1. A heavy nucleus (U235 or Pu239), when bombarded by slow moving neutrons, split into two
or more nuclei.
2. Two or more neutrons are produced by fission of each nucleus.
3. Huge amount of energy is produced as a result of nuclear fission.
4. All the fission fragments are radioactive, giving off β and radiations.
<span>5. The atomic weights of fission products range from about </span>70 to 160.
6. The nuclear chain reactions can be controlled and maintained steadily by absorbing a
desired number of neutrons. This process is used in nuclear reactor.
<span>7. All the fission reactions are self-propagating chain-reactions because fission products contain </span>
neutrons (secondary neutrons) which further cause fission in other nuclei.
8. Every secondary neutron, released in the fission process, does not strike a nucleus, some
escape into air and hence a chain reaction cannot be maintained.
<span>9. The number of neutrons, resulting from a single fission, is known as the multiplication factor. </span>
When the multiplication factor is less than 1, a chain reaction does not take place.
<span>10. The control of chain reaction is necessary in order to maintain a steady reaction. This is </span>
carried out by absorbing a desired number of neutron by employing materials like
percentage of Cd, B or steel.
11. In a nuclear reactor, the multifactor is one. This is achieved by proper arrangement of
<span>fissionable materials.</span>
The correct answer is A. All electrons become free and separate from the nuclei. In metallic bonds, the electrons of the metal atoms are delocalized. The electron in the electron sea can freely roam around or are free to flow.
Answer:
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
Explanation:
Step 1: Data given
A sample of sulfur hexafluoride gas occupies a volume of 5.10 L
Temperature = 198 °C = 471 K
The volume will be reduced to 2.50 L
Step 2 Calculate the new temperature via Charles' law
V1/T2 = V2/T2
⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L
⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K
⇒with V2 = the reduced volume of the gas = 2.50 L
⇒with T2 = the new temperature = TO BE DETERMINED
5.10 L / 471 K = 2.50 L / T2
T2 = 2.50 L / (5.10 L / 471 K)
T2 = 230.9 K = -42.1
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
5.18mL i hope this helps i hope this does to!
