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elena55 [62]
3 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]3 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!!!

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The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C

Explanation:

Here we make use of the Clausius-Clapeyron equation;

ln\left (\frac{p_{2}}{p_{1}}  \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}}  \right )

Where:

P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K

P₂ = 0.2 atm = The substance vapor pressure at temperature T₂

\Delta H_{vap} = The heat of vaporization = 28.5 kJ/mol

R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

ln\left (\frac{0.2}{1}  \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15}  \right )

\therefore T_2 = \frac{-3427.95}{ln(0.2)-6.175}

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To convert to Celsius degree temperature, we subtract 273.15 as follows

T₂ in °C = 440.37 - 273.15 = 167.22 °C

Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.

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