Answer:
Hydrofluoric acid.
Explanation:
To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:
1. Acetic acid
Ka = 1.8x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 1.8x10^-5
pKa = 4.74
2. Benzoic acid
Ka = 6.5x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 6.5x10^-5
pKa = 4.18
3. Hydrofluoric acid.
Ka = 6.8x10^-4
pKa =..?
pKa = –logKa
pKa = –Log 6.8x10^-4
pKa = 3.17
4. Hypochlorous acid
Ka = 3.0x10^-8
pKa =..?
pKa = –logKa
pKa = –Log 3.0x10^-8
pKa = 7.52
Note: the smaller the pKa value, the stronger the acid.
The pka of the various acids as calculated above is given below:
Acid >>>>>>>>>>>>>>>>>> pKa
1. Acetic acid >>>>>>>>>> 4.74
2. Benzoic acid >>>>>>>> 4.18
3. Hydrofluoric acid >>>> 3.17
4. Hypochlorous acid >> 7.52
From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.
<u>Answer:</u> C) be hypertonic to Tank B.
<u>Explanation: </u>
<u>
The ability of an extracellular solution to move water in or out of a cell by osmosis</u> is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the <u>total concentration of all the solutes in the solution.
</u>
Three terms (hypothonic, isotonic and hypertonic) are used <u>to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane</u>. When we use these terms, we only take into account solutes that can not cross the membrane, which in this case are minerals.
- If the liquid in tank A has a lower osmolarity (<u>lower concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypotonic with respect to the latter.
- If the liquid in tank A has a greater osmolarity (<u>higher concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypertonic with respect to the latter.
- If the liquid in tank A has the same osmolarity (<u>equal concentration of solute</u>) as the liquid in tank B, the liquid in tank A would be isotonic with respect to the latter.
In the case of the problem, option A is impossible because the minerals can not cross the membrane, since it is permeable to water only. There is no way that the concentration of minerals decreases in tank A, so <u>the solution in this tank can not be hypotonic with respect to the one in Tank B. </u>
Equally, both solutions can not be isotonic and neither we can say that the solution in tank A has more minerals that the one in tank B because the liquid present in tank B is purified water that should not have minerals. Therefore, <u>options B and D are also not correct.</u>
Finally, the correct option is C, since in the purification procedure the water is extracted from the solution in tank A to obtain a greater quantity of purified water in tank B. In this way, the solution in Tank A would be hypertonic to Tank B.
Answer:
Mass of sea food = 30.98 Kg
Mass of sea food in pound = 68.31 lbs
Explanation:
Salmon, crab and oysters all are sea food.
Mass of sea food = Mass of salmon + Mass of crab + mass of oyster
Mass of salmon = 22 kg
Mass of crab = 5.5 kg
Mass of oysters = 3.48 kg
Mass of sea food = Mass of salmon + Mass of crab + mass of oyster
= 22 + 5.5 + 3.48
= 30.98 Kg
1 Kg = 2.205 lbs
Therefore, 30.98 kg = 30.98 × 2.205
= 68.31 lbs
Answer:
It's compound? It's chemical compound would be represented my letters or numbers
