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Irina-Kira [14]
3 years ago
5

Why is studying light and electrons important for chemistry and understanding the universe?

Chemistry
1 answer:
Brut [27]3 years ago
7 0

Explanation :

The study of light and electrons are important for chemistry & universe.

   As we know, everything is made up of particles even light & universe also. The particles are made of atoms and that atoms are made up of electrons, protons & neutrons.

   An electron is one of the most important type of subatomic particles. Electrons combine with the proton to form atom. These electrons, protons & neutrons play a vital role in chemistry and universe.

   The importance of an electron is that when two atoms react with each other or approach each other, their outermost shells come into contact first and these outermost shells electrons are involved in the chemical reaction.

    Also important for chemical reactions, creating bonds, for electricity and the understanding of electrons has allowed for the better understanding of some forces which are in our universe such as electromagnetic force.

    We know that the light is act as particle nature and it is a part of universe. As we know that chemists are only responsible for the chemical's study. Chemists cannot study the whole universe that is why they study light and the light is important because nothing would be able to survive.

    The studying of light is important for the photoelectric effect, absorption, bio-molecules synthesis, vitamin D synthesis, vision, colors, drying & evaporation, sterilization, solar energy, spectroscopy, signal system and also for electron excitation.

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4 years ago
How many grams of sodium fluoride should be added to 300. mL of 0.0310 M of hydrofluoric acid to produce a buffer solution with
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Explanation : Given,

The dissociation constant for HF = K_a=6.8\times 10^{-4}

Concentration of HF (weak acid)= 0.0310 M

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (6.8\times 10^{-4})

pK_a=4-\log (6.8)

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Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[NaF]}{[HF]}

Now put all the given values in this expression, we get:

2.60=3.17+\log (\frac{[NaF]}{0.0310})

[NaF]=0.00834M

Now we have to calculate the moles of NaF.

\text{Moles of NaF}=\text{Concentration of NaF}\times \text{Volume of solution}=0.00834M\times 0.300L=0.0025mole

Now we have to calculate the mass of NaF.

\text{Mass of }NaF=\text{Moles of }NaF\times \text{Molar mass of }NaF=0.0025mole\times 42g/mole=0.105g

Therefore, the mass of sodium fluoride added should be 0.105 grams.

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3 years ago
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