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ss7ja [257]
2 years ago
15

What are the similarities between DIFFUSION, FACILITATED DIFFUSION and OSMOSIS?

Chemistry
2 answers:
sattari [20]2 years ago
8 0

Answer:

1)Both osmosis and diffusion equalises the concentration of two solutions.

2)Both are passive transport.

3)Do not require energy.

4)In both processes particle move or flow from the higher concentration to lower concentration until equilibrium achieved.

Thank you.

nalin [4]2 years ago
6 0
Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.
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Compare amplitudes, wavelengths, and frequencies of waves
mr_godi [17]

Answer:

The answer is option b.

Explanation:

Amplitude is the distance apart each wave is.

8 0
2 years ago
determine the frequency and wavelength (in nm) of the light emitted when the e- fell from n=4 and n=2
Lostsunrise [7]

Answer:

Frequency = 6.16 ×10¹⁴ Hz

λ = 4.87×10² nm

Explanation:

In case of hydrogen atom energy associated with nth state is,

En =  -13.6/n²

For n = 2

E₂ = -13.6 / 2²

E₂ = -13.6/4

E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

E₄ = -13.6 / 4²

E₄ = -13.6/16

E₄ = -0.85 ev

Kinetic energy of electron = -E₄ = 0.85 ev

Wavelength of radiation emitted:

E = hc/λ = E₄ - E₂

hc/λ = E₄ - E₂

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev   - (-3.4ev )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J

λ = 4.87×10⁻⁷ m

m to nm:

4.87×10⁻⁷ m ×10⁹nm/1 m

4.87×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /4.87×10⁻⁷ m

Frequency = 6.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 6.16 ×10¹⁴ Hz

3 0
3 years ago
What happens in a reduction half-reaction?
Marta_Voda [28]
The answer is elements gain electrons. Oxidation reduction is elements lose electrons. And oxygen is added/lost can be a type of oxidation/reduction reaction.
3 0
3 years ago
Read 2 more answers
Consider the following electron configurations to answer the questions that follow: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (ii
Ahat [919]

Option (i) would have the highest 2nd Ionization Energy.

Option (i) is Sodium.

Can be Written as 2, 8 , 1

For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.

Now After Removing that Electron...

Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.

Neon has a very stable Octet.

It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.

So

Option (i) [Sodium] has the highest 2nd Ionization Energy

6 0
2 years ago
How many phosphorous atoms are found in three molecules of magnesium phosphate, 3Mg3(PO4)2?
aleksandr82 [10.1K]

Answer: Option (d) is the correct answer.

Explanation:

Since the given formula is 3Mg_{3}(PO_{4})_{2}. According to cross method formula, magnesium has +2 charge so, PO_{4} is multiplied by 2.

Thus, 1 molecule of magnesium phosphate will contain 2 atoms of phosphorus.

Therefore, three molecules of magnesium phosphate contains following number of atoms.

  • Mg = 9
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  • O = 24

Hence, we can conclude that there are 6 atoms of phosphorus in three molecules of magnesium phosphate, 3Mg_{3}(PO_{4})_{2}.

4 0
3 years ago
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