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notsponge [240]
3 years ago
7

a 20.0 liter flask contains a mixture of argon at 0.72 atmosphere and oxygen at 1.65 atmosphere. what is the total pressure in t

he flask
Chemistry
1 answer:
RoseWind [281]3 years ago
7 0

Answer:

Total pressure in flask is 2.37 atm.

Explanation:

According to the Dalton's law of partial pressure, the total pressure of the flask would be the sum of partial pressure of the gases present in mixture.

P(total) = P1+ P2+P3+...Pn

n= number of gases present

Given data:

Pressure of argon gas = 0.72 atm

Pressure of oxygen gas = 1.65 atm

Total pressure in flask = ?

Solution:

P(total) = P (argon) + P (oxygen)

P(total) = 0.72 + 1.65

P(total) =  2.37 atm

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Answer:

1.18x10⁸L of SO₂ and 2.36x10⁸L of H₂S

Explanation:

The balanced reaction is:

8SO₂(g) + 16H₂S(g) → 16H₂O(l) + 3S₈(s)

To solve this question we must find the moles of S₈ in 4.50x10⁵kg. With these moles and the reaction we can find the moles of SO₂ needed to react (Twice these moles = Moles Of H₂S needed). Using PV = nRT we can find the volume of the gas required:

<em>Moles S₈ - molar mass: 256.52g/mol-</em>

4.50x10⁵kg = 4.50x10⁸g * (1mol / 256.52g) =

1.75x10⁶ moles S₈

<em>Moles SO₂:</em>

1.75x10⁶ moles S₈ * (8mol SO₂ / 3mol S₈) = 4.68x10⁶ moles SO₂

<em>Moles H₂S:</em>

4.68x10⁶ moles SO₂ * 2 = 9.36x10⁶ moles H₂S

The volume could be obtained as follows:

PV = nRT

V = nRT / P

<em>V is volume in liters</em>

<em>n are moles: 4.68x10⁶ moles SO₂ and 9.36x10⁶ moles H₂S</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 22°C + 273.15 = 295.15K</em>

<em>P is pressure = 0.961atm</em>

<em />

Replacing:

Volume SO₂ and H₂S:

4.68x10⁶ moles * 0.082atmL/molK * 295.15K / 0.961atm =

<h3>1.18x10⁸L of SO₂ and:</h3>

<em>9.36x10⁶ moles H₂S</em> * 0.082atmL/molK * 295.15K / 0.961atm =

<h3>2.36x10⁸L of H₂S</h3>

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3 years ago
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If an automobile air bag has a volume of 11.7 L, what mass of NaN3 (in g) is required to fully inflate the air bag upon impact
Usimov [2.4K]

Answer:

33.95 grams of NaN3

Explanation:

Number of moles of NaN3 = mass (m)/MW = m/65 mole

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5 0
3 years ago
How many distinct dichlorination products can result when isobutane is subjected to free radical chlorination?
Elenna [48]

Answer:

c. 3

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When Isobutane is subjected to free radicals chlorination, three distinct dichlorination can be formed.

The mechanism of the formation of these products can be seen in the image attached below.

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