Step 1: Count atoms from structure
Carbon (C) = 15 atoms
Hydrogen (H) = 11 atoms
Oxygen (O) = 4 atoms
Nitrogen (N) = 1 atom
Iodine (I) = 4 atoms
Step 2: Compute mass contribution
\[ \begin{aligned} \text{Mass of C} &= 15 \times 12 = 180 \, \text{g/mol} \\ \text{Mass of H} &= 11 \times 1 = 11 \, \text{g/mol} \\ \text{Mass of O} &= 4 \times 16 = 64 \, \text{g/mol} \\ \text{Mass of N} &= 1 \times 14 = 14 \, \text{g/mol} \\ \text{Mass of I} &= 4 \times 127 = 508 \, \text{g/mol} \end{aligned} \]
Step 3: Total molar mass of thyroxine
\[ \text{Molar mass} = 180 + 11 + 64 + 14 + 508 = 777 \, \text{g/mol} \]
Step 4: Percentage of Iodine
\[ \text{Percentage of Iodine} = \frac{508}{777} \times 100 \approx 65.37\% \]
Correct Answer: 65%
Step 1:
Thyroxine contains the elements carbon (C), hydrogen (H), oxygen (O), nitrogen (N), and iodine (I). To calculate the percentage of iodine, we first need to determine the total molar mass of thyroxine and the mass of iodine present in the compound.
Let's assume the chemical formula of thyroxine is C₁₄H₁₈I₄N₂O₄ (based on the provided structure in the image).
Step 2:
To find the molar mass of thyroxine, we add the molar masses of each element based on its quantity in the formula:
- Carbon (C): 12 × 14 = 168 g/mol
- Hydrogen (H): 1 × 18 = 18 g/mol
- Iodine (I): 127 × 4 = 508 g/mol
- Nitrogen (N): 14 × 2 = 28 g/mol
- Oxygen (O): 16 × 4 = 64 g/mol
Now, sum these values:
\[
\text{Molar mass of thyroxine} = 168 + 18 + 508 + 28 + 64 = 786 \, \text{g/mol}
\]
Step 3:
The mass of iodine in thyroxine is 508 g/mol (from 4 iodine atoms). Now, calculate the percentage of iodine in the compound:
\[
\text{Percentage of iodine} = \left( \frac{508}{786} \right) \times 100 = 64.6\%
\]
Step 4:
Rounding this value to the nearest integer gives the percentage of iodine in thyroxine as 65%.
Final Answer:
The percentage of iodine in thyroxine is 65%.

The correct structures of A and B formed in the following reactions are : 
The major product formed in the following reaction is : 
For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
| \([A]\) (mol/L) | \(t_{1/2}\) (min) |
|---|---|
| 0.100 | 200 |
| 0.025 | 100 |
A. The order of the reaction is \( \frac{1}{2} \).
B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.