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zavuch27 [327]
2 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
Chemistry
1 answer:
qwelly [4]2 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 milligrams of product can be produced from the complete Diels-Alder reaction of 180. mg of anthracene and 100. mg of ma
ZanzabumX [31]

Answer: It will be produced 276,3 mg of product

Explanation: The reaction of anthracene (C14H10) and maleic anhydride (C4H2O3) produce a compound named 9,10-dihydroanthracene-9,10-α,β-succinic anhydride (C18H12O3), as described below:

C14H10 + C4H2O3 → C18H12O3

The reaction is already balanced, which means to produce 1 mol of C18H12O3 is necessary 1 mol of anthracene and 1 mol of maleic anhydride.

1 mol of C14H10 equals 178,23 g. As it is used 180 mg of that reagent, we have 0,001 mol of anthracene. With it, the reaction produces 0,001 mol of C18H12O3.

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3 years ago
How many molecules of ethane are present in 64.28 liters of ethane gas (C2H6) at STP
Rom4ik [11]

Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, 17.277\times 10^{23} molecule.

Solution :

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22.4 L volume of ethane present in 1 mole of ethane gas

64.28 L volume of ethane present in \frac{64.28L}{22.4L}\times 1mole=2.869moles of ethane gas

And, as we know that

1 mole of ethane molecule contains 6.022\times 10^{23} molecules of ethane

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3 years ago
Read 2 more answers
A solution containing a mixture of 0.0351 M potassium chromate ( K 2 CrO 4 ) and 0.0537 M sodium oxalate ( Na 2 C 2 O 4 ) was ti
ki77a [65]

Answer:

Explanation:

As it was stated in the comments by other user, the question is incomplete but luckily it was posted the rest of the question, so, I'm gonna answer it with the data of that. If you have another questions there, please submit it again or put it in the comments.

<u>a) Which barium salt will precipite first?</u>

In order to know this, we need to take a look at the Ksp of both salts. The given Ksp are:

Ksp 1: 2.1x10^-10 (BaCrO4)

Ksp 2: 1.3x10^-6 (BaC2O4)

Now, we can see that Ksp1 < Ksp 2, but what's this Ksp value means? a Ksp value means that an aqueous solution will form a precipite (So the solid formed, it's not soluble in water), As Ksp 1 is a smaller value than Ksp2, means that the concentrations of Ba and CrO4 are too small, and therefore, it takes more time to precipite. <em>Therefore, the BaC2O4 will precipite first.</em>

<u>b) Concentration of Ba2+ present to BaCrO4 precipite</u>

In this, we know that the reaction taking place is the following:

BaCrO4(s) <--------> Ba^2+ + CrO4^2-   Ksp = 2.1x10^-10

The expression for Ksp is:

Ksp = [Ba][CrO4]

We know the concentration of CrO4 cause this comes from the K2CrO4 so, replacing here, we solve for Ba:

[Ba^2+] = Ksp / [CrO4^2-]

[Ba^2+] = 2.1x10^-10 / 0.0351

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<u>c) Concentration of Ba^2+ to reduce 10% oxalate concentration</u>

The 10% concentration of oxalate is:

[C2O4] = 0.0537 * 0.1 = 0.00537 M

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BaC2O4(s) <--------> Ba^2+ + C2O4^2-   Ksp = 1.3x10^-6

[Ba^2+] = 1.3x10^-6 / 0.00537

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<u>d) ratio of oxalate and chromate when Ba concentration is 0.005 M</u>

In this case, we calculate concentration of CrO4 and C2O4 with the value of Ba and it's respective Ksp, and then, calculate the ratio:

[CrO4^2-] = Ksp1 / [Ba^2+] = 2.1x10^-10 / 0.0050 = 4.2x10^-8 M

[C2O4^2-] = Ksp2 / [Ba^2+] = 1.3x10^-6 / 0.0050 = 2.6x10^-4 M

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[C2O4^2-] / [CrO4^2-] = 2.6x10^-4 / 4.2x10^-8

[C2O4^2-] / [CrO4^2-] = 6.19x10^3

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