The dipole moment depends on the electronegativity difference and the molecular geometry. Here’s the analysis:
Thus, the decreasing order of dipole moments is:
\[ {H}_2{O} > {NH}_3 > {H}_2{S} > {NF}_3 \]Final Answer:
\( {H}_2{O} > {NH}_3 > {H}_2{S} > {NF}_3 \)
Observe the following reactions:
\( AB(g) + 25 H_2O(l) \rightarrow AB(H_2S{O_4}) \quad \Delta H = x \, {kJ/mol}^{-1} \)
\( AB(g) + 50 H_2O(l) \rightarrow AB(H_2SO_4) \quad \Delta H = y \, {kJ/mol}^{-1} \)
The enthalpy of dilution, \( \Delta H_{dil} \) in kJ/mol\(^{-1}\), is:
In the given circuit, if the potential at point B is 24 V, the potential at point A is:
