The problem asks for the calculation of the retardation factor (Rf) of an organic compound based on the distances it and the solvent moved on a thin layer chromatographic plate. The final answer needs to be expressed in a specific format.
In thin layer chromatography (TLC), the retardation factor, denoted as \( R_f \), is a ratio that quantifies the movement of a compound relative to the movement of the solvent front. It is a characteristic property of a compound for a given stationary phase (the TLC plate coating) and mobile phase (the solvent). The formula to calculate the retardation factor is:
\[ R_f = \frac{\text{Distance travelled by the compound (solute)}}{\text{Distance travelled by the solvent front}} \]The \( R_f \) value is always between 0 and 1, as the compound can never travel farther than the solvent front.
Step 1: Identify the given values from the problem statement.
The distance moved by the organic compound (solute) is given as 3.5 cm.
The distance moved by the solvent (mobile phase) is given as 5 cm.
Step 2: Substitute these values into the formula for the retardation factor \( R_f \).
\[ R_f = \frac{\text{Distance travelled by the compound}}{\text{Distance travelled by the solvent front}} = \frac{3.5 \text{ cm}}{5.0 \text{ cm}} \]Step 3: Calculate the numerical value of the retardation factor.
\[ R_f = \frac{3.5}{5.0} = 0.7 \]Step 4: Express the calculated \( R_f \) value in the required format of \( \_\_\_\_ \times 10^{-1} \).
To convert 0.7 into the desired format, we can write it as:
\[ 0.7 = 7 \times 0.1 \]Since \( 0.1 \) is equal to \( 10^{-1} \), we have:
\[ R_f = 7 \times 10^{-1} \]The retardation factor is calculated as the ratio of the distance moved by the compound to the distance moved by the solvent.
\[ R_f = \frac{3.5}{5.0} = 0.7 \]Expressing this in the specified format:
\[ 0.7 = 7 \times 10^{-1} \]Thus, the retardation factor of the organic compound is 7 \( \times 10^{-1} \).

Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: