greenhouse gases can cause Earth's atmosphere to trap more and more heat. This causes Earth to warm up.
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>
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
![[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%20%5D%20%3D%20%5Csqrt%7BCa%20%5Ctimes%20Ka%20%7D%20%3D%20%5Csqrt%7B0.249%20%5Ctimes%204.50%20%5Ctimes%2010%5E%7B-4%7D%20%20%7D%20%3D%200.0106%20M)
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
![\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D%20%5Ctimes%20100%5C%25%20%3D%20%5Cfrac%7B0.0106M%7D%7B0.249%7D%20%5Ctimes%20100%5C%25%20%3D%204.26%5C%25)
Electron, because they’re about 1/2000 the mass of a proton or nuetron
Answer:
on the surface of the cathode