The heat of hydration of CuSO4 to CuSO4 · 5H2O is given by the difference between the heat of solution of anhydrous CuSO4 and CuSO4 · 5H2O.
Heat of hydration = Heat of solution of anhydrous CuSO4 − Heat of solution of CuSO4 · 5H2O.
Substitute the values:
\( x = \left| -70 \, \text{kJ mol}^{-1} - \left( +12 \, \text{kJ mol}^{-1} \right) \right| \).
\( x = \left| -70 - 12 \right| = \left| -82 \right| = 82 \).
Final Answer: 82 kJ.
Consider the following statements:
(A) Availability is generally conserved.
(B) Availability can neither be negative nor positive.
(C) Availability is the maximum theoretical work obtainable.
(D) Availability can be destroyed in irreversibility's.
List-I (Details of the processes of the cycle) | List-II (Name of the cycle) |
---|---|
(A) Two adiabatic, one isobaric and two isochoric | (I) Diesel |
(B) Two adiabatic and two isochoric | (II) Carnot |
(C) Two adiabatic, one isobaric and one isochoric | (III) Dual |
(D) Two adiabatics and two isothermals | (IV) Otto |
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is: