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4vir4ik [10]
3 years ago
11

what is the period of oscillation of an electron that is bouncing up and down in response to the passage of a packet of red lgih

t?
Chemistry
1 answer:
Talja [164]3 years ago
7 0

Answer:

2.33*10^-15 s

Explanation:

We need to estimate the period of oscillation of an electron that bounced up/down due to the effects of a passing packet of red light. The period of oscillation is the wavelength of the red light divided by the speed of the red light. We know that:

V = f*λ

where V = speed of the red light = 3*10^8 m/s

f = frequency = 1/T where T is the period of oscillation in seconds (s)

λ = the wavelength of the red light = 0.7*10^-6 m

Thus:

V = λ/T

T = λ/V = (0.7*10^-6)/(3*10^8) = 2.33*10^-15 s

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HELP PLEASE
zmey [24]

The speed of sound depends upon

a) air density

b) temperature

c) pressure

d) humidity

so there can be different reasons that the instrument goes out of tune outside.

However in the given options the best is The speed of sound decreases with a decrease in temperature.

4 0
3 years ago
A rate of change is it velocity or acceleration .​
Lorico [155]

Answer:

Acceleration is the correct answer. Hope this helps!

Explanation:

3 0
3 years ago
When the pH of a solution is changed from 4 to 3,
Ivenika [448]
The correct answer is option 2. When the pH of a solution is changed from 4 to 3, the hydronium ions increases by a factor of 10.  pH is a scale used to measure the acidity or alkalinity of an aqueous solution. It can be used to calculate the amount of hydronium ions in the solution. It is expressed as: pH = -log [H+].

 

7 0
3 years ago
The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?
n200080 [17]

<u>Answer:</u> The moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of atmosphere = 5.00\times 10^{18}kg=5.00\times 10^{21}g

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = \frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol

Moles of carbon dioxide in air = \frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol

Hence, the moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

6 0
3 years ago
An alternative definition of electronegativity is Electronegativity= constant (I.E — E.A.) where I.E. is the ionization energy a
Kazeer [188]

Answer:

The electronengativity values of given elements is as follows.

Fluorine - 4

Chlorine -3

Bromine - 2.9

Iodine- 2.5

Explanation:

Electronegativity  =consant (I.E-E.A)

The electron affinity and ionization energy values of the given elements is as follows.

(In attachment)

First we have to find the value of constant by using the fluorine atom to whom the electronengativity taken as "4".

<u>Fluorine:</u>

4=constant[1678-(-327.8)]

Constant=0.0019942168

By using this constant values we can find electronegatvity values of remaining elements.

<u>Chlorine:</u>

Electronegativity=0.0019942168[1255+348.7]=3.1980\sim 3

Therefore, electronegativity of chlorine is 3.

<u>Bromine:</u>

Electronegativity=0.0019942168[1138+324.5]=2.91\sim 2.9

Therefore, electronegativity of bromine is 2.9.

<u>Iodine:</u>

Electronegativity=0.0019942168[1007+295.7]=2.59\sim 2.5

Therefore, electronegativity of iodine is 2.5.

8 0
3 years ago
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