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kolbaska11 [484]
3 years ago
14

A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 42.9 kPa. What is the partial pressure of

carbon dioxide if the partial pressure of oxygen is 6.6 kPa and the partial pressure of nitrogen is 23.0 kPa?
Chemistry
2 answers:
Lady bird [3.3K]3 years ago
4 0
The total Pressure equals the sum of all pressures contained 

<span>Since total pressure and the pressure of nitrogen and oxygen is given, finding the pressure of carbon dioxide is given by: </span>

<span>Pressure of Carbon dioxide = 42.9- 6.6- 23.0 </span>
<span>=13.3kPa </span>
timofeeve [1]3 years ago
4 0

Answer : The partial pressure of the carbon dioxide gas is, 13.3 kPa

Solution :

According to the Dalton's law of partial pressures, the total pressure of an ideal gas mixture is equal to the sum of the partial pressures of the gases in the mixture.

P_T=P_{O_2}+P_{N_2}+P_{CO_2}

where,

P_T = total pressure = 42.9 kPa

P_{O_2} = partial pressure of oxygen gas = 6.6 kPa

P_{N_2} = partial pressure of nitrogen gas = 23.0 kPa

P_{CO_2} = partial pressure of carbon dioxide gas = ?

Now put all the given values in the above formula, we get the partial pressure of carbon dioxide gas.

42.9=6.6+23.0+P_{CO_2}

P_{CO_2}=13.3kPa

Therefore, the partial pressure of the carbon dioxide gas is, 13.3 kPa

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Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

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Explanation:

Based on the reaction:

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<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>

Initial moles of H₂SO₄ and KOH are:

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0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.

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