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VMariaS [17]
3 years ago
13

which statement is true about the part of the electromagnetic spectrum that is visible to the human eye?

Chemistry
2 answers:
IgorLugansk [536]3 years ago
6 0
Hello @Lucysrv18,

How are you doing? In this case, we know that Electromagnetic waves that are visible to the human eye is neither made up of two electromagnetic waves nor three. Therefore B and C is incorrect.

These electromagnetic waves are divided and they are our final two options. For sure, i can say that the wavelengths in a visible electromagnetic wave are not divided into nine, but instead it is divided into 7. 

The answer to your question is A. 

Thank you,
Darian D. 

mezya [45]3 years ago
5 0

Answer: a

i JUST TOOK THE TEST

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How do the names of covalent compounds differ from the names of ionic compounds?
Tresset [83]
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5 0
3 years ago
Consider two gaseous systems: one for which the Vrms of the particles is low and another for
inessss [21]

Answer:

See explanation

Explanation:

According to chemistry libretexts(2020); "The rms velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molar mass. Thus quadrupling the temperature of a given gas doubles the rms velocity of the molecules. ... As the temperature of a gas is increased, the velocity of the molecules is also increased."

Hence considering two gases for which the  Vrms of the particles of one is high and that of the other is low, we can conclude that the gas having the higher  Vrms is at a higher temperature than the gas having a lower  Vrms according to the foregoing.

8 0
3 years ago
Calculate the mass of 1.0 L of helium (He), 1.0 L of chlorine gas (Cl2), and 1.0 L of air (79% N2, 21% O2 by volume) at 25°C and
Zigmanuir [339]

To calculate the mass we use the following formulas:

PV=nRT     (1)

and

n = m / M    (2)

where:

P - pressure (atm)

V - volume (L)

n - moles

R - gas constant = 0.082 (L × atm) / (mol × K)

T - temperature (°K) (25°C + 273 = 298°K)

m - mass (g)

M - molecular mass (g/mole)

Now we rewrite equation (1):

n = PV / RT

And replace n with m / M from equation (2):

m / M = PV / RT

m = (P × V × M) / (R ×T)

1 L of He will have a mass of:

m = (1 × 1 × 4) / (0.082 × 298) = 0.1637 g

1 L of Cl₂ will have a mass of:

m = (1 × 1 × 71) / (0.082 × 298) = 2.9055 g

1.0 L of air will contain 0.79 L of N₂ and 0.21 L of O₂

0.79 L of N₂ will have a mass of:

m = (1 × 0.79 × 28) / (0.082 × 298) = 0.9052 g

0.21 L of O₂ will have a mass of:

m = (1 × 0.21 × 32) / (0.082 × 298) = 0.2750 g

mass of air = mass of N₂ + mass of O₂

mass of air = 0.9052 + 0.2750 = 1.1802 g

A balloon filed with helium will rise because as you see 1 L of helium is lighter than 1 L of air.

Chlorine gas is dangerous because chlorine is very toxic for human life and more of that is heavier than the air so will diffuse very hard from the area where the leak appeared.

7 0
4 years ago
Help ASAP the question is in the picture
QveST [7]

Answer:

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5 0
1 year ago
What does this do to the electrons outside the nucleus in the gaseous atoms
AleksandrR [38]

Answer:

Explanation:

As you know, ionization energy is the energy needed to remove one mole of electrons from one mole of atoms in the gaseous state

X

+

energy

→

X

+

+

e

−

Right from the start, you can tell that the harder it is to remove an electron from an atom, the higher the ionization energy will be.

Now, the periodic trends for ionization energy can be describe as follows

ionization energy increases as you move from left to right across a period

ionization energy decreases as you go down a group

As you mentioned, if you compare the first ionization energies for oxygen and chlorine using these two trends, you will get conflicting results.

If you follow the way ionization energy increases across period, chlorine would have a higher ionization energy, since it's closer to the noble gases.

On the other hand, if you go by how ionziation energy decreases from top to bottom in a group, oxygen would have higher ionization energy, since it's located in period 2, as compared with period 3 for chlorine.

As it turns out, the trend for groups overpowers the trend for periods. As aresult, oxygen will have a higher ionization energy than chlorine.

This happens because the smaller oxygen atom has its outermost electrons held tighter by the nucleus. By comparison, chlorine's outermost atoms are located further away from the nucleus.

Not only that, but they are screened from the charge of the nucleus better, since they're located on the third energy level.

Oxygen's outermost electrons are screened by

2

electrons, while chlorine's are screened by

8

electrons.

All these factors will make chlorine's outermost electrons a little easier to remove, which implies a smaller ionization energy than that of oxygen.v

6 0
3 years ago
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