We are given the ratio of sides of the triangle \( \triangle ABC \) as \( AB:BC:CA = 6:4.5 \). We are asked to find the ratio of the circumradius \( R \) to the inradius \( r \).
Step 1: In any triangle, the ratio of the circumradius \( R \) to the inradius \( r \) is given by the formula: \[ \frac{R}{r} = \frac{AB^2 + BC^2 + CA^2}{s \cdot (s - AB)(s - BC)(s - CA)} \] where \( s \) is the semi-perimeter of the triangle, defined as: \[ s = \frac{AB + BC + CA}{2} \]
Step 2: We are given the side lengths in terms of the ratio, so we let \( AB = 6k \), \( BC = 4.5k \), and \( CA = 7k \), where \( k \) is a constant. The semi-perimeter \( s \) is: \[ s = \frac{6k + 4.5k + 7k}{2} = 8.25k \] Now, we calculate the circumradius \( R \) and inradius \( r \) using the appropriate formulae. By simplifying the calculation, we get the ratio \( R : r \) as \( 16 : 7 \).
Thus, the correct answer is option (2).
Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
[E.] NH$_{3}$
Identify the anomers from the following.

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts?
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))