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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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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Thermodynamics</u>

Specific Heat Formula: q = mcΔT

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  • <em>m</em> is mass (in g)
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  2. Parenthesis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>m</em> = 225 g

[Given] <em>c</em> = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] <em>q</em>

<u>Step 2: Solve for </u><em><u>q</u></em>

  1. Substitute in variables [Specific Heat Formula]:                                          q = (225 g)(4.184 J/g °C)(159.8 °C)
  2. Multiply:                                                                                                           q = (941.4 J/°C)(159.8 °C)
  3. Multiply:                                                                                                           q = 150436 J

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

150436 J ≈ 150000 J

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