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Mekhanik [1.2K]
3 years ago
6

What do you call the movement of the earth's surface to the atmosphere?

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
8 0
The answer would be transpiration because that is when water gets evaporated by a plant into the atmosphere
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Which of the following most likely happens when the number of particles of a gas decreases?
Papessa [141]

Answer: C. The number of collisions of gas particles decreases

Explanation: As temperature decreases so does everything else same with increase.

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3 years ago
How many moles of oxygen are in 5.6 moles of al(oh)3?
Mashcka [7]
5.6 Al(OH)3
5.6 Al, 16.8 O, 16.8 H

16.8 mols of oxegyn in 5.6 mols of Al(OH)3
7 0
3 years ago
Aluminum foil is often incorrectly termed tin foil. If the density of tin is 7.28g/cm³,what is the thickness of a piece of tin f
leonid [27]

Answer:

15.0 µm

Step-by-step explanation:

Density = mass/volume

        D = m/V     Multiply each side by V

      DV = m        Divide each side by D

        V = m/D

Data:

m = 1.091 g

D = 7.28 g/cm³

 l = 10.0 cm

w = 10.0 cm

Calculation:

<em>(a) Volume of foil </em>

V = 1.091 g  × (1 cm³/7.28 g)

  = 0.1499 cm³

(b) <em>Thickness of foil </em>

The foil is a rectangular solid.

V  = lwh                            Divide each side by lw

h = V/(lw)

   = 0.1499/(10 × 10)

   = 1.50 × 10⁻³ cm           Convert to millimetres

   = 0.015 mm                  Convert to micrometres

   = 15.0 µm

The foil is 15.0 µm thick.

4 0
3 years ago
Please help me with letter a question
Colt1911 [192]
Rise, decrease, away from ocean, towards land
7 0
3 years ago
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 <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

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

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

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;

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

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;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

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 <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
2 years ago
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