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Fofino [41]
3 years ago
14

How does evaporation show interactions among the hydrosphere and atmosphere?

Chemistry
2 answers:
Vladimir79 [104]3 years ago
6 0

Answer:

water evaporates from the ocean into atmosphere. water vapour condenses to form clouds. clouds produce rain. rainwater needed for plant growth.

Explanation:

blsea [12.9K]3 years ago
5 0
Water evaporates from the ocean into atmosphere. water vapour condenses to form clouds. clouds produce rain. rainwater needed for plant growth.
You might be interested in
What does mass change to and what does it do
Nataly [62]

Answer:

Mass is the amount of matter in an object and does not change with location.

Explanation:

4 0
3 years ago
NEED HELP ASAP WITH THESE QUESTIONS GIVING FAIR AMOUNT OF POINTS IF HELPED WITH ALL QUESTIONS Violet light has a wavelength of 4
scoray [572]

Answer:

a) 7.14e19 Hz

b) 2.298e-27 J

c) 2.793e-19 J; 7.117e9 nm

d) 7.5e14 Hz; 4.96e-19 J

e) 6.2947e-18 J; 31.6 nm

f) 2.21e-22 J

g) 7.1e-19 J; 1.1e15 Hz

h) 3.422e-19 J; 581 nm

i) 4.2e14 Hz

j) 1.92e8 m

k) 7.14e16 Hz; Ultraviolet

Explanation:

Frequency: ν       Wavelength: λ       Energy: E       Speed of light: C (3.00e8)       Planck's Constant: h (6.626e-34)

ν -> λ    λ = C/ν

λ -> ν    ν = C/λ

For either of these equations, wavelength must be converted to meters or nanometers, depending on the equation.

For ν -> λ, after doing the equation, convert the wavelength into nanometers by dividing by 1e-9.

For converting λ -> ν, convert the wavelength into meters by multiplying by 1e-9.

For energy: E = hν = hc/λ

Now that the setup is out of the way:

a) Violet light has a wavelength of 4.20 x 10-12 m. What is the frequency?

λ -> ν    ν = C/λ

\frac{3.00e8}{4.20e-12} = 7.14e19 Hz

b) A photon has a frequency (n) of 3.468 x 106 Hz. Calculate its energy

E = hν = hc/λ

(6.626e-34) (3.468e6) = 2.298e-27 J

c) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 4.215 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (4.215e14) = 2.793e-19 J

ν -> λ    λ = C/ν

\frac{3.00e8}{4.215e14} = 7.117 m

\frac{7.117m}{1}*\frac{1nm}{1e-9m} = 7.117e9 nm

d) Calculate the frequency and the energy of blue light that has a wavelength of 400 nm  (h = 6.62 x 10-34 J-s).

λ -> ν    ν = C/λ

\frac{400 nm}{1} *\frac{1e-9m}{1nm} = 4e-7 m

\frac{3.00e8}{4e-7} = 7.5e14 Hz

E = hν = hc/λ

(6.626e-34) (7.5e14) = 4.96e-19 J

e) Calculate the wavelength and energy of light that has a frequency of 9.5 x 1015 Hz.

ν -> λ    λ = C/ν

\frac{3.00e8}{9.5e15} = 3.16e-8 m

\frac{3.16e-8m}{1}*\frac{1nm}{1e-9m} = 31.6 nm

E = hν = hc/λ

(6.626e-34) (9.5e15) = 6.2947e-18 J

f) A photon of light has a wavelength of 0.090 cm. Calculate its energy.

E = hν = hc/λ

Convert the wavelength from cm to meters:

\frac{0.090cm}{1} *\frac{1m}{100cm} = 9e-4 m

\frac{(6.626e-34)(3.00e8)}{9e-4} = 2.21e-22 J

g) Calculate the energy and frequency of red light having a wavelength of 2.80 x 10-5 cm.

E = hν = hc/λ

Convert the wavelength from cm to meters:

\frac{2.80e-5cm}{1} *\frac{1m}{100cm} = 2.8e-7 m

\frac{(6.626e-34)(3.00e8)}{2.8e-7} = 7.1e-19 J

λ -> ν    ν = C/λ

Convert the wavelength from cm to meters:

\frac{2.80e-5cm}{1} *\frac{1m}{100cm} = 2.8e-7 m

\frac{3.00e8}{2.8e-7} = 1.1e15 Hz

h) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 5.165 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (5.165e14) = 3.422e-19 J

ν -> λ    λ = C/ν

\frac{3.00e8}{5.165e14} = 5.81e-7 m

\frac{5.81e-7m}{1}*\frac{1nm}{1e-9m} = 581 nm

i) The wavelength of green light from a traffic signal is centered at 7.20 x 10-5 cm. Calculate the frequency.

λ -> ν    ν = C/λ

Convert the wavelength from cm to meters:

\frac{7.20e-5 cm}{1} *\frac{1m}{100cm} = 7.2e-7 m

\frac{3.00e8}{7.2e-7} = 4.2e14 Hz

j) If it takes 1.56 seconds for radio waves (which travel at the speed of light) to reach the moon from Earth, how far away is the moon?

  All we want to do here is to convert frequency (speed) to wavelength (distance). This problem requires a bit of thought, but it isn't bad once you realize that frquency is speed and wavelength is distance. It becomes just like the other problems after that. Also, I'll leave this distance in meters, but I think you can figure out how to convert it if it wants it in another unit.

  One second is equal to 1 Hertz, so our frequency is 1.56 Hz.

ν -> λ    λ = C/ν

\frac{3.00e8}{1.56} = 1.92e8 m

  The actual distance from the earth to the moon via google is 3.84e7, but sometimes problems like this will mess with the numbers to make sure that you didn't just look up the answer. I'm still pretty sure that this is right, however.

k) Calculate the frequency of light that has a wavelength of 4.20 x 10-9m. Identify the type of electromagnetic radiation.

First, we convert wavelength to frequency, as normal:

λ -> ν    ν = C/λ

\frac{3.00e8}{4.20e-9} = 7.14e16 Hz

Then we identify the electromagnetic wave type. You can look up a conversion chart for these on google, but since our frequency is in the e15 - e17 range, this light is considered ultraviolet.

5 0
4 years ago
Writing a pressure equilibrium constant expression Ammonia and oxygen react to form nitrogen monoxide and water, like this: 4 NH
Sliva [168]

<u>Answer:</u> The equilibrium constant expression for the given equation is written below.

<u>Explanation:</u>

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{p}

For the general chemical equation:

aA+bB\rightleftharpoons cC+dD

The expression of K_p follows:

K_p=\frac{p_C^c\times p_D^d}{p_A^a\times p_B^b}

For the given chemical equation:

4NH_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_2O(g)

The expression of K_p for above equation follows:

K_p=\frac{(p_{H_2O}^5)\times (p_{NO}^4)}{(p_{NH_3}^4)\times (p_{O_2}^5)}

Hence, the equilibrium constant expression for the given equation is written above.

5 0
3 years ago
How are reactivity and electronegativity related? 1 point)
Svetach [21]

Elements which have high electron negativity are more reactive than elements with low electron negativity.

This is because elements with high electron negativity will gain more electrons to become more stable.

When electron negativity of an element decrease it becomes more into its element form.

There will be more ionization energy of the elements who have more electron negativity.

Learn more at brainly.com/question/24372525

7 0
3 years ago
In order to complete his research project, Roger needs to make a mixture of 86 mL of a 36% acid solution from a 39% acid
Zina [86]

The volume of the 39% acid solution that Roger needs to use to make a mixture of 86 mL of a 36% acid solution is 74.27 mL.

The mixture of the acids solutions is given by:

CV = C_{1}V_{1} + C_{2}V_{2}    (1)      

Where:

C: is the concentration if the mixture = 36%  

V: is the total volume of the mixture = 86 mL

C₁: is the concentration of acid 1 = 39%

V₁: is the volume if acid 1 =?

C₂: is the concentration of acid 2 = 17%

V₂: is the volume of acid 2

The sum of V₁ and V₂ must be equal to V, so:

V = V_{1} + V_{2}

V_{2} = V - V_{1}  (2)

By entering equation (2) into (1), we have:

CV = C_{1}V_{1} + C_{2}(V - V_{1})

36\%*86 mL = 39\%*V_{1} + 17\%(86 mL - V_{1})

Changing the percent values to decimal ones:

0.36*86 mL = 0.39*V_{1} + 0.17(86 mL - V_{1})  

Now, by solving the above equation for V₁:

V_{1} = 74.27 mL  

Therefore, the volume of the 39% acid solution is 74.27 mL.

       

To learn more about mixture and solutions, go here: brainly.com/question/6358654?referrer=searchResults

I hope it helps you!  

         

7 0
3 years ago
Read 2 more answers
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