Question

Question was provided via image upload.

Explanation:

Step1: Balance mass/atomic numbers (Q7)

Mass: $A = 0 + 40 = 40$; Atomic: $Z = -1 + 20 = 19$ → $\ce{^{40}_{19}K}$

Step2: Balance mass/atomic numbers (Q8)

Mass: $A = 4 + 226 = 230$; Atomic: $Z = 2 + 88 = 90$ → $\ce{^{230}_{90}Th}$

Step3: Balance mass/atomic numbers (Q9)

Mass: $A = 35 - 0 = 35$; Atomic: $Z = 14 - (-1) = 15$ → $\ce{^{35}_{15}P}$

Step4: Balance mass/atomic numbers (Q10)

Mass: $A = 140 - 140 = 0$; Atomic: $Z = 56 - 57 = -1$ → $\ce{^{0}_{-1}e}$

Step5: Balance mass/atomic numbers (Q1[LLM SSE On Failure]

Answer:

1. $\ce{^{40}_{19}K}$
2. $\ce{^{230}_{90}Th}$
3. $\ce{^{35}_{15}P}$
4. $\ce{^{0}_{-1}e}$
5. $\ce{^{234}_{90}Th}$
6. $\ce{^{110}_{53}I}$
7. $\ce{^{198}_{86}Rn}

ightarrow \ce{^{4}_{2}He} + \ce{^{194}_{84}Po}$; YES

1. $\ce{^{237}_{92}U}

ightarrow \ce{^{0}_{-1}e} + \ce{^{237}_{93}Np}$; YES

1. $\ce{^{229}_{91}Pa} \xrightarrow{\text{alpha}} \ce{^{225}_{89}Ac} + \ce{^{4}_{2}He}$; $\ce{^{225}_{89}Ac} \xrightarrow{\text{alpha}} \ce{^{221}_{87}Fr} + \ce{^{4}_{2}He}$