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elena55 [62]
2 years ago
5

In order for fission reactions to be successful, they must be self-perpetuating, meaning they must be able to keep themselves go

ing.
What do you call the minimum amount of material that is needed for the reaction to keep going?

A) isotope
B) mass defect
C) critical mass
D) neutron
Chemistry
1 answer:
aleksandrvk [35]2 years ago
3 0

Answer:

Option C is correct.

The minimum amount of material that is needed for a fission reaction to keep going is called the critical mass.

Explanation:

Nuclear fission is the term used to describe the breakdown of the nucleus of a parent isotope into daughter nuclei.

Normally, the initial energy supplied for nuclear fission is the energy to initiate the first breakdown of the first set of radioactive isotopes that breakdown. Once that happens, the energy released from the first breakdown is enough to drive further breakdown of numerous isotopas in a manner that leads to more energy generation.

But, for this to be able to be sustained and not fizzle out, a particular amount of radioactive material to undergo nuclear fission must be present. This particular amount is termed 'critical mass'

Hope this Helps!!!

You might be interested in
Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll
makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature T_1 = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}

where; B = - 152.5 \ cm^3 /mol   C = -5800 cm^6/mol^2

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

4.138*10^{-4}  \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

Multiplying through with V² ; we have

4.138*10^4  \ V ^3 = V^2 - 152.5 V - 5800 = 0

4.138*10^4  \ V ^3 - V^2 + 152.5 V + 5800 = 0

V = 2250.06  cm³ mol⁻¹

Z = \frac{PV}{RT}

Z = \frac{1800*2250.06}{8.314*10^3*523.15}

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

T_c = 647.1 \ K \\ \\ P_c = 22055 \  kPa  \\ \\ \omega = 0.345

T__{\gamma}} = \frac{T}{T_c}

T__{\gamma}} = \frac{523.15}{647.1}

T__{\gamma}} = 0.808

P__{\gamma}} = \frac{P}{P_c}

P__{\gamma}} = \frac{1800}{22055}

P__{\gamma}} = 0.0816

B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}

B_o = 0.083 - \frac{0.422}{0.808^{1.6}}

B_o = 0.51

B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}

B_1 = -0.282

The compressibility is calculated as:

Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}

Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}

Z = 0.9386

V= \frac{ZRT}{P}

V= \frac{0.9386*8.314*10^3*523.15}{1800}

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At T_1 = 523.15 \  K \ and  \ P = 1800 \ k Pa

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = \frac{PV}{RT}

Z = \frac{1800*10^3 *0.1249}{729.77*523.15}

Z = 0.588

3 0
2 years ago
What reaction would take place if a hot tungsten filament bulb was surrounded by air
expeople1 [14]
I don't think it wont be a big explosion 
4 0
2 years ago
Read 2 more answers
Use the internet or your textbook as a reference to compare and contrast the Arrhenius Theory of acids and bases vs. the Brønste
Readme [11.4K]

Answer and Explanation:

1. Arrhenius Theory which describes the concept protonic. The substance that gives H+ ions when diluted in water is called as an acid (e.g. HCl) and the substance that dissociates OH-ions whenever it is diluted in water is called as the base (e.g. NaOH)

on the other hand

Bronsted Lowery Theory describes the concept of a proton donor-acceptor. The proton-donating species is an acid and the proton-accepting species is known as a base.

2. The Chemical name and nature of acid is shown below:-

Nature Chemical Name

a. HCl Acidic Hydrochloric Acid

b. KOH Basic Potassium hydroxide

c. HNO Acidic Nitric Acid

d. Mg(OH)2 Basic Magnesium hydroxide

3 0
2 years ago
A chemical engineer calculated that 15.0 mol H2 was needed to react with excess N2 to prepare 10.0 mol NH3. But the actual yield
rjkz [21]

Answer:

The actual number of moles is 9 moles.

It is less than 15

Number of moles needed is 9 moles

Explanation:

15H2 + 10N2 ——-> 10NH3

Now from the question, we can see that the percentage yield is 60%

The percentage yield can be calculated as actual moles of H2/Theoretical moles of H2 * 100%

From the equation, we can see that the theoretical number of moles of hydrogen is 15.

Now to get the actual : 60 = x/15 * 100

x = 9

The actual number of moles is 9 moles.

It is less than 15

Number of moles needed is 9 moles

8 0
2 years ago
One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori
Igoryamba

Answer:

P_2=0.398bar=39800Pa

T_2=118.7K\\

Q=-3729.9J

W=-61753.24J

ΔU_T=0J

ΔH_T=0J

Explanation:

Hello,

At the first state, the molar volume is:

v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3

The volume in both the second and third state:

v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3

Now, as it is about an adiabatic process, one remembers the following relationships:

PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa

- And the temperature:

T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J

Then the total heat:

Q=Q_1+Q_2=0-3729.9J=-3729.9J

- The work for the first process is:

W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

W=W_1+W_2=-61753.24J+0J=-61753.24J

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\

Best regards.

8 0
2 years ago
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