<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:
d. The gold(III) ion is most easily reduced.
Explanation:
The standard reduction potentials are
Au³⁺ + 3e⁻ ⟶ Au; 1.50 V
Hg²⁺ + 2e⁻ ⟶ Hg; 0.85 V
Zn²⁺ + 2e⁻ ⟶ Zn; -0.76 V
Na⁺ + e⁻ ⟶ Na; -2.71 V
A <em>more positive voltage</em> means that there is a <em>stronger driving force</em> for the reaction.
Thus, Au³⁺ is the best acceptor of electrons.
Reduction Is Gain of electrons and, Au³⁺ is gaining electrons, so
Au³⁺ is most easily reduced.
Answer:
The correct answer is: pH= 4.70
Explanation:
We use the <em>Henderson-Hasselbach equation</em> in order to calculate the pH of a buffer solution:
![pH= pKa + log \frac{ [conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%3D%20pKa%20%2B%20log%20%20%20%5Cfrac%7B%20%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
Given:
pKa= 4.90
[conjugate base]= 4.75 mol
[acid]= 7.50 mol
We calculate pH as follows:
pH = 4.90 + log (4.75 mol/7.50 mol) = 4.90 + (-0.20) = 4.70
The answer is 2
Hope this helped ??
The concentration of the hydroxide ion given that the concentration of the hydronium ion 1.0 times 10^-14 M. The reverse mathematical method used to determine the pOH can be used to get the hydroxide ion concentration from the pOH. How many hydroxide ions are there in a solution with a pOH of 5.70, for instance.
Calculate 10-5.70, or "inverse" log, on a calculator (- 5.70). It indicates that one hydroxide ion is produced by one part of the NaOH solution. Because of this, the molar concentration of hydroxide ions in the solution is the same as the molar concentration of the NaOH.
To learn more about hydroxide, click here.
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