to the balanced chemical equation 4) 5 Cz04(aq) + 2 MnO4-(aq) + 6He(a)-10CO2() .
ID: 591956 • Letter: T
Question
to the balanced chemical equation 4) 5 Cz04(aq) + 2 MnO4-(aq) + 6He(a)-10CO2() . 2 Mn2+(aq) . 8 H2on 0.3500 grams of oxalic acid, H2C204 will react with permanganate, KMnO4 solution. mL of 0.100 M potassium A) 15.5 ml B) 389 mL C) 972 mL D) 7.7 ml Consider the following galvanic cell. Na No Sn NO, NO. Sn Fe' No, NO 1B) 5) Identify the anode and cathode, and indicate the direction of Nat ion and NO3- ion flow from the 5) salt bridge. A) Sn is the anode and Fe is the cathode; NOs"ions flow into half-cell compartment (A) and B) Sn is the anode and Fe is the cathode; Na+ ions flow into half-cell compartment (A) and C) Fe is the anode and Sn is the cathode; Na* ions flow into half-cell compartment (A) and D) Fe is the anode and Sn is the cathode; NO3 ions flow into half-cell compartment (A) and Nat ions flow into half-cell compartment (B). NO3"ions flow into half-cell compartment (B). NO3- ions flow into half-cell compartment (B). Nat ions flow into half-cell compartment (B). 6) 6) Which of the following statements must be true for the entropy of a pure solid to be zero? L The temperature must be O K IL. The solid must be crystalline, not amorphous The solid must be crystalline, not 11. The solid must be perfectly ordered. IV. The solid must be an element. B) I and II C) L II, and II D)11, Ill, and A)IExplanation / Answer
4)
Molar mass of H2C2O4 = 2*MM(H) + 2*MM(C) + 4*MM(O)
= 2*1.008 + 2*12.01 + 4*16.0
= 90.036 g/mol
mass of H2C2O4 = 0.35 g
we have below equation to be used:
number of mol of H2C2O4,
n = mass of H2C2O4/molar mass of H2C2O4
=(0.35 g)/(90.036 g/mol)
= 3.887*10^-3 mol
From balanced chemical reaction, we see that
when 5 mol of H2C2O4 reacts, 2 mol of MnO4- reacts
mol of MnO4- reacted = (2/5)* moles of H2C2O4
= (2/5)*3.887*10^-3
= 1.555*10^-3 mol
This is number of moles of MnO4-
we have below equation to be used:
M = number of mol / volume in L
0.1 = 1.555*10^-3/ volume in L
volume = 0.0155 L
volume = 15.5 mL
Answer: A
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