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11111nata11111 [884]
2 years ago
5

Lexus.com/assessments/learnosity.aspx?idAssessment=1824894&idWebuser=4331214&idSection=1621362&idHtmllet=12178858&am

p;close=true&popup=true
Waves Unit Test
Waves Unit Test
The formula vv = f. describes the relationship between the speed of a wave (vw), and its frequency (f) and wavelength (A).
For example, if a wave has a frequency of 120 Hz and a wavelength of 5m, it would have a speed of 600 m/s.
In one or two sentences, describe what would happen to the frequency of this wave if the wavelength is increased to 10m but the
speed stays the same at 600 m/s.
points)
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Short Answer Rubric (2 points)

Chemistry
1 answer:
Kobotan [32]2 years ago
6 0

From the relationship between wavelength and frequency, the frequency of the wave is reduced by half to 60 Hz as the wavelength is increased twice to 10 m at constant speed.

<h3>What is the relationship between wavelength, velocity and frequency of a wave?</h3>

The velocity, frequency and wavelength of a wave are related by the formula given below:

  • v = fλ

where

  • v is velocity
  • f is frequency
  • λ is wavelength of the wave

From the above relationship:

If the wavelength of a wave is increased to 10m but the speed stays the same at 600 m/s, the frequency can be calculated as follows:

f = v/λ

f = 600/10

f = 60 Hz

Therefore, the frequency of the wave is reduced by half as the wavelength is increased twice at constant speed.

Learn more about frequency and wavelength of a wave at: brainly.com/question/4386945

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Naturally occurring zirconium exists are five isotopes. axe with a mass of 89.905 u(51.45%); Zr with a mass of 90.906 u (11.22%)
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8 0
3 years ago
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.

4 0
3 years ago
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