Thats allllllll otttt but ill help
step by step
The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M
calculation
concentration = moles /volume in liters
volume in liters = 150 /1000= 0.15 L
number of moles calculation
write the equation for dissociation of Al2Cl3
that is AlCl3 ⇔ Al^3+ + 3 Cl ^-
find the moles of Cl^- formed
moles =mass/molar mass
mass in grams= 550/ 1000 =0.55 grams
molar mass of Cl^- =35.5 g/mol
moles is therefore= 0.55/35.5 =0.0155 moles
by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles
concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M
Answer:
5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O
Explanation:
Fe⁺² + MnO₄⁻ + H⁺ => Mn⁺² + Fe⁺³ + H₂O
5(Fe⁺² => Fe⁺³ + 1e⁻) => 5Fe⁺² => 5Fe⁺³ + 5e⁻
<u>MnO₄⁻ + 5e⁻ => Mn⁺² => MnO₄⁻ + 8H⁺ + 5e⁻ => Mn⁺² + 4H₂O</u>
=> 5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O
Which type of solution are you talking about?