D. Yep, D is the answer, alright.
Answer:
Polyhydroxyl alcohols
Explanation:
Whenever we have several C-OH bonds, we have a polyhydroxyl alcohol. For example, if we have just one alcohol group, that is, an R-OH group, then the naming is simple, say, we have EtOH, it's ethanol.
The problem becomes more complicated when we have several hydroxyl groups present in the alcohol. Let's say we have an ethane molecule and we replace the hydrogen atoms of carbon 1 and 2 with hydroxyl groups. In that case, we have 1,2-ethanediol. Similarly, we can have triols etc.
That said, we have poly (several) hydroxyl groups and we can generalize this to having polyhydroxyl alcohols.
Answer: amu
Explanation: it stands for atomic mass unit and it is the measurement that’s used to measure the mass of atoms
The greatest molar amount of metal product is obtained from silver (1) nitrate.
Let us determine the mass of metal obtained in each case.
For La^3+;
La^3+ + 3e ----> La
1 mole of La is deposited by 3F
x moles of La is deposited by 10 F
x = 1 × 10/3
x = 3.33 moles
For Zn^2+;
1 mole Zn^2+ is deposited by 2F
x moles of Zn^2+ is deposited by 10 F
x = 5 F
For Ag^+
1 mole of Ag^+ is deposited by 1 F
x moles of Ag^+ is deposited by 10 F
x = 10 moles
For Ba^2+;
1 mole of Ba^2+ is deposited by 2 F
x moles of Ba^2+ is deposited by 10 F
x = 5 moles
Learn more: brainly.com/question/967776
Answer: Dissociation constant of the acid is
.
Explanation: Assuming the acid to be monoprotic, the reaction follows:

pH of the solution = 6
and we know that
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
![[H^+]=antilog(-pH)](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dantilog%28-pH%29)
![[H^+]=antilog(-6)=10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dantilog%28-6%29%3D10%5E%7B-6%7DM)
As HA ionizes into its ions in 1 : 1 ratio, hence
![[H^+]=[A^-]=10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BA%5E-%5D%3D10%5E%7B-6%7DM)
As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:
![[HA]=[HA]-[H^+]](https://tex.z-dn.net/?f=%5BHA%5D%3D%5BHA%5D-%5BH%5E%2B%5D)
![[HA]=0.1M-10^{-6}M](https://tex.z-dn.net/?f=%5BHA%5D%3D0.1M-10%5E%7B-6%7DM)
![[HA]=0.09999M](https://tex.z-dn.net/?f=%5BHA%5D%3D0.09999M)
Dissociation Constant of acid,
is given as:
![K_a=\frac{[A^-][H^+]}{HA}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7BHA%7D)
Putting values of
in the above equation, we get


Rounding it of to one significant figure, we get
