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3 years ago

13

A manganese electrode was oxidized electrically. If the mass of the electrode decreased by 225 mg during the passage of 1580 cou

lombs, what was the oxidation state of the manganese produced

1 answer:

qwelly [4]3 years ago

8 0

Answer:

<u>Oxidation state of Mn = +4</u>

Explanation:

Atomic mass of Mn = 55g/mol

From Faraday's law of electrolysis,

Electrochemical equivalent = $\frac{mass}{charge}$

i.e Z = $\frac{m}{Q}$ = $\frac{0.225g}{1580C}$ = 0.0001424 g/C

But Equivalent weight, E = atomic mass ÷ valency = Z × 96,485

⇒ $\frac{55}{valency}$ = 0.0001424 × 96,485

<u>∴ Valency of Mn = +4</u>

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How many moles are in 6.5g of carbonate

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Answer:

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Explanation:

Given data:

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3 years ago

Calculate the energy released in each of the following fusion reactions. Give your answers in MeV. (a) 2H + 3H → 4He + n 17.54 C

Vikentia [17]

Answer:

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b) E = 18.99 MeV

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d) You can use the methods applied for the other parts to solve this, the equation is not properly written

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Energy released = 17.55 MeV

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Energy released,

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c) $2H + 2H \rightarrow 3 He$ + n

$\triangle M = (M_{3He} + M_{n} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.008664) - (2.0141 + 2.0141 )\\\triangle M = -0.003536 u$

Energy released,

$E = -0.003536 * 931.5\\E = -3.29 Mev$

E = 3.29 MeV(Energy is released)

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

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$\triangle M = (M_{3H} + M_{1H} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.007825) - (2.0141 + 2.0141 )\\\triangle M = -0.00435 u$

$E = -0.00435 * 931.5\\E =-4.075 Mev$

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