Since HCl is a strong acid, it readily dissociates, so we can assume that there are 1.0x10^-3 M of H+ ions. We can use the formula:
pH=-log[H+]
pH=-log(1.0x10^-3)
pH=3
We can find the pOH by subtracting 14-pH:
14-3=11
pOH=11
The [OH-] can be found by using the formula:
pOH=-log[OH-]
11=-log[OH-]
10^-11=[OH-]
[OH-]=1.0x10^-10
Answer:any substance which may be harmful to the environment or hazardous to your health if inhaled, ingested or absorbed through the skin.
Explanation:
Answer:
None of these
Explanation:
Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.
Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :
Step -1 : Generation of stable carbocation.
Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.
Step-2: Attack of the ring to the carbocation
The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.
The product formed is shown in mechanism does not mention in any of the options.
So, None of these is the answer
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
Dozen = 12,
ii. 1 score = 20
iii. 1 ream = 500
iv. 1 gross = 1.44