Answer:
HCN
Explanation:
Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.
When an acid donates a proton, it changes into a base which is known as its conjugate base.
Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.
The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.
Also, the strongest acid leads to the weakest conjugate base and vice versa.
Thus, Out of HI, HCN, 
, 
and HCl , the weakest acid is:- HCN
<u>Thus, HCN corresponds to the strongest conjugate base. </u>
A) They are different states of matter.
Answer:
[OH-] for this solution is 4.255*10^-12
Explanation:
We are given
[H+] = 2.35 × 10-3 M
we need to find the concentration of [OH-]
we know from Equilibrium
[H+][OH-] = 10^-14
[OH-] = 10^14/2.35*10^10^-3
[OH-] = 0.4255*10^-11
[OH] = 4.255*10^-12
Therefore the Concentration of [OH-] for this solution is 4.255*10^-12
The key results explain how you'll achieve a certain goal. They have to be quantitative, but always measurable in terms of their progress. This has to be considered when it comes to key results, because only results count for each milestone on the way to achieving the goal in question.
You have the stoichiometric equation. This tells you unequivocally that an
18
⋅
g
mass of water, 1 mole, reacts with a
56.07
⋅
g
mass of quicklime to form a
74.09
⋅
g
mass of slaked lime.
If you don't from where I am getting these numbers, you should know, and someone will be willing to elaborate.
Here, you have formed
6.21
⋅
m
o
l
of quicklime which requires stoichiometric lime AND water. And thus you need a mass of
6.21
⋅
m
o
l
×
18.01
⋅
g
⋅
m
o
l
−
1
water
≅
88
⋅
g
.
In practice, of course I would not weigh out this mass. I would just pour
100
−
200
⋅
m
L
of water into the lime.
