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Mariulka [41]
2 years ago
8

Use the PhET simulation to identify what happens to the concentrations of a dilute Drink mix mixture or a pure Drink mix solutio

n for the following scenarios. A pure Drink mix solution can be obtained by draining all the mixture, then adding Drink mix to the container as a Solution (radio button in the upper right area of the simulation). Note that the Concentration meter can be placed toward the bottom of the container or in the stream of the Drink mix. Drag the appropriate items to their respective bins.

Chemistry
1 answer:
elena-s [515]2 years ago
7 0

Hello user, you did not add a picture of an attachment to enable me help you solve this problem. Please check the attachment I added i believe it is the question that needs to be solved.

Explanation:

1. In the first column

<u>INCREASE CONCENTRATION</u>: these are the items that needs to be placed in this bin

a. Add drink mix solid to a diluted mixture of drink mix in pure water

b. Add drink mix solution to a diluted mixture of drink mix in pure water

c. Add drink mix solid to pure drink mix solution

d Evaporate water from a diluted mixture of drink mix in pure water

e Evaporate water from pure drink mix solution

2. The items below are what should be placed in this bin,<u>DECREASE CONCENTRATION:</u>

a. Add water to a diluted mixture of drink mix in pure water

b. Add water to pure drink mix solution

3. These items here should be placed here. <u>DOES NOT AFFECT CONCENTRATION:</u>

a. Add pure drink mix solution to pure drink mix solution

b. Drain the pure drink mix solution

c. Drain the diluted mixture

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Answer : The initial temperature of the limestone is 1.05\times 10^{2}^oC.

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Mass of limestone = 62.6 g

Mass of water = 75 g

Final temperature of limestone = 51.9^oC

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Initial temperature of water = 23.1^oC

The specific heat capacity of limestone = 0.921J/g^oC

The specific heat capacity of water = 4.186J/g^oC

The formula used :

q = m × c × ΔT

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q_{system}=-q_{surrounding}

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62.6g\times 0.921J/g^{0}C \times (51.9^{0}C-T_{\text{ initial limestone}})= -[75g\times 4.186J/g^{0}C \times (51.9^{0}C-23.1^oC)]

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T_{\text{ initial limestone}}=104.926^{0}C=1.05\times 10^{2}^oC

Therefore, the initial temperature of the limestone is 1.05\times 10^{2}^oC.




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