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Olin [163]
3 years ago
10

Tell me some 4 interesting facts about oxygenthanksss​

Physics
1 answer:
mezya [45]3 years ago
8 0

Answer:

Hey

Oxygen gas is colorless, odorless, and tasteless.

Oxygen supports combustion and is required for fire, but itself does not burn and is not flammable.

Common uses of oxygen include production of steel, plastics and textiles, brazing.

Oxygen fuels our cells and helps provide the basic building blocks that our bodies need to survive. Our cells combine oxygen with nitrogen and hydrogen to produce various proteins that build new cells. ... Oxygen is also necessary for constructing replacement cells for our bodies.

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In this case, lithium would be classified as a(n)
Black_prince [1.1K]

Lithium is to be considered as an ION

Atom is defined as its natural state where it will have all its electrons present in its shell

Since in this case Lithium is giving its electron so after giving the electron it will become an ion

Compound is the combination of two or more different atoms

Molecules is the combination of same type of atoms

So here answer must be

<em>C) ion</em>

5 0
4 years ago
Read 2 more answers
A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr
slega [8]
<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.  

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a <u>kinetic energy. </u>

This is what Einstein proposed:  

Light behaves like a stream of particles called photons with an energy  E:

E=h.f (1)  

So, the energy E of the incident photon must be equal to the sum of the Work function \Phi of the metal and the kinetic energy K of the photoelectron:  

E=\Phi+K (2)  

Where \Phi is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal.  </u>

In this case \Phi=2eV  and K_{1}=4eV

So, for the first light source of wavelength \lambda_{1}, and  applying equation (2) we have:

E_{1}=2eV+4eV   (3)  

E_{1}=6eV   (4)  

Now, substituting (1) in (4):  

h.f=6eV (5)  

Where:  

h=4.136(10)^{-15}eV.s is the Planck constant

f is the frequency  

Now, the <u>frequency has an inverse relation with the wavelength </u>

\lambda_{1}:  

f=\frac{c}{\lambda_{1}} (6)  

Where c=3(10)^{8}m/s is the speed of light in vacuum  

Substituting (6) in (5):  

\frac{hc}{\lambda_{1}}=6eV (7)  

Then finding \lambda_{1}:  

\lambda_{1}=\frac{hc}{6eV } (8)  

\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}  

We obtain the wavelength of the first light suorce \lambda_{1}:  

\lambda_{1}=2.06(10)^{-7}m   (9)

Now, we are told the second light source \lambda_{2}  has the double the wavelength of the first:

\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)   (10)

Then: \lambda_{2}=4.12(10)^{-7}m   (11)

Knowing this value we can find E_{2}:

E_{2}=\frac{hc}{\lambda_{2}}   (12)

E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}   (12)

E_{2}=3.011eV   (13)

Knowing the value of E_{2} and \lambda_{2}, and knowing we are working with the same work function, we can finally find the maximum kinetic energy K_{2} for this wavelength:

E_{2}=\Phi+K_{2} (14)  

K_{2}=E_{2}-\Phi (15)  

K_{2}=3.011eV-2eV  

K_{2}=1.011 eV  This is the maximum kinetic energy for the second light source

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4 years ago
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Fan blades would be an analogy for which atomic model? Thomson's Bohr's electron cloud Dalton's
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Pulling the two ends of a rubber band further and further apart from each other.
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