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.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32