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2 years ago

Which quantity is equivalent to the product of the absolute index of refraction of water and the speed of light in water?

1 answer:

3 0

The quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water system is the speed of light in vacuum.

<h3>What is the speed of light?</h3>

Speed of light is the rate of speed though the light travels. To find the speed of light in any medium, the following formula is used.

$v=\dfrac{c}{n}$

Here, (n) is the index of reaction and (c) is the speed of light in the vacuum. The speed of light in the vacuum is almost equal to the 3.0×10⁸ m/s.

Now the quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water has to be find out.

The above formula can be written as,

$v=\dfrac{c}{n}\\vn=c\\c=vn$

Here, the product of index of refraction and speed of light is equal to the speed of light in vacuum. This will be true for water as well.

Thus, the quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water system is the speed of light in vacuum.

Learn more about the speed of light here;

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At standard temperature and pressure. 0.500 mole of xenon gas occupies

ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Mole ratio
Dimensional Analysis

Explanation:

Step 1: Define

Identify.

0.500 mole Xe (g)

Step 2: Convert

[DA] Set up: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)$
[DA] Evaluate: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}$

Topic: AP Chemistry

Unit: Stoichiometry

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2 years ago

If the solubility of a gas in water at 25°C is 22.25 mg/L at 2.5 atm of pressure, what is its solubility (in mg/L) at 25°C

tamaranim1 [39]

Answer:

8.9 mg/l

Explanation: Temp doesnt matter so throw that out automatically then your equation is;

S1/P1=S2/P2

We are looking for S2 and that equation is;

S2=S1*P2/P1 and that is S2=22.25*1/2.5

A little bit of simple math and you get your answer: 8.9 mg/l

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Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r

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Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)

1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑ H bonds broken - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .

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