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MakcuM [25]
3 years ago
14

As particles move faster it would feel cool to touch true or false brainly

Chemistry
1 answer:
svetlana [45]3 years ago
6 0

Answer: False

Explanation: The kinetic energy of the molecules is the energy possessed by virtue of motion of particles.

Kinetic energy of the particles is directly proportional to the temperature of the gas.

K.E=\frac{3RT}{2}

where T= temperature

R= gas constant

Thus if the particles are moving faster that means the kinetic energy is high, the temperature must be higher and the particles must feel hot to touch.


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Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate
Andrew [12]

Answer:

(2R,3S)-2-ethoxy-3-methylpentane

and

(2S,3S)-2-ethoxy-3-methylpentane

Explanation:

For this case, we will have  CH_3CH_2O^- as nucleophile. Also, this compound is also in excess. So, we will have as solvent CH_3CH_2OH a protic solvent. Therefore the Sn1 reaction would be favored.

The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).

7 0
3 years ago
A student dissolves equal amounts of salt in equal amounts of warm water, room-temperature water, and ice water. Which result is
Tanya [424]

doesnt salt desolve ice? so wouldn't the salt dissolve in the ice water?

6 0
3 years ago
Read 2 more answers
The solubility of Cr(NO3)3⋅9H2O in water is 208 g per 100 g of water at 15 ∘C. A solution of Cr(NO3)3⋅9H2O in water at 35 ∘C is
Zinaida [17]

Answer:

102g of crystals

Explanation:

When the Cr(NO₃)₃⋅9H₂O is dissolved in water at 15°C, the maximum mass that water will dissolve in the equilibrium is 208 g per 100g of water. When you heat the water, this mass will increases.

In this problem, at 35°C the water dissolves 310g in 100g of water, as in the equilibrium at 15°C the maximum mass is 208g, the mass of crystals that will form is:

310g - 208g = <em>102g of crystals</em>

<em>-Crystals are the Cr(NO₃)₃⋅9H₂O that is not dissolved-.</em>

I hope it helps!

5 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
3.) All matter has both physical and chemical properties. A physical property is one that does not change the chemical nature of
dexar [7]

Answer:

D

Explanation:

It would be D because you are observing the reaction and don’t change anything

8 0
2 years ago
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