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kaheart [24]
3 years ago
8

A 10 liter flask at 298 K contains a gaseous mixture of O2 and CO2 at 1 atmosphere. Which statement is true for the partial pres

sures of O2 and CO2 if 0.2 mole of O2 is present in the flask? (Given the universal gas constant R = 0.082 L∙atm/K∙mol)
Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

The partial pressures of O₂ and CO₂ are 0.489 atm and 0.511 atm respectively.

Explanation:

From the Questions we are given;

Volume = 10 Liter

Temperature = 298 K

Pressure = 1 atm

We need to calculate the partial pressures of O₂ and CO₂

Step 1 : Number of moles of gaseous mixture

Using the ideal gas equation;

PV =nRT, where P is the pressure, V is the volume, n is the number of moles, T is the temperature in K and R is the ideal gas constant (0.082 L∙atm/K∙mol)

Therefore;

n =\frac{PV}{RT}

n = \frac{(10)(1)}{(298)(0.082)}

Solving for n

n = 0.409 moles

Step 2: Moles of CO₂

Total number of moles of the mixture = 0.409 moles

Moles of Oxygen = 0.2 moles

Therefore;

Moles of CO₂ = 0.409 moles - 0.2 moles

                      = 0.209 moles

Step 3: Partial pressures of O₂ and CO₂

Partial pressure = \frac{No. of moles}{Total number of moles}(total pressure)

Therefore;

Partial pressure of Oxygen gas

= \frac{number of moles of Oxygen}{Total number of moles} (Total pressure)

= \frac{0.2}{0.409}(1)\\= 0.489 atm

Partial pressure of CO₂

= \frac{number of moles of CO₂}{Total number of moles} (Total pressure)

= \frac{0.209}{0.409}(1)\\= 0.511 atm

Thus, the partial pressures of O₂ and CO₂ are 0.489 atm and 0.511 atm respectively.

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B is the right answer "right"
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3 years ago
Which of the following are not single-displacement reactions?
Goshia [24]

Answer:

\boxed{\text{B and C }}

Explanation:

In a single-displacement reaction, one element exchanges partners with another element in a compound.

\textbf{A. } \rm Fe + 2HCl \longrightarrow FeCl_2 + H_2

This is a single-displacement reaction, because the element Fe exchanges partners with H in HCl.

\textbf{B. } \rm KOH + HNO_3 \longrightarrow H_2O + KNO_3

This is not a single-displacement reaction, because it is a reaction between two compounds.

This is a double displacement reaction in which the K⁺ and H⁺ cations change partners with the anions.

\textbf{C. } \rm Na_2S + 2HCl \longrightarrow 2NaCl + H_2S

This is not a single-displacement reaction. It is another double displacement reaction, in which the Na⁺ and H⁺ cations change partners with the anions.

\textbf{D. } \rm Ca + 2HOH \longrightarrow Ca(OH)_2 + H_2

This is a single-displacement reaction, because the element Ca exchanges partners with H in H₂O.

\boxed{\textbf{B and C }} are not single-displacement reactions.

6 0
3 years ago
Phosphorous would gain or lose electrons?
devlian [24]

Answer: the valence electron for phosphorus is 5. To achieve an octet electron arrangement, it needs to lose 5 electrons or gain 3 electrons. It is easier to gain 3 electrons than to lose 5 electrons. So phosphorus has to gain 3 electrons.

Explanation:

Hope it helps sorry if it doesn't

4 0
3 years ago
. A container with a volume of 0.50 L contains air at a pressure of 1.0 atm. If the volume is reduced to 0.10 L at constant temp
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Answer:

The answer to your question is Pressure = 5 atm

Explanation:

Data

Volume 1 = V1 = 0.5 l

Pressure 1 = P1 = 1 atm

Volume 2 = V2 = 0.1 l

Pressure 2 = P2 = x

Formula

To solve this problem use the Boyle's equation

                      V1P1 = V2P2

Solve for P2

                     P2 = V1P1/V2

Substitution

                     P2 = (0.5 x 1) / 0.1

Simplification

                     P2 = 0.5/0.1

Result

                     P2 = 5 atm          

3 0
3 years ago
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