Standard enthalpy of formation of Graphite, Diamond and Fullerenes is (in KJ/mole)
(a) 0,19.0, 381
(b) 0,1.90, 38.1
(c) 0,1.90, 381
(d) 1.90, 0, 38.1

Standard enthalpy of formation of water is -286 kJ mol^-1 . Calculate the enthalpy change for formation of 0.018 kg of water.

The direction cosines of x-axis are (A) 0,0,1 (B) 1,0,0 (C) 0,1,0 (D) 0,1,1

The abscissa and ordinate of the origin are (a) (0, 0) (b) (1, 0) (c) (0, 1) (d) (1, 1)

The matrix of the transformation reflection in the line x+y=0 is (A) [(-1,0),(0,-1)] (B) [(1,0),(0,-1)] (C) [(0,1),(1,0)] (D) [(0,-1),(-1,0)]

Since their discovery in 1985 , fullerenes have received the attention of many chemical researchers. In a recentrly reported data, the specific internal energy of combustion of crystalline C_(60) is found to be -36kJ g^(-1) at 298K . Compute the standard enthalpy of combustion and formation for the same. Standard enthalpy of combustion of graphite is -395kJ mol^(-1) . If the standard enthalpy of formation of diamond is +2kJ per mol of C- atom, which is more stable :C_(60) or diamond ?

Standard molar enthalpy of combustion of glucose is -2880 kJ. If only 25% of energy is available for muscular work and 1.0 km walk consumes 90 kJ of energy, what maximum distance (in km) a person can walk after eating 90 g of glucose.

If a random variable x assumes the values 0,1 and 2 with probabilities 0.30,0.50 and 0.20, then its expected value is (A) 1.50(B)3(C)0.90 (D) 1