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Amiraneli [1.4K]
3 years ago
11

Describe how you would prepare 500.0mL of an aqueous solution that is 30.0% isopropyl alcohol by volume

Chemistry
1 answer:
Artyom0805 [142]3 years ago
5 0

Answer:

The solution can be prepared by adding 150.0 mL of isopropyl alcohol to 350.0 mL of water to make 30.0% isopropyl alcohol aqueous solution.

Explanation:

∵ The volume% of a solution = [(volume of the solute(isopropy)/(volume of the solution)] x 100.

The volume % = 30.0%, volume of the solution = 500.0 mL.

∴ Volume of the solute = (the volume% of a solution)(volume of the solution)/100 = (30.0%)(500.0 mL)/100 = 150.0 mL.

∵ The volume of solution = volume of the solute + volume of water

∴ Volume of water = volume of the solution - volume of the solute = (500.0 mL) - (350.0 mL) = 150.0 mL.

  • <em>So, the solution can be prepared by adding 150.0 mL of isopropyl alcohol to 350.0 mL of water to make 30.0% isopropyl alcohol aqueous solution.</em>
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To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.

In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.

Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.

35.0° C - 25.0° C = 10.0° C

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q = 523 J/°C × 10.0° C

q = 5230 J

Therefore, it will take 5230 joules (J) to raise the temperature of the water.

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a . 4

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