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Nadya [2.5K]
2 years ago
12

What are atoms in general

Chemistry
2 answers:
AVprozaik [17]2 years ago
6 0

Atoms are the basic building blocks of ordinary matter. Atoms can join together to form molecules, which in turn form most of the objects around you.

miss Akunina [59]2 years ago
5 0

atoms are the building blocks of life used to create everything including you

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A buffer solution contains 0.20 mol of propionic acid (CH3CH2COOH) and 0.25 mol of sodium propionate (CH3CH2COONa) in 1.50 dm3.
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3 years ago
What action leads to crystal formation in minerals?
ArbitrLikvidat [17]
I believe c is the right answer.
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O que são processos físicos de separação das misturas? Dê 3 exemplos
chubhunter [2.5K]

Answer:

Explanation:

The physical methods of separating mixtures are used in sorting a mixture of substances.

It requires no chemical changes occurring between their components and parts in any significant way.

Examples are:

  1. Decantation
  2. Filtration
  3. Sublimation
  4. Magnetism
  5. Centrifugation

The methods simply relies on the physical properties of matter.

8 0
3 years ago
Two isotopes of hydrogen fuse to form a neutron plus the larger element,A) beryllium.B) carbon.C) deuterium.D) helium.
Nezavi [6.7K]

Answer:  D) helium.

Explanation:

Nuclear fission is a process which involves the conversion of a heavier nuclei into two or more small and stable nuclei along with the release of energy.

_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{56}^{143}Ba+_{36}^{90}Kr+3_0^1\textrm{n}

Nuclear fusion is a process which involves the conversion of two small nuclei to form a heavy nuclei along with release of energy.

Example: _1^2\textrm{H}+_1^3\textrm{H}\rightarrow _2^4\textrm{He}+_0^1\textrm{n}+\text{energy}

Thus when deuterium and tritium , the two isotopes of hydrogen are fused, a heavier nuclei helium is being formed from two smaller nuclei releasing a neutron.

3 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
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