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max2010maxim [7]
3 years ago
6

How many liters of oxygen are in 8.32 moles of oxygen at STP

Chemistry
1 answer:
Jlenok [28]3 years ago
6 0

Answer:

186 Liters at STP conditions

Explanation:

1 mole of any gas at STP conditions occupies 22.4 Liters.

Therefore, 8.32 moles O₂(g) = 8.32 moles x 22.4Liters/mole = 186 Liters (3 sig.figs.)

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3 years ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

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3 years ago
How many electrons does phosphorous (P) need to gain to have a stable outer electron level?
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The correct answer is 3.
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Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr
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Charles’ Law........

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3 years ago
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Calculate the energy of an electron in the n = 2 level of a hydrogen atom.
ANEK [815]

Answer: The energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.

Explanation:

Given: n = 2

The relation between energy and n^{th} orbit of an atom is as follows.

E = - \frac{13.6}{n^{2}} eV

Substitute the values into above formula as follows.

E = - \frac{13.6}{n^{2}} eV\\= - \frac{13.6}{(2)^{2}}\\= - 3.40 eV

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Thus, we can conclude that the energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.

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