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Anna11 [10]
2 years ago
8

Current of 250. a flows for 24.0 hours at an anode where the reaction occurring is as follows: mn2+(aq) + 2h2o(l) → mno2(s) + 4h

+(aq) + 2e–what mass of mno2 is deposited at this anode?
Chemistry
2 answers:
Tju [1.3M]2 years ago
7 0
From Faraday's 1st law of electrolysis,
Total electricity passed into system = Q = IT = 250 X 24 X 60 X 60
                                                                       = 2.16 X 10^7 C

We know that, 96500 C = 1 F
∴ 2.16 X 10^7 C =  <span>223.8 F
</span>
Now, number of moles of<span> MnO2  deposited = 223.8/2=111.9 
</span>
Finally, 1 mole of MnO2 ≡ 86.94 g
∴ 111.9 mole of MnO2 ≡  111.9 X 86.94 = 9728 g

Thus, <span>mass of MnO2 that will be deposited at anode = 9728 g</span>
Colt1911 [192]2 years ago
7 0

Answer:

The mass of mno2 is deposited at this anode is 9.732x10³ g

Explanation:

please look at the solution in the attached Word file

Download docx
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4. Convert the following: a. 4g mol of MgCl2 to g b. 2 lb mol of C3H8 to g c. 16 g of N2 to lb mol d. 3 lb of C2H6O to g mol
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Answer:

a) 381.2 g

b) 39916 g

c) 0.0013 lb mol

d) 29.6 g mol

Explanation:

The molecular weight (mw) of a compound is the mass of it per mole, so it's the ratio of the mass (m) per mole (n).

a) The molecular weight of one mol is found at the periodic table. So, for Mg, mw = 24.3 g/mol, for Cl = 35.5 g/mol, so for MgCl2, mw = 24.3 + 2*35.5 = 95.3 g/mol. The g mol is the mass divided by the molecular weight:

g mol = m/mw

4 = m/95.3

m = 381.2 g

b) The pound (lb) is a unity of mass, and the lb mol is a unity of the mass divided by the molecular weight. So, by the periodic table, the molecular weight of C3H8 is 3*12 (of C) + 8*1 (of H) = 44 lb/mol.

lb mol = m/mw

2 = m/44

m = 88 lb

1 lb = 453.592 g

So, m = 88*453.592 = 39916 g

c) The molecular weight of N2 is 2*14 (of N) = 28 lb/mol.

m = 16/453.592 = 0.0353 lb

lb mol = m/mw

lb mol = 0.0353/28

lb mol = 0.0013 lb mol

d) The molecular weight is 2*12 (of C) + 6*1(of H) + 1*16(of O) = 46 g/mol

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g mol = 1360.78/46

g mol = 29.6 g mol

6 0
2 years ago
A 0.530 M Ca(OH)2 solution was prepared by dissolving 36.0 grams of Ca(OH)2 in enough water. What is the total volume of the sol
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Answer:

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