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2 years ago

The image below shows a certain type of global wind:

Chemistry

1 answer:

forsale [732]2 years ago

6 0

Answer:

Number 3

It is the way the earth stay a good tempercher for us.

Explanation:

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Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th

lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

"Your question is not complete, it seems to be missing the diagram of the emission spectrum"

the diagram of the emission spectrum has been added.

From the given chart;

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

c is the speed of light = 3 x 10⁸ m/s
f is the frequency of the wave
λ is the wavelength

$f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz$

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0

2 years ago

Photosynthesis produces A. energy B. glucose. c. carbon dioxide. D. water.

marysya [2.9K]

The answer is glucose

4 0

2 years ago

Read 2 more answers

Balance the following reaction. A.) 4 B.) 2 C.) 5 D.) blank E.) 3

slega [8]

Answer:

C.5

Hope this helps

7 0

2 years ago

What changes chemical energy to something you can use

lutik1710 [3]

based on the law of conservation of energy its it atoms hold by strong chemical bonds

6 0

3 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$