91影视

(Write the content here) Given a 0.35 M K鈧働O鈧� solution, we have the following dissociation reaction: \[K_{3}PO_{4} \rightarrow 3K^{+} + PO_{4}^{3-}\] The stoichiometry shows that for each K鈧働O鈧�, we get 3 K鈦� ions and 1 PO鈧劼斥伝 ion. Therefore, the concentrations are: \[ [K^+] = 3 \times 0.35 \mathrm{M} = 1.05 \mathrm{M} \] \[ [PO_{4}^{3-}] = 0.35 \mathrm{M} \]

(Write the content here) Given a \(5 \times 10^{-4}\) M CuCl鈧� solution, we have the following dissociation reaction: \[ CuCl_{2} \rightarrow Cu^{2+} + 2Cl^{-} \] The stoichiometry shows that for each CuCl鈧�, we get 1 Cu虏鈦� ion and 2 Cl鈦� ions. Therefore, the concentrations are: \[ [Cu^{2+}] = 5 \times 10^{-4} \mathrm{M} \] \[ [Cl^{-}] = 2 \times (5 \times 10^{-4}) \mathrm{M} = 1 \times 10^{-3} \mathrm{M} \]

(Write the content here) This is a simple molecular solution. There are no ions formed when ethanol (CH鈧僀H鈧侽H) is dissolved in water. So, the concentration of CH鈧僀H鈧侽H in the solution is 0.0184 M.

(Write the content here) First, let's find the moles and then the new concentration when the solutions of Na鈧侰O鈧� and K鈧係O鈧� are mixed together. As the volumes are additive, we will consider the total volume for calculating new concentrations. Initial moles of Na鈧侰O鈧�: \[ n_{Na_2CO_3} = C_{1}V_{1} = 0.010 \mathrm{M} \times 0.035 \mathrm{L} = 3.5 \times 10^{-4} \text{moles} \] Initial moles of K鈧係O鈧�: \[ n_{K_2SO_4} = C_{2}V_{2} = 0.200 \mathrm{M} \times 0.050 \mathrm{L} = 0.01 \text{moles} \] The total volume after mixing: \[ V_{\text{total}} = V_{1} + V_{2} = 35.0 \mathrm{~mL} + 50.0 \mathrm{~mL} = 85.0 \mathrm{~mL} = 0.085 \mathrm{L} \] Now let's find the new concentrations: Concentration of Na鈧侰O鈧�: \[ C'_{Na_2CO_3} = \frac{n_{Na_2CO_3}}{V_{total}} = \frac{3.5 \times 10^{-4} \text{moles}}{0.085 \text{L}} = 0.0041 \mathrm{M} \] Concentration of K鈧係O鈧�: \[ C'_{K_2SO_4} = \frac{n_{K_2SO_4}}{V_{total}} = \frac{0.010 \text{moles}}{0.085 \text{L}} = 0.118 \mathrm{M} \] Now we have the new concentrations of both salts in the mixed solution. For the Na鈧侰O鈧� dissociation reaction: \[ Na_{2}CO_{3} \rightarrow 2Na^{+} + CO_{3}^{2-} \] We now calculate the ion concentrations: \[ [Na^+] = 2 \times 0.0041 \mathrm{M} = 0.0082 \mathrm{M} \] \[ [CO_{3}^{2-}] = 0.0041 \mathrm{M} \] For the K鈧係O鈧� dissociation reaction: \[ K_{2}SO_{4} \rightarrow 2K^{+} + SO_{4}^{2-} \] We now calculate the ion concentrations: \[ [K^+] = 2 \times 0.118 \mathrm{M} = 0.236 \mathrm{M} \] \[ [SO_{4}^{2-}] = 0.118 \mathrm{M} \]