Base+salt > acid +alkali > neutralization i think this is the reaction
Answer:
Explanation:
By definition, <em>half neutralization</em> is the point at which half of the acid has been neutralized.
The neutralization reaction that you are studying is the acid-base reaction:
- HCl (aq) + NaOH (aq) → NaCl(aq) + H₂O (aq)
Then, since the starting molarity of the acid (HCl) is 0.2 M, you just need to find half of that concentration:
- Half molarity = M / 2 = 0.2 M / 2 = 0.1 M
So, the answer is the first choice: a. 0.1 M.
Answer:
Water serves to suspend the red blood cells to carry oxygen to the cells. It is the solvent for the electrolytes and nutrients needed by the cells, and also the solvent to carry waste material away from the cells. With water as the solvent, osmotic pressure acts to transport the needed water into cells.
Explanation:
Answer:
Electrons in a hydrogen atom must be in one of the allowed energy levels. If an electron is in the first energy level, it must have exactly -13.6 eV of energy.
...
Energy Levels of Electrons.
Energy Level Energy
1 -13.6 eV
2 -3.4 eV
3 -1.51 eV
4 -.85 eV
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 ∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
= 2x10^-8 Mc) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M