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4vir4ik [10]
3 years ago
11

Yasmin's teacher asks her to make a supersaturated saline solution. Her teacher tells her that the solubility of the salt is 360

g/L at room temperature (25 °C).
How can Yasmin make a supersaturated saline solution?


She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature (25 °C).

She can add 360 g of salt to 1 L of room temperature water (25 °C) and stir the solution until all the salt dissolves.

She can add 380 g of salt to 1 L of cold water (5 °C) and stir the solution until most of the salt dissolves. Then, she can carefully heat the solution to room temperature (25 °C).

She can add 380 g of salt to 1 L of room temperature water (25 °C) and let the solution sit for 24 hours, so the salt dissolves.
Chemistry
2 answers:
KatRina [158]3 years ago
6 0

B just took test and got the correct answer.

Aleks [24]3 years ago
3 0

Answer:

She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.  

Explanation:

A supersaturated solution contains more salt than it can normally hold at a given temperature.

A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.

Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.

B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.

C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.

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Answer:

The volume of cupboard is 2.0043 m³.

Explanation:

Given data:

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4 0
3 years ago
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Answer:

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Explanation:

Hello,

The suitable differential equation for this case is:

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As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

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Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

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