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KATRIN_1 [288]
3 years ago
13

A light strikes the boundary of a medium at a 45 degree angle. Instead of entering the second medium, the light reflects back in

to the first medium. What can you conclude about the index of refraction of the second medium?
A it is much smaller than the index of refraction of the first medium

B It must equal to zero

C It is much larger than the index of refraction of the first medium

D It must be equal to 1.00
Chemistry
1 answer:
Dahasolnce [82]3 years ago
6 0

Answer:

D It must be equal to 1.00

Explanation:

The refraction in the second medium must be equal to 1. The refraction index is given as the ration of the angle of incidence to the angle of refraction. This is given as a fraction. In other words:

\frac{sin\theta _{i} }{sin\theta _{r} }  = n

where Θ₁ and Θ₂ are angles of incidence an refraction, and n is the refractive index.

At a critical angle, the refraction is equal to the reflection inside the medium. This results in a phenomenon called total internal reflection where light is reflected internally in the medium.

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So I am lost and I don't know how what any of those conversion factor things are and I'm wondering how to do #2,3, and 4. (I did
dalvyx [7]

Explanation:

#2.

A centigram is 1/100 of a gram, so that means a gram equals 100 centigrams.

Therefore you multiply 72.4 grams by 100/1 (or just 100), and get 7240 cg.

You did that one right but put the wrong unit in the answer. It is is cg ( centigrams).

#3.

1 liter is equal to 1000 milliliters, and I kiloliter is equal to 1000 liters. So one kiloliter is 1000*1000 milliliters or 1,000,000 milliliters.

The conversion factor would be

1/1000000

#4.

1 gigabyte is equal to 10^9 bytes.

I byte is equal to 10^9 bytes.

So 1 gigabyte is 10^9 * 10^9 nanobytes, or 10^18.

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6 0
3 years ago
At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249
oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= \frac{1249 g}{396.35 g/mol}

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

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Explanation:

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