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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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A piece of unknown metal weighs 348g. When the metal piece absorbs 6.64kj of heat , its temperature increases from 24.4C to 43.6
Morgarella [4.7K]

Answer:

This metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

Explanation:

A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

H = \frac{E}{M*(T2-T1)}

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

H = \frac{6640 J}{0.348 Kg*(43.6-24.4) C} =\frac{6640 J}{6.686 KgC}

Finally, the specific heat of this metal is

H = 993 \frac{J}{kgC}

The aluminium could be the metal, its specific heat is similar to that found in this problem.

Finally,  we can conclude that this metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

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3 years ago
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