A solution was prepared by dissolving 1210 mg of K3Fe(CN)6 (329.2 g/mol) in sufficient water to give 775 mL. Calculate (a) the m

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A solution was prepared by dissolving 1210 mg of K3Fe(CN)6 (329.2 g/mol) in sufficient water to give 775 mL. Calculate (a) the m

olar analytical concentration of K3Fe(CN)6. (b) the molar concentration of K1. (c) the molar concentration of Fe(CN)6 32. (d) the weight/volume percentage of K3Fe(CN)6. (e) the number of millimoles of K1 in 50.0 mL of this solution. (f ) ppm Fe(CN)6 32. (g) pK for the solution. (h) pFe(CN)6 for the solution

Chemistry

1 answer:

SashulF [63] · Answer 1 · 2023-12-28T16:41:20+03:00

Answer:

I am using ur calculated value

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 M

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 mol

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 molMass of Fe(CN)63− = moles * molar mass of Fe(CN)63−

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 molMass of Fe(CN)63− = moles * molar mass of Fe(CN)63−=3.674⋅10−3 mol * 212.0 g/mol

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 molMass of Fe(CN)63− = moles * molar mass of Fe(CN)63−=3.674⋅10−3 mol * 212.0 g/mol= 0.7788g

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 molMass of Fe(CN)63− = moles * molar mass of Fe(CN)63−=3.674⋅10−3 mol * 212.0 g/mol= 0.7788gppm = mass of solute / mass of solution ∗106

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 molMass of Fe(CN)63− = moles * molar mass of Fe(CN)63−=3.674⋅10−3 mol * 212.0 g/mol= 0.7788gppm = mass of solute / mass of solution ∗106= 0.7788g / 775 g ⋅106 (for water 1ml = 1g)

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 molMass of Fe(CN)63− = moles * molar mass of Fe(CN)63−=3.674⋅10−3 mol * 212.0 g/mol= 0.7788gppm = mass of solute / mass of solution ∗106= 0.7788g / 775 g ⋅106 (for water 1ml = 1g)= 1005 ppm

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 molMass of Fe(CN)63− = moles * molar mass of Fe(CN)63−=3.674⋅10−3 mol * 212.0 g/mol= 0.7788gppm = mass of solute / mass of solution ∗106= 0.7788g / 775 g ⋅106 (for water 1ml = 1g)= 1005 ppm= 1.005 mg/ L

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 molMass of Fe(CN)63− = moles * molar mass of Fe(CN)63−=3.674⋅10−3 mol * 212.0 g/mol= 0.7788gppm = mass of solute / mass of solution ∗106= 0.7788g / 775 g ⋅106 (for water 1ml = 1g)= 1005 ppm= 1.005 mg/ LHope its clear to U!

I am using ur calculated valueMolarity of Fe(CN)63−=4.74⋅10−3 MMoles = molarity * V(L)=4.74⋅10−3M⋅0.775L=3.674⋅10−3 molMass of Fe(CN)63− = moles * molar mass of Fe(CN)63−=3.674⋅10−3 mol * 212.0 g/mol= 0.7788gppm = mass of solute / mass of solution ∗106= 0.7788g / 775 g ⋅106 (for water 1ml = 1g)= 1005 ppm= 1.005 mg/ LHope its clear to U!for the molarity of Fe(CN)63−. , i got 4.74x10−3 M