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%
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).