<u>Answer:</u> The original element is ![_{84}^{212}\textrm{Po}](https://tex.z-dn.net/?f=_%7B84%7D%5E%7B212%7D%5Ctextrm%7BPo%7D)
<u>Explanation:</u>
Alpha decay is defined as the process in which alpha particle is emitted. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units.
The released alpha particle is also known as helium nucleus.
![_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha](https://tex.z-dn.net/?f=_Z%5EA%5Ctextrm%7BX%7D%5Crightarrow%20_%7BZ-2%7D%5E%7BA-4%7D%5Ctextrm%7BY%7D%2B_2%5E4%5Calpha)
For the given alpha decay process of an isotope:
![^{Z}_{A}\textrm{X}\rightarrow ^{208}_{82}\textrm{Pb}+_2^4\alpha](https://tex.z-dn.net/?f=%5E%7BZ%7D_%7BA%7D%5Ctextrm%7BX%7D%5Crightarrow%20%5E%7B208%7D_%7B82%7D%5Ctextrm%7BPb%7D%2B_2%5E4%5Calpha)
<u>To calculate A:</u>
Total mass on reactant side = total mass on product side
A = 208 + 4
A = 212
<u>To calculate Z:</u>
Total atomic number on reactant side = total atomic number on product side
Z = 82 + 2
Z = 84
The isotopic symbol of unknown element is ![_{84}^{212}\textrm{Po}](https://tex.z-dn.net/?f=_%7B84%7D%5E%7B212%7D%5Ctextrm%7BPo%7D)
Hence, the original element is ![_{84}^{212}\textrm{Po}](https://tex.z-dn.net/?f=_%7B84%7D%5E%7B212%7D%5Ctextrm%7BPo%7D)
Answer:
Kc = 9.52.
Explanation:
<em>A + 2B ⇌ C,</em>
Kc = [C]/[A][B]²,
Concentration: [A] [B] [C]
At start: 0.3 M 1.05 M 0.55 M
At equilibrium: 0.3 - x 1.05 - 2x 0.55 + x
0.14 M 1.05 - 2x 0.71 M
- For the concentration of [A]:
∵ 0.3 M - x = 0.14 M.
∴ x = 0.3 M - 0.14 M = 0.16 M.
∴ [B] at equilibrium = 1.05 - 2x = 1.05 M -2(0.16) = 0.73 M.
<em>∵ Kc = [C]/[A][B]²</em>
∴ Kc = (0.71)/(0.14)(0.73)² = 9.5166 ≅ 9.52.
0.753 is the answer for this question
10^(-0.123)=0.753
Nh4, nigrogen needs 4 Electrons and each
Answer : The steps for balancing a redox reaction using half reaction method will include the following steps :
1) Write the basic ionic form of the reaction equation.
2) Divide the equation into two separate half reaction which will have oxidation half and reduction half reactions.
3) Balance the atoms present in each half of the reactions except O and H atoms in the reaction.
4) Balance the oxygen atoms in the reduction half of the reaction by the addition of two water molecules in the given conditions of the reaction.
5) Balance the charges present in both the half reactions.
6) The two halves of the equations reduction and oxidation are added to complete the overall reactions. After the addition of two reaction halves, cancel the same number electrons on both sides. The net ionic equation can be written after this step.
7) Finally, Verify whether the equation consists of the same type, number of co-efficients, and charges on both sides of the equation.