -㏒[OH-]
7 = -㏒ [OH-]
∴[OH-] = 10^-7
by using ICE table:
Mn(OH)2(s) ⇄ Mn2+ (aq) + 2OH-(aq)
initial 0 10^-7
change +X +2X
Equ X (10^-7 + 2X)
when Ksp = [Mn2+][OH-]^2
when Ksp of Mn(OH)2 = 4.6 x 10^-14
by substitution:
4.6 x 10^-14 = X*(10^-7+2X)^2 by solving this equation for X
∴ X =2.3 x 10-5 M
∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol
= 0.002 g/ L
For this question you need to find out all of the stages of a star and the process of a nuclear fusion. Then compare what's different about it and what's the same.
Answer:
Option D. KBr < KCl < NaCl
Explanation:
We'll begin by calculating the number of mole of each sample.
This can be obtained as follow:
For NaCl:
Mass = 1 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mole of NaCl =?
Mole = mass /Molar mass
Mole of NaCl = 1/58.5
Mole of NaCl = 0.0171 mole
For Kbr:
Mass = 1 g
Molar mass of KBr = 39 + 80 = 119 g/mol
Mole of KBr =?
Mole = mass /Molar mass
Mole of KBr = 1/119
Mole of KBr = 0.0084 mole
For KCl:
Mass = 1 g
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mole of KCl =?
Mole = mass /Molar mass
Mole of KCl = 1/74.5
Mole of KCl = 0.0134 mole
Summary
Sample >>>>>>>> Number of mole
NaCl >>>>>>>>>> 0.0171
KBr >>>>>>>>>>> 0.0084
KCl >>>>>>>>>>> 0.0134
Arranging the number of mole of the sampl in increasing order, we have:
KBr < KCl < NaCl
