1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
AlexFokin [52]
2 years ago
8

Suppose that Daniel has a 3.00 3.00 L bottle that contains a mixture of O 2 O2 , N 2 N2 , and CO 2 CO2 under a total pressure of

4.80 4.80 atm. He knows that the mixture contains 0.230 0.230 mol N 2 N2 and that the partial pressure of CO 2 CO2 is 0.350 0.350 atm. If the temperature is 273 273 K, what is the partial pressure of O 2 O2
Chemistry
2 answers:
vazorg [7]2 years ago
7 0

Answer:

The partial pressure of O2 is 2.73 atm

Explanation:

Step 1: Data given

Volume of the bottle = 3.00 L

Total pressure = 4.80 atm

Moles N2 = 0.230 moles

Partial pressure CO2 = 0.350 atm

Temperature = 273 K

Step 2: Calculate total number of moles

p*V = nRT

⇒ p= the pressure = 4.80 atm

⇒V = the volume = 3.00 L

⇒ n = the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 273 K

n = (pV)/(RT)

n = (4.80*3.00)/(0.08206*273)

n = 0.643 moles

Step 3: Calculate mol fraction CO2

Mol fraction CO2 = partial pressure/ total pressure

Mol fraction CO2 = 0.350 atm / 4.80 atm

Mol fraction CO2 = 0.0729

Step 4: Calculate moles CO2

Moles CO2 = mol fraction * total moles

Moles CO2 = 0.0729 * 0.643 moles

Moles CO2 = 0.0469 moles

Step 5: Calculate moles O2

moles O2 = total moles - moles N2 - moles CO2

moles O2 = 0.643 - 0.230 - 0.0469

moles O2 = 0.3661 moles

Step 6: Calculate mol fraction O2

Mol fraction O2 = 0.3661 / 0.643

mol fraction O2 = 0.569

Step 7: Calculate partial pressure O2

Partial pressure O2 = mol fraction * total pressure

Partial pressure O2 = 0.569 * 4.80 atm

Partial pressure O2 = 2.73 atm

The partial pressure of O2 is 2.73 atm

Alenkinab [10]2 years ago
5 0

Answer:

Partial pressure O₂ → 2.74 atm

Explanation:

Let's analyse the data given:

Volume → 3L

In the bottle there is a mixture of gases that contains, O₂, N₂ and CO₂.

Total pressure is 4.80 atm

Let's apply the Ideal Gases Law to determine the total moles of the mixture

P . V = n .  R. T

4.80 atm . 3L = n . 0.082 . 273K

n = 4.80 atm . 3L / 0.082 . 273K → 0.643 moles

We apply the concept of mole fraction:

Mole fraction of a gas X = moles of gas X / Total moles

Mole fraction of a gas X = Partial pressure X / Total pressure

In a mixture, sum of mole fraction of each gas = 1

We determine mole fraction of N₂ → 0.230 / 0.643 = 0.357

We determine mole fraction of CO₂ → 0.350 atm / 4.80 atm = 0.0729

1 - mole fraction N₂ - mole fraction CO₂ = mole fraction O₂

1 - 0.357 - 0.0729 = 0.5701 → mole fraction O₂

We replace in the formula: Mole fraction O₂ = Partial pressure O₂ / 4.80 atm

0.5701 . 4.80 atm = Partial pressure O₂ → 2.74 atm

You might be interested in
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
2 years ago
Read 2 more answers
What happened to the reaction below
Citrus2011 [14]

c the temperature is increased?5)

3 0
3 years ago
(a) the characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustio
ella [17]

By stoichiometry and assume that:

CxH2xOy + zO2 -> xCO2 + xH2O 

<span>
CO2: 9.48/44 = 0.215 mmol 
H2O: 3.87/18 = 0.215 mmol 
mass of C = 0.215 * 12 = 2.58 mg 
mass of H = 0.215 * 2 * 1 = 0.43 mg 
mass of O in ethylbutyrate = 4.17 - 2.58 - 0.43 = 1.11 mg 
So C/O = 2.58/1.11 ≈ 3 </span>

<span>
Thus we have C3H6O</span>

<span> </span>

8 0
2 years ago
Read 2 more answers
Which statement is true about a liquid but not a gas?
vovangra [49]

Answer:

I think its D

Explanation:

.........................

3 0
3 years ago
Read 2 more answers
The pressure on a sample of gas is increased from 1.0 atm to 3.0 atm. If the new volume is 0.52 L, find the original volume.
swat32

Answer: 1.56 ATM

Explanation: if we assume temperature is constant, gas obeys

Boyles law pV= constant. Then p1·V1= p2·V2. And V1 = p2V2/p1

= 3.0 atm·0,52 l / 1.0 atm

3 0
3 years ago
Other questions:
  • What is this question in with the question tell me a funny joke cuz I'm having a bad day ​
    10·2 answers
  • What are the empirical formulas of a) ethylene glycol, a radiator antifreeze, molecular formula C2H6O2 b) peroxodisulfuric acid,
    12·1 answer
  • What gas is used in freezers and fridges? ​
    12·1 answer
  • For what reasons would modern scientists accept dalton's atomic theory but not the idea of an atom proposed by the greek philoso
    13·1 answer
  • The common laboratory solvent ethanol is often used to purify substances dissolved in it. The vapor pressure of ethanol , CH3CH2
    13·1 answer
  • In the Haber Process, hydrogen (H2) and nitrogen (N2) react to produce
    11·1 answer
  • What element does not follow the horizontal and verticals trend?
    7·1 answer
  • The compound KOH is called
    14·2 answers
  • N2(g) + 3H2(g) -&gt; 2NH 3(g)
    10·1 answer
  • Which layer of the soil profile would be affected the most by weathering and erosion?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!