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Zepler [3.9K]
3 years ago
10

Increasing order of unsaturation

Chemistry
1 answer:
Basile [38]3 years ago
7 0

Whats the question? Im not sure what your asking

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Why is studying light and electrons important for chemistry and understanding the universe?
Brut [27]

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.

7 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
What is the Law of Conservation of energy?
ladessa [460]

Answer:

<u>~</u><u>Law of Conservation of </u><u>energy~</u>

The law of conservation of energy states that energy can neither be created nor destroyed, only energy can be converted from one form to another.

3 0
3 years ago
If a reaction has an equilibrium constant just greater than 1 what type of reaction is it?
SashulF [63]

Answer:

c. reversible favoring products, equilibrium constants are calculated in an equation of the type aA + bB = cC + dD, were Keq = ([C][D]) / ([A][B]).

Explanation:

If the equilibrium constant is just greater than 1, that means that the products are favored.

8 0
3 years ago
Give an example of when<br> carbon can displace a metal<br> from its compounds and when it cannot?
RUDIKE [14]

Answer:

If a metal is less reactive than carbon, it can be extracted from its oxide by heating with carbon. The carbon displaces the metal from the compound, and removes the oxygen from the oxide. This leaves the metal.

Explanation:

5 0
3 years ago
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