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timofeeve [1]
2 years ago
10

How many moles are in 68.5 liters of oxygen gas at STP?

Chemistry
2 answers:
mr Goodwill [35]2 years ago
8 0

Answer:

3.0580 moles are in 68.5 liters of oxygen gas at STP

Explanation:

Let the number of moles of oxygen gas be n

Volume occupied by n moles of oxygen gas  at STP = 68.5 L

At STP, 1 mol of gas occupies 22.4 L

Then 68.4 L of volume will be occupied by:

\frac{1}{22.4} \times 68.5 L=3.0580 mol

3.0580 moles are in 68.5 liters of oxygen gas at STP

givi [52]2 years ago
3 0
Since we know that one mole of any gas at STP is equal to 22.4 L we can multiply 135L by the following conversion: 1 mole/22.4L. When you set up the problem it looks like this…: (135L)x 1 mole/22.4L =6.03 moles of oxygen gas The liters cancel out and you are left with moles as your units.


So your answer is then 3.058
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Mass of CO₂ produced : 58.67 g

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

CS₂ + 3O₂  -------> CO₂ + 2SO₂

mol of CO₂ based on mol of O₂ as a limiting reactant(CS₂ as an excess reactant)

From the equation, mol ratio of mol CO₂ : mol O₂ = 1 : 3, so mol CO₂  :

\tt \dfrac{1}{3}\times 4=\dfrac{4}{3}

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4 0
2 years ago
A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o
julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

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Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

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