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WINSTONCH [101]
3 years ago
12

If one neutron initiates a fission event that produces two neutrons in the products, how many new reactions can now be initiated

? 9 If each of the neutrons produced in the first fission event then initiates a fission event that produces one neutron in the products, how many new reactions can now be initiated by each neutron? How many neutrons in total were produced by the two fission events described?
Chemistry
1 answer:
alexdok [17]3 years ago
7 0

Answer:

See Explanation

Explanation:

A fission reaction is a chain reaction. The neutrons that are produced in one reaction leads to further chain reaction in the system.

If one neutron initiates fission that leads to the production of two other neutrons,two more new reactions are initiated.

If these reactions each produce one neutron, then another two new reactions are initiated. This makes a total of four new reactions from the two fission events described.

A total of four neutrons is produced from the two fission events.

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What volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below
mina [271]

Answer:

37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.

Explanation:

Equation for the reaction:

2 CO + 2 NO ------> N2 + 2 CO2

2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen

At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.

So therefore, we can say:

2 * 22.4 L of CO produces  22.4 L of N2

44.8 L of CO produces 22.4 L of N2

Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:

44.8 L of CO = 22.4 L of N

x L = 18.9 L

x L = 18.9 * 44.8 / 22.4

x L = 18.9 * 2

x = 37.8 L

The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L

8 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
3 years ago
Which best describes the structure of 2-butene
lyudmila [28]
The 2 represents that it is a double carbon bond
it looks like..
C-C = C-C
6 0
3 years ago
Read 2 more answers
After Callie obtained her original results, she wanted to dig deeper. She determined that the germinating corn seed had utilized
Airida [17]

Answer:

Callie expect 600 molecules of CO2 to have been released as a waste during the same amount of time.

Explanation:

During cellular respiration 1 molecule of glucose undergoes oxidation to form 6 molecules of CO2 as a waste product.

     According to the question callie determined that the germinating corn seed had utilized 100 molecules of glucose.

    So 100 molecules of glucose will release 100×6=600 molecules of CO2 as a waste product.

8 0
3 years ago
Read 2 more answers
QUICK! WILL MARK BRAINLIEST
nalin [4]

Answer:

Ability to be bent = Malleability

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Color, Phase, or Hardness = Physical Property

5 0
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