91影视

First, we calculate the moles of each solute in the given solutions. For each ion, we'll apply the formula: \[\text{moles} = \text{concentration (M)} \times \text{volume (L)}\](a) For \(32.0 \, \text{mL}\) of \(0.30 \, \text{M} \, \text{KMnO}_4\), the moles are:\[0.30 \, \text{mol/L} \times 0.0320 \, \text{L} = 0.0096 \, \text{mol}\]For \(15.0 \, \text{mL}\) of \(0.60 \, \text{M} \, \text{KMnO}_4\), the moles are:\[0.60 \, \text{mol/L} \times 0.0150 \, \text{L} = 0.0090 \, \text{mol}\](b) For \(60.0 \, \text{mL}\) of \(0.100 \, \text{M} \, \text{ZnCl}_2\), the moles are:\[0.100 \, \text{mol/L} \times 0.0600 \, \text{L} = 0.0060 \, \text{mol}\]For \(5.0 \, \text{mL}\) of \(0.200 \, \text{M} \, \text{Zn(NO}_3)_2\), the moles are:\[0.200 \, \text{mol/L} \times 0.0050 \, \text{L} = 0.0010 \, \text{mol}\](c) The moles of \(\text{CaCl}_2\):\[\text{mass} = 4.2 \, \text{g}, \text{molar mass of CaCl}_2 = 111 \, \text{g/mol}\]\[\text{moles of CaCl}_2 = \frac{4.2 \, \text{g}}{111 \, \text{g/mol}} = 0.0378 \, \text{mol}\]

Assume the volumes are additive:(a) The total volume after mixing is:\[32.0 \, \text{mL} + 15.0 \, \text{mL} = 47.0 \, \text{mL} = 0.0470 \, \text{L}\](b) The total volume after mixing is:\[60.0 \, \text{mL} + 5.0 \, \text{mL} = 65.0 \, \text{mL} = 0.0650 \, \text{L}\](c) The total volume is:\[150.0 \, \text{mL} = 0.150 \, \text{L}\] (No need to add, already given as total volume with \(\text{KCl}\)).

Using the moles calculated and the total volume, we find the concentration of each ion:(a) Combine moles of \(\text{KMnO}_4\):\[\text{total moles} = 0.0096 + 0.0090 = 0.0186 \, \text{mol}\]\[\text{Concentration of } \text{K}^+\text{ and } \text{MnO}_4^- = \frac{0.0186 \, \text{mol}}{0.0470 \, \text{L}} = 0.396 \, \text{M}\](b) For \(\text{ZnCl}_2\), \(\text{Zn}^{2+}\) is:\[\text{total } \text{Zn}^{2+} = 0.0060 \, \text{mol} + 0.0010 \, \text{mol} = 0.0070 \, \text{mol}\]\[\text{Concentration of } \text{Zn}^{2+} = \frac{0.0070 \, \text{mol}}{0.0650 \, \text{L}} = 0.108 \, \text{M}\]For \(\text{Cl}^-\), from \(\text{ZnCl}_2\) only:\[\text{moles of } \text{Cl}^- = 2 \times 0.0060 = 0.0120 \, \text{mol}\]\[\text{Concentration of } \text{Cl}^- = \frac{0.0120 \, \text{mol}}{0.0650 \, \text{L}} = 0.185 \, \text{M}\]For \(\text{NO}_3^-\), from \(\text{Zn(NO}_3)_2\) only:\[\text{moles of } \text{NO}_3^- = 2 \times 0.0010 = 0.0020 \, \text{mol}\]\[\text{Concentration of } \text{NO}_3^- = \frac{0.0020 \, \text{mol}}{0.0650 \, \text{L}} = 0.0308 \, \text{M}\](c) For \(\text{CaCl}_2\), \(\text{Ca}^{2+}\) is:\[\text{Concentration of } \text{Ca}^{2+} = \frac{0.0378 \, \text{mol}}{0.150 \, \text{L}} = 0.252 \, \text{M}\]For \(\text{Cl}^-\) from both sources (\(\text{CaCl}_2\) and \(\text{KCl}\)):\[\text{From } \text{CaCl}_2: 2 \times 0.0378 = 0.0756 \, \text{mol}\]\[\text{From } \text{KCl}: 0.02 \, \text{mol/L} \times 0.150 \, \text{L} = 0.003 \, \text{mol}\]\[\text{Total moles } \text{Cl}^- = 0.0756 + 0.003 = 0.0786 \, \text{mol}\]\[\text{Concentration of } \text{Cl}^- = \frac{0.0786 \, \text{mol}}{0.150 \, \text{L}} = 0.524 \, \text{M}\]

After finding all concentrations for each part:- (a) \(\text{K}^+\) and \(\text{MnO}_4^-\) are both at \(0.396 \, \text{M}\).- (b) \(\text{Zn}^{2+}\) is at \(0.108 \, \text{M}\), \(\text{Cl}^-\) at \(0.185 \, \text{M}\), and \(\text{NO}_3^-\) at \(0.0308 \, \text{M}\).- (c) \(\text{Ca}^{2+}\) is at \(0.252 \, \text{M}\) and \(\text{Cl}^-\) at \(0.524 \, \text{M}\).