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OLEGan [10]
3 years ago
10

At STP (101.3 kPa, 0 °C), what volume would 2.00 moles of oxygen gas occupy? The gas constant is 8.31

Chemistry
1 answer:
lakkis [162]3 years ago
4 0

Answer:

44.8 L

Explanation:

Ideal Gas Equation -

i.e.,

PV = nRT

where,

P = pressure

V = volume

n = moles

R = universal gas constant

T = temperature

Using the information given in the question, Volume of the gas can be calculated -

P = 101.3 kPa

V = ?

n = 2.00 moles

R = 8.31

T = 0 degree C = 273.15 K

Using the above data, and putting the data in the respective formula -

PV = nRT

101.3 kPa * V = 2.00 moles * 8.31 * 273.15 K

V = 44.8 L

Hence, the volume of the given gas = 44.8 L

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<u>Answer:</u> The lewis dot structure is attached below.

<u>Explanation:</u>

A Lewis dot structure is defined as the representation of atoms having electrons around the atom where electrons are represented as dots.

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3 years ago
How many moles of H2O are in 64.0 g of H2O
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Answer:

Explanation:

stoichiometry of C₂H₂ to H₂O is 2:2.

Number of moles of C₂H₂ = molar mass of C₂H₂  

Since the molar mass of C₂H₂  is 26 g/mol.

Number of C₂H₂  moles reacted = 64.0 g / 26 g/mol = 2.46 mol.

according to a molar ratio of 2:2.

the number of H₂O moles formed = a number of C₂H₂  moles reacted.

Therefore the number of H₂O moles produced = 2.46 mol

8 0
2 years ago
What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)
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Answer : The volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

Explanation :

The balanced chemical reaction will be:

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First we have to calculate the moles of Al.

\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}

Molar mass of Al = 27 g/mole

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Now we have to calculate the moles of O_2.

From the reaction we conclude that,

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Now we have to calculate the volume of O_2 consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of O_2 occupies 22.4 liter volume of O_2 gas

So, 0.75\times 2 mole of O_2 occupies 0.75\times 2\times 22.4 liter volume of O_2 gas

Therefore, the volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

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