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Salsk061 [2.6K]
3 years ago
5

One of the emission lines that provides the green color in fireworks, which com s from barium salts, has a wavelength of 505nm.

a. Determine the frequency of this green line. b. Determine the energy associated with this green line. c. Arrange the visible spectrum in order of increasing wavelength, frequency, and energy
Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
6 0

Explanation:

(a)

The relation between frequency and wavelength is shown below as:

c=frequency\times Wavelength


c is the speed of light having value 3\times 10^8\ m/s

Given, Wavelength = 505 nm

Also, 1 m = 10^{-9} nm

So,  

Wavelength = 505\times 10^{-9} m

Thus, Frequency is:

Frequency=\frac{c}{Wavelength}

Frequency=\frac{3\times 10^8}{505\times 10^{-9}}\ Hz

Frequency=5.94\times 10^{14}\ Hz

(b)

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,

E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{505\times 10^{-9}}\ J

E=\frac{10^{-26}\times \:19.878}{10^{-9}\times \:505}\ J

E=3.94\times 10^{-19}\ J

(c)

Electromagnetic spectrum is range of the frequencies and their respective wavelengths of the various type of the electromagnetic radiation.

In order of the increasing frequency and the photon energy and the decreasing wavelength the spectrum are:  

Red , Orange, Yellow, Green, Blue, Indigo, Violet

Increasing wavelength is the opposite trend. Thus, The longest visible wavelength is red and the shortest is violet.

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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

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