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3 years ago

How many moles of aluminum oxide can be made if 5.23 mol Al completely react

Chemistry

2 answers:

fenix001 [56]3 years ago

5 0

4Al + 3O2 --> 2Al2O3

(5.23mol Al)(2mol Al2O3/4mol Al) = 2.62mol Al2O3

vodka [1.7K]3 years ago

3 0

Answer:

2..615moles of Aluminium oxide

Explanation:

4Al + 3O2 -------> 2Al2O3

(5.23mol Al)×(2molaAl2O3) /(4molAl)

2.615moles of Aluminium oxide

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When metal atoms combine chemically with atoms of other elements, what do they usually do?

Fiesta28 [93]

Answer:

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The answer you'll be expected to give depends on the subject of the chapter you're studying. If you're studying covalent bonds, then the answer will probably be "form covalent bonds". If you're studying ionic bonds, then the answer will be "lose electrons".

Explanation:

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8 0

3 years ago

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.892 M and [Fe2 ] = 0.0150

Effectus [21]

Given:

Concentration of Cr2+ = 0.892 M

Concentration of Fe2+ = 0.0150 M

To determine:

The cell potential, Ecell

Explanation:

The half cell reactions for the given cell are:

Anode: Oxidation

Cr(s) ↔ Cr2+(aq) + 2e⁻ E⁰ = -0.91 V

Cathode: Reduction

Fe2+ (aq) + 2e⁻ ↔ Fe (s) E⁰ = -0.44 V

------------------------------------------

Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)

E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V

The cell potential can be deduced from the Nernst equation as follows:

Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]

Here, n = number of electrons = 2

Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V

Ans: The cell potential is 0.418 V

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3 years ago

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