Question

Short answer

a.The balanced reaction is ${}_2{}^4\mathrm{He}+{}_{18}{}^{40}\mathrm{Ar}\to {}_{19}{}^{43}\mathrm{K}+{}_1{}^1\mathrm{H}$

b. The balanced reaction is ${}_0{}^1\mathrm{n}+{}_{92}{}^{238}\mathrm{U}\to {}_{92}{}^{239}\mathrm{U}$

c.The balanced reaction is${}_0{}^1\mathrm{n}+{}_7{}^{14}\mathrm{N}\to {}_6{}^{14}\mathrm{C}+{}_1{}^1\mathrm{H}$

d. The balanced reaction is ${}_{83}{}^{209}\mathrm{Bi}+{}_{28}{}^{64}\mathrm{Ni}\to {}_{111}{}^{272}\mathrm{Rg}+{}_0{}^1\mathrm{n}$

Step-by-step solution

Part(a) Step 1 : Given information

We need to complete the uncomplete reaction

$?+{}_{18}{}^{40}\mathrm{Ar}\to {}_{19}{}^{43}\mathrm{K}+{}_1{}^1\mathrm{H}$

Part(a) Step 2 : Simplify

Let the missing element be 'X' whose atomic mass is 'x' and atomic number be 'y' respectively.

by using the conservation of mass we get

$$\begin{gathered} \Rightarrow x+40=43+1 \\ \Rightarrow x+40=44 \\ \Rightarrow x=44-40 \\ \Rightarrow x=4 \end{gathered}$$

so the atomic mass of the element will be 4.

now by balancing the atomic number before and after the reaction, we get

$$\begin{gathered} \Rightarrow y+18=19+1 \\ \Rightarrow y+18=20 \\ \Rightarrow y=20-18 \\ \Rightarrow y=2 \end{gathered}$$

so the atomic number of the element is $2$.

From the element periodic element table, we can easily find the element whose atomic mass is $4$ans atomic number is $2$.

so the missing element is${}_2{}^4\mathrm{He}$

Part(b) Step 1 : Given information

We need to complete the reaction

${}_0{}^1\mathrm{\eta }+{}_{92}{}^{238}\mathrm{U}\to ?$

Part(b) Step 2 : Simplify

We have to find the atomic number and atomic mass of a given reaction.

Let the missing element be 'X' whose atomic mass is 'x' and atomic number is 'y' respectively .

so by using the conservation of mass we get

$$\begin{gathered} \Rightarrow 1+238=x \\ \Rightarrow x=239 \end{gathered}$$

so the atomic mass of the missing element is $239$.

Now calculation of atomic number,

the atomic number of neutron + the atomic number of uranium = the atomic number of element 'X'

$$\begin{gathered} \Rightarrow 0+92=y \\ \Rightarrow y=92 \end{gathered}$$

so the atomic number of missing will be $92$.

from periodic table , we can easily find out the element whose atomic number is $92$and atomic mass is $239$

the missing element is ${}_{92}{}^{239}\mathrm{U}$

Part(c) Step 1 : Given information

We need to complete the given uncompleted reaction

$?+{}_{28}{}^{64}\mathrm{Ni}\to {}_{111}{}^{272}\mathrm{Rg}+{}_0{}^1\mathrm{\eta }$ ${}_0{}^1\mathrm{\eta }+?\to {}_6{}^{14}\mathrm{C}+{}_1{}^1\mathrm{H}$

Part(c) Step 2 : Simplify

Let the missing element be 'X' whose atomic mass is 'x' and atomic number is 'y' respectively

by conservation of mass, we get

$$\begin{gathered} \Rightarrow 1+x=14+1 \\ \Rightarrow 1+x=15 \\ \Rightarrow x=15-1 \\ \Rightarrow x=14 \end{gathered}$$

so the atomic mass of the missing element is $14$.

Now, for atomic number

the atomic number of element neutron + atomic number of 'X'= the atomic number of 'C' + the atomic number of hydrogen

$$\begin{gathered} \Rightarrow 0+y=6+1 \\ \Rightarrow y=7 \end{gathered}$$

so the atomic number of the missing element is $7$.

From the element periodic table, we can easily find out the element whose atomic number is $7$and atomic mass is $14$.

The missing element is nitrogen.

Part(d) Step 1 : Given information

We need to complete the Reaction

$?+{}_{28}{}^{64}\mathrm{Ni}\to {}_{111}{}^{272}\mathrm{Rg}+{}_0{}^1\mathrm{\eta }$

Part(d) Step 2 : Simplify

Let the missing element be 'X' whose atomic mass is 'x' and atomic number is 'y' respectively.

so by using the conservation of the mass , we get

$$\begin{gathered} \Rightarrow x+64=272+1 \\ \Rightarrow x+64=273 \\ \Rightarrow x=273-64 \\ \Rightarrow x=209 \end{gathered}$$

the atomic mass of the missing element is $209$

Now for finding the atomic number of the missing element

the atomic number of element 'X' + atomic number of 'Ni'= the atomic number of 'Rg' + the atomic number of neutron

$$\begin{gathered} \Rightarrow y+28=111+0 \\ \Rightarrow y+28=111 \\ \Rightarrow y=111-28 \\ \Rightarrow y=83 \end{gathered}$$

so the atomic number of the missing element is $83$.

From the element periodic table, we can easily find out the element whose atomic number is $83$and atomic mass is $209$

The missing element is the ${}_{83}{}^{209}\mathrm{Bi}$