Answer:
2, strong acid
Explanation:
Data obtained from the question. This includes:
[H+] = 0.01 M
pH =?
pH of a solution can be obtained by using the following formula:
pH = –Log [H+]
pH = –Log 0.01
pH = 2
The pH of a solution ranging between 0 and 6 is declared to be an acid solution. The smaller the pH value, the stronger the acid.
Since the pH of the above solution is 2, it means the solution is a strong acid.
Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.
D = m / V
D = 2790 g / 205 mL
D = 13.60 g/mL
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
