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Marat540 [252]
3 years ago
7

An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a

t atmospheric pressure and room temperature (25∘C)? Assume ideal behavior.What is the partial pressure of oxygen in air at atmospheric pressure (1 atm)? Assume ideal behavior
Chemistry
1 answer:
Lisa [10]3 years ago
3 0

Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.  

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):  

V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol

V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21

P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.  

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3 years ago
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What does it mean to dilute a solution?
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Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to the solution

7 0
3 years ago
Suppose that 3.33 g of acetone at 25.0 °C condenses on the surface of a 44.0-g block of aluminum that is initially at 25 °C. If
jeyben [28]

Answer:

68.6 °C

Explanation:

From conservation of energy, the heat lost by acetone, Q = heat gained by aluminum, Q'

Q = Q'

Q = mL where Q = latent heat of vaporization of acetone, m = mass of acetone = 3.33 g and L = specific latent heat of vaporization of acetone = 518 J/g

Q' = m'c(θ₂ - θ₁) where m' = mass of aluminum = 44.0 g, c = specific heat capacity of aluminum = 0.9 J/g°C, θ₁ = initial temperature of aluminum = 25°C and θ₂ = final temperature of aluminum = unknown

So, mL = m'c(θ₂ - θ₁)

θ₂ - θ₁ = mL/m'c

θ₂ = mL/m'c + θ₁

substituting the values of the variables into the equation, we have

θ₂ = 3.33 g × 518 J/g/(44.0 g × 0.9 J/g°C) + 25 °C

θ₂ = 1724.94 J/(39.6 J/°C) + 25 °C

θ₂ = 43.56 °C + 25 °C

θ₂ = 68.56 °C

θ₂ ≅ 68.6 °C

So, the final temperature (in °C) of the metal block is 68.6 °C.

5 0
3 years ago
An ideal diatomic gas at 80 k is slowly compressed adiabatically and reversibly to half its volume. what is its final temperatur
Yanka [14]

An ideal diatomic gas at 80 k is slowly compressed adiabatically and reversibly to half its volume. The final temperature is 104.8k

<h3>Calculation of final temperature </h3>

The formula used for compression is:-

TV^(γ-1)=C

where;

T= temperature=80k

V=volume(given volume is half of its original volume i.e v/2)

γ=CP/CV= ( 7.R/2)/(5R/2)=7/5

C= constant

using  the given  values in the formula;

80 x V^[(7/5)-1]=T(final) (V/2)^[(7/5)-1]

80=T(final) x (V/V2)^(2/5)

80=T(final)  x (1/2)^(2/5)

T(final)= 80 x  (2)^2/5

T(final)= 80 x 1.31

final temperature  =104.8k

Learn more about Temperature here:-

brainly.com/question/14466404

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6 0
1 year ago
Calculate the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 22
steposvetlana [31]

Answer : The value of \Delta G_{rxn} is -8.65 kJ/mol

Explanation :

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln Q    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -16.7 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient = 22.7

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-16.7kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (310K)\times \ln (22.7)

\Delta G_{rxn}=-8.65kJ/mol

Therefore, the value of \Delta G_{rxn} is -8.65 kJ/mol

8 0
4 years ago
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