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zepelin [54]
3 years ago
6

Calculate the electrical energy per gram of anode material for the following reaction at 298 K:

Chemistry
2 answers:
ale4655 [162]3 years ago
8 0
The substance oxidized is MnO2 which means that the oxidizing agent and at the same time the anode material is Li. The standard cell potential is already given and it is given in terms of per mole of reactant. So, the electrical energy per gram of anode is
3.15 / 7 = 0.45 V / g of Li
antiseptic1488 [7]3 years ago
6 0

The answer is:

E per gram = 0.45 V

The explanation:

when MnO2 is the substance who oxidized here so, the oxidizing agent and the anode here is Li.

and when the molar mass of Li is = 7 g/mol

and in our reaction equation we have 1 mole of Li will give 3.15 V of the electrical energy

that means that :

7 g of Li gives → 3.15 V

So 1 g of Li will give→ ???

∴ The E per gram = 3.15 V / 7 g of Li

= 0.45 V

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A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma
-BARSIC- [3]

Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:

  • 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
  • n = 0.307 mol

So now we know that

  • MolH₂ + MolN₂ = 0.307 mol

and

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
  • (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
  • 0.614 - 2MolN₂ + 28molN₂ = 3.49
  • 0.614 + 26MolN₂ = 3.49
  • MolN₂ = 0.111 mol

Now we calculate MolH₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ + 0.111 = 0.307
  • MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:

  • N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
  • H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

5 0
3 years ago
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djyliett [7]

Answer:

318 g / 19.32 g     v = 16. your volume is 16 hope this helps

Explanation:

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I recently did this assignment!

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Answer:

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