To find the total charge \( Q \) drawn during the process, we use the formula:
\( Q = I \times t \), where \( I \) is the current in Amperes and \( t \) is the time in seconds.
Given \( I = 1.25 \) A and \( t = 1.5 \) h, we need to convert the time from hours to seconds. There are 3600 seconds in an hour, so \( t = 1.5 \times 3600 = 5400 \) s.
Substitute these values into the formula:
\( Q = 1.25 \, \text{A} \times 5400 \, \text{s} = 6750 \, \text{C} \).
The calculated charge is \( 6750 \, \text{C} \)


One mole of a monoatomic ideal gas starting from state A, goes through B and C to state D, as shown in the figure. Total change in entropy (in J K\(^{-1}\)) during this process is ............... 
The number of chiral carbon centers in the following molecule is ............... 
A tube fitted with a semipermeable membrane is dipped into 0.001 M NaCl solution at 300 K as shown in the figure. Assume density of the solvent and solution are the same. At equilibrium, the height of the liquid column \( h \) (in cm) is ......... 