Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,

pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
Answer:
1. 374.69 K
Explanation:
Hello,
In this case, pure water's boiling point is 373.15 K, thus by considering the boiling point increase equation:

Whereas i=2 since two ionic species are formed,actually, the experimental value is 2.42 so better work with it, thus:

Thus, the required boiling point is:

Regards.
Answer:
0%
Explanation:
- The probability that both of the parents are homozygous recessive is zero.
- In our case, The gene for tallness is dominant over the gene for dwarfism
- Therefore; TT is homozygous dominant while tt is homozygous recessive and Tt is heterozygous.
- Heterozygous genotype (Tt) will exhibit a tall phenotype.
- When two parents are crossed and all the offsprings are tall then the possible genotype of the parents is either;
- Homozygous dominant for both parents or
TT x TT
2. One parent is homozygous dominant and the other is homozygous recessive.
TT x Tt
- Therefore, the probability of having both parents as homozygous recessive when all the offspring are tall is Zero.