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slega [8]
2 years ago
8

Describe the nature and purpose of these components of a nuclear reactor:(b) moderator;

Chemistry
1 answer:
sertanlavr [38]2 years ago
8 0

The moderator of a nuclear reactor is a substance that slows neutrons down. In traditional nuclear reactors, the moderator is the same thing as the coolant: it’s water! When fast neutrons strike the hydrogen atoms in H2O, they slow down a lot (like a billiard ball striking another). There are other good moderators like graphite, beryllium, and more.

<h3>What is moderator ?</h3>

Other neutrons are released at extremely high speeds when an incoming neutron splits an atom's nucleus. A sustained chain reaction is achievable if at least one (on average) of these neutrons can be made to split another fuel atom. In fact, if a neutron is traveling slowly, fuel atoms (like uranium) are more likely to absorb it (see Figure 1). In order to make a chain reaction easier to achieve, moderators are used in many reactor designs.

Neutrons can be slowed down more effectively in some materials than others. Because of the rules of conservation of energy and momentum, a neutron (mass 1) colliding with a heavy nucleus like a fuel atom cannot significantly slow down (mass 235).

To learn more about moderator  from the given link:

brainly.com/question/27936671

#SPJ4

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1-ethyl-3-methylcyclohexane
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How many nitrogen atoms are represented in 2Ca(NO3)2?
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Explanation:

there is 2 nitrogen but if you mean nitrate is 6

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What would happen to the rate of a reaction with rate law rate = k [NO]^2[H2] if
Ede4ka [16]

The rate of a reaction would be one-fourth.

<h3>Further explanation</h3>

Given

Rate law-r₁ = k [NO]²[H2]

Required

The rate of a reaction

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.  

Can be formulated:  

Reaction: aA ---> bB  

\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}

or  

\large{\boxed{\boxed{\bold{v~=~+\frac{\Delta B}{\Delta t}}}}

The concentration of NO were halved, so the rate :

\tt r_2=k[\dfrac{1}{2}No]^2[H_2]\\\\r_2=\dfrac{1}{4}k.[No]^2[H_2]\\\\r_2=\dfrac{1}{4}r_1

3 0
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You are working the grill at a restaurant and are having trouble getting the grill lit through electrical means. You light a mat
never [62]

Answer:

in this situation I would a little bold

Explanation:

first I don't know what extinguisher I would use pretty much any that helps with fires. I'll back people up, take the hood and put it on the small fire that way it will light out more and if I open the hood and there still a little fire I would use the extinguisher and no one gets hurt :)

6 0
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Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container
Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

Explanation:

Here Pressure in a container is given as

P=\frac{1}{3} \rho

Here

  • P is the pressure which is to be calculated
  • ρ is the density of the gas which is to be calculated as below

                                         \rho =\frac{mass}{Volume of container}

        Here

                mass is to be calculated for 75 gas phase molecules as

                      m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times  10^{-21} g

              Volume of container is 0.5 lts

     So density is given as

                         \rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3

  • is the mean squared velocity which is given as

                                        =RMS^2

      Here RMS is the Root Mean Square speed given as 605 m/s so

                                      =RMS^2\\=(605)^2\\=366025

Substituting the values in the equation and solving

P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa

So the average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

6 0
4 years ago
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