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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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The hydrogen atom is not actually electronegative enough to form bonds to xenon. were the xenon-hydrogen bond to exist, what wou
Marat540 [252]

Answer : The structure of XeH_4 will be square-planar.

Explanation :

In the given molecule XeH_4, 'Xe' is the central atom and 'H' is the terminal atom.

Xenon has 8 valence electrons and hydrogen has 1 valence electron. Therefore, the total number of valence electrons are 8 + 4(1) = 12 electrons.

The number of electrons used in Xe-H bonding = 8 electrons

The remaining electrons which are used as lone pair on central atom (Xe) = 12 - 8 = 4 electrons

There are 4 bonding pairs and 2 lone pairs of electrons, they will be arranged in the octahedral arrangement around the central atom with 2 lone pairs of electrons on central atom. The lone pairs are arranged linearly across the central atom. The resulting structure will be square-planar.

The structure of XeH_4 is shown below.

5 0
3 years ago
Perform the following calculations and give your answer with the correct number of significant figures:
GarryVolchara [31]
I think this is the answer try it 

172.3995<span>
</span>
4 0
3 years ago
Read 2 more answers
what mass of carbon dioxide will be produced when 12.9 g of butane reacts with an excess of oxygen in the following reaction?
Nadya [2.5K]
39.1 gCO2
I think if that’s an option
3 0
3 years ago
__Pb(NO3)2 + __NaCI = __NaNO3 + __PbCI2
Elena L [17]

Answer:

Pb(NO3)2 + 2NaCl = 2NaNO3 + PbCl2

6 0
2 years ago
If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of
Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.1\ g}{186.04\ g/mol}

Moles\ of\ ferrocene= 0.0005375\ mol

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

\rho=\frac{Mass}{Volume}

or,  

Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.044\ g}{78.49\ g/mol}

Moles\ of\ acetyl\ chloride= 0.0005606\ mol

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms  0.0005375 mole of monoacetylferrocene

Moles of product formed =  0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

5 0
3 years ago
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