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vekshin1
3 years ago
10

What can you conclude about red visible light waves compared to blue visible light waves?

Chemistry
2 answers:
lawyer [7]3 years ago
7 0

Explanation:

On comparing blue visible light with red visible light, there is difference in the wavelength, frequency, and energy. The difference between both red and blue visible lights are as follows.

Blue visible light:

  • It has low wavelength.
  • High frequency.
  • More energy.              

Red visible light:

  • It has high wavelength.
  • Low frequency.
  • Less energy.
motikmotik3 years ago
3 0

Answer:

The red light and the blue light are the two visible lights from the light spectrum. The wavelength of red light ranges from 635 to 700 nm. It has lower frequency ranging from 430 to 480 THz (Trillion hertz).

On the other hand, the wavelength of blue light ranges from 490 to 450 nm. It has higher frequency ranging from 610 to 670 THz.

Thus, it can be concluded that blue light has higher frequency and higher energy but lower wavelength in comparison to the red light.

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If a hydrogen atom and a helium atom have the same kinetic energy:________
Nuetrik [128]

Answer: If a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.

Explanation:

The relation between energy and wavelength is as follows.

E = \frac{hc}{\lambda}\\

This means that energy is inversely proportional to wavelength.

As it is given that energy of a hydrogen atom and a helium atom is same.

Let us assume that E_{hydrogen} = E_{helium} = E'. Hence, relation between their wavelengths will be calculated as follows.

E_{hydrogen} = \frac{hc}{\lambda_{hydrogen}}    ... (1)

E_{helium} = \frac{hc}{\lambda_{helium}}         ... (2)

Equating the equations (1) and (2) as follows.

E_{hydrogen} = E_{helium} = E'\\\frac{hc}{\lambda_{hydrogen}} = \frac{hc}{\lambda_{helium}} = E'\\\lambda_{helium} = \lambda_{hydrogen} = E'

Thus, we can conclude that if a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.

7 0
3 years ago
How many bromine atoms are present in 39.0 g of ch2br2?
Mandarinka [93]
Answer is: there is  2,69·10²³ atoms of bromine.
m(CH₂Br₂) = 39,0 g.
n(CH₂Br₂) = m(CH₂Br₂) ÷ M(CH₂Br₂).
n(CH₂Br₂) = 39 g ÷ 173,83 g/mol.
n(CH₂Br₂) = 0,224 mol.
In one molecule of CH₂Br₂, there is two bromine atoms, so:
n(CH₂Br₂) : n(Br) = 1 : 2.
n(Br) = 0,448 mol.
N(Br) = n(Br) · Na.
N(Br) = 0,448 mol · 6,022·10²³ 1/mol.
n(Br) = 2,69·10²³.
7 0
3 years ago
How do you know that oxygen was the gas removed from the volume of air and not another
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8 0
3 years ago
Metals have low ionization energies and readily share their ________ or outer electrons with each other to form an electron ____
Dmitry [639]

Explanation:

Metals are the species which readily lose electrons in order to attain stability. This electron lost by the atom is actually present in its outermost shell which is also known as valence shell.

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period.  

But when we move down a group then there occurs an increase in atomic size of the atoms due to addition of number of electrons in the atoms. Hence, ionization energy decreases along a group.

Thus, we can conclude that metals have low ionization energies and readily share their valence or outer electrons with each other to form an electron sea. These electrons are delocalized or shared among all the atoms that are bonded together and can therefore move freely throughout the metal structure.

3 0
3 years ago
Read 2 more answers
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