A current of 250. A flows for 24.0 hours at an anode where Mn^2+(aq) is being co

Question

A current of 250. A flows for 24.0 hours at an anode where Mn^2+(aq) is being converted to MnO_2(s). What mass of MnO_2 is deposited at this anode? a. 19.5 kg b. 12.9 kg c. 4.87 kg d. 2.43 kg e. none of these is within 10% of the answer 41. What is the pH at the half-stoichiometric point for the titration of 0.20 M NH(aq) with 0.50 M HNO_2 (aq)? For NH_3, K_p = 1.8 Times 10^-5. a) 9.26 b) 4.74 c) 11.13 d) 8.56 e) 7.00 42. For the titration of 50.0 mL of 0.020 M aqueous salicylic acid with 0.020 M KOH(aq). calculate the pH after the addition of 30.0 mL of KOH(aq). For salycylic acid. pK_a = 2.97. a) 3.15 b) 2.12 c) 2.97 d) 2.33 e) 7.00 43. Calculate the [OH] in an aqueous solution which is 0.125 M NH_3 and 0.300 M NH_4 Cl. For NH_3 K = 1.8 Times 10^-5. a) 0.425 M b) 0.125 M c) 15 Times 10^-6 M d) 1.8 Times 10^-5 M e) 43 Times 10^-5 M

Explanation / Answer

40.

We know that

Q, or the Quantity of charge transferred (Coulombs, C) = current flow (amperes, A) x time (seconds, s)

therefore in this case, Q= 250 A * 24*(60*60) seconds = 21600000 Coulombs

Also, 1 Faraday (F) = 96 500 Coulombs (C) = 1 mole of electrons (Because Faraday's constant= 96500 C/mol)

therefore moles of electrons passed through circuit = 21600000/96500 = 223.834

As it takes two moles of electrons to form one mole of MnO2 from Mn2+

therefore moles of MnO2 formed = 223.834/2 = 111.917

mass of MnO2 = moles of MnO2 x atomic mass of MnO2 = 111.917 * 87 g

= 9736.779 grams = 9.737 kg

Correct option is (e): None of these is within 10% of the answer.

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