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goldenfox [79]
3 years ago
8

A cylinder at left with balls evenly spaced throughout the cylinder has an arrow leading to a cylinder at right cylinder with ba

lls in a layer at its bottom. Which change of state is shown in the model?
Chemistry
2 answers:
Lelechka [254]3 years ago
7 0

gas to liquid

Explanation:

The change of state indicated by this analogy is from gas to liquid.

Cylinder to the left is filled with gases

Cylinder to the right is made up of liquid.

  • Gases occupy the volumes of containers they are introduced into.
  • They are random and possess a high kinetic energy.
  • Liquids have definite volume and flow with one another.
  • The gases in A are dispersed and in random motion.
  • This phase change is called condensation

learn more:

Phase change brainly.com/question/1875234

#learnwithBrainly

Aliun [14]3 years ago
7 0

Answer:

Sublimation

Explanation:

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Note.... the colliding molecules must possess certain minimum energy (threshold energy)to make the collision effective n..the ad
Oliga [24]

Answer:

<h3>I think, answer is threshold energy.</h3>

OR

<h3>activation energy. </h3>

Explanation:

<h3>Hope it helps you....</h3>

<h2>Thank you..</h2>
4 0
3 years ago
Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has
Ymorist [56]

Answer:

\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

Explanation:

Given that:

The Half-life of ^{235}U = 7.13 \times 10^8 \ years is less than that of ^{238} U = 4.51 \times 10^9 \ years

Although we are not given any value about the present weight of ^{235}U.

So, consider the present weight in the percentage of ^{235}U to be  y%

Then, the time elapsed to get the present weight of ^{235}U = t_1

Therefore;

N_1 = N_o e^{-\lambda \ t_1}

here;

N_1 = Number of radioactive atoms relating to the weight of y of ^{235}U

Thus:

In( \dfrac{N_1}{N_o}) = - \lambda t_1

In( \dfrac{N_o}{N_1}) =  \lambda t_1 --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of ^{235}U to be = t_2

Then:

In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

here;

N_2 =  Number of radioactive atoms of ^{235}U relating to 3.0 a/o weight

Now, equating  equation (1) and (2) together, we have:

In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) =  \lambda( t_1-t_2)

replacing the half-life of ^{235}U = 7.13 \times 10^8 \ years

In( \dfrac{N_2}{N_1})  = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)      ( since \lambda = \dfrac{In 2}{t_{1/2}} )

∴

\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

8 0
3 years ago
A reaction produces 0.883 moles of H₂O. How many molecules are produced?
prisoha [69]
The answer is 5.32 × 10²³ molecules

<span>Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance:
</span>6.023 <span>× 10²³ units per 1 mole

We have 0.883 moles.
If 1 mole has </span>6.023 × 10²³ molecules, 0.883 moles will have x molecules:
1 mole : 6.023 × 10²³ molecules = 0.883 moles : x

x = 6.023 × 10²³ molecules * 0.883 moles : 1 mole = 5.32 × 10²³ molecules
6 0
3 years ago
Whats the complete chemical symbol for the ion that has : 13 neutrons 11 protons , and 9 electrons.
myrzilka [38]
The answer gonna be Na2+
7 0
3 years ago
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ExtremeBDS [4]
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6 0
3 years ago
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