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Elena L [17]
3 years ago
9

Look at the diagram below.

Chemistry
2 answers:
Ronch [10]3 years ago
8 0

Answer:

<h3>Ray 3 is the reflected ray</h3>

=>reflection of light may occur whenever light travels from a medium of a given refractive index into a medium with a different refractive index. In the most general case, a certain fraction of the light is reflected from the interface, and the remainder is refracted.

Kazeer [188]3 years ago
5 0

Answer:

3rd statment

Explanation:

ray 1 and 2 are same vertical line

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2 Calculate What is the density of a liquid with a mass of 17.4g<br> and a volume of 20 mL?
jolli1 [7]

Answer:

P= 0.87g/mL or 0.87g/cm^3

Explanation:

P=m/v

P=density

P=17.4g/20mL

P= 0.87g/mL

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8 0
4 years ago
A solution contains 0.021 M Cl and 0.017 M I. A solution containing copper (I) ions is added to selectively precipitate one of t
lidiya [134]

<u>Answer:</u> Copper (I) iodide will precipitate first.

<u>Explanation:</u>

We are given:

K_{sp} of CuCl = 1.0\times 10^{-6}

K_{sp} of CuI = 5.1\times 10^{-12}

Concentration of Cl^-\text{ ion}=0.021M

Concentration of I^-\text{ ion}=0.017M

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

  • <u>For CuCl:</u>

K_{sp}=[Cu^+][Cl^-]

Putting values in above equation, we get:

1.0\times 10^{-6}=[Cu^+]\times 0.021

[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M

Concentration of copper (I) ion = 4.76\times 10^{-5}M

  • <u>For CuI:</u>

K_{sp}=[Cu^+][I^-]

Putting values in above equation, we get:

5.1\times 10^{-12}=[Cu^+]\times 0.017

[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M

Concentration of copper (I) ion = 3.00\times 10^{-10}M

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.

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Which situation shows a nonrenewable resource in use?
miv72 [106K]

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B

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