What information is given in a chemical formula?

Americans spend up to $100 billion annually for bottled water (41 billion gallons). The only beverages with higher sales are car

grigory [225]

Answer: C) be hypertonic to Tank B.

Explanation:

The ability of an extracellular solution to move water in or out of a cell by osmosis is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the total concentration of all the solutes in the solution.

Three terms (hypothonic, isotonic and hypertonic) are used to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane. 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 (lower concentration of solute) 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 (higher concentration of solute) 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 (equal concentration of solute) 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 the solution in this tank can not be hypotonic with respect to the one in Tank B.

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, options B and D are also not correct.

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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3 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut

Irina-Kira [14]

Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

It is a stichiometry problem.

We should write the balance equation of the mentioned chemical reaction:

2Al + 3CuCl₂ → 3Cu + 2AlCl₃.

It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.

Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

n = mass / molar mass

The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.

The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.

From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.

∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with (2:3) ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.