g bereaksi dengan NaOH, sehingga: M1 × V1 = M2 × V2 0,1 × 25 = M2 × 22,5 M2 = (0,1 × 25) / 22,5 = 0,111 M Jadi, molaritas larutan NaOH sebenarnya adalah 0,111 M. Interpretasi hasil ini penting agar larutan NaOH yang digunakan
r 2.717 kJ Given the moles of water formed (0.05 mol), the enthalpy change per mole is: ΔH_neut = - (q / moles) = - (2.717 kJ / 0.05 mol) = -54.34 kJ/mol This value is close to the expected standard enthalpy change, confirming
grams (assuming density is 1 g/mL) Temperature change: ΔT = T_f - T_initial Determining Enthalpy Change (ΔH) To find the molar enthalpy change of the neutralization: Calculate the total heat released (q) for the reaction. Divide by the number of moles of limiting reactant (H
