0.34 V - (8.314 J/(mol·K) * 298.15 K / (2 * 96485 C/mol)) * ln(4.48 / 3.2×10−3)

51. What is the ΔH° at 298 K of the given reaction?*- 3.8 kJ/mol3.8 J/mol0Cannot defined

The value of log K for the reaction A⇌B at 298 K is _______. (Nearest integer)Given: ΔH°=−54.07 kJ mol−1ΔS°=10JK−1 mol−1(Taken 2.303×8.314×298=5705 )

The Cv of a Vander Waal’s gas is 0.096 dm3/mol. Find the volume of a single molecule

Using Ohm's Law, find V when the current I =1.2 milliamps and R = 5.8 Kilohms

Calculate ΔG for the cell reaction at 298 KZn(s)/Zn (0.0004 M) || Cd (0.2 M)/Cd(s)Given= + 0.76 V= –0.403 V(1) –42960 J (2) –84500 J(3) +84920 J (4) +42960 J+2 +2E oZn/Zn+2E oC /Cd