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cluponka [151]
2 years ago
9

If 2.0 mL of 6.0M HCl is used to make a 500.0-mL aqueous solution, what is the molarity?

Chemistry
2 answers:
RideAnS [48]2 years ago
5 0

Answer:

0.024M

Explanation:

Data obtained from the question include:

C1 = 6M

V1 = 2mL

C2 =?

V2 = 500mL

The molarity of the diluted solution can be obtained as follows:

C1V1 = C2V2

6 x 2 = C2 x 500

Divide both side by 500

C2 = (6 x 2) /500

C2 = 0.024M

The molarity of the diluted solution is 0.024M

pentagon [3]2 years ago
4 0

Answer:

The molarity is 0.024 mol / L

Explanation:

Molarity = Number of moles / Volume of solution

= (6.0 mol/L x 2.0 x 10^{-3} L) / 0.500 L

= 0.024 mol / L

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grigory [225]

<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.

  1. 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.
  2. 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.
  3. 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.

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U-238 + ₁⁰ n ⇒ U-239 ⇒ Np -239 +  ₋₁⁰β ⇒ Pu-239 +  ₋₁⁰β

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