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Kazeer [188]
3 years ago
12

An empty water bottle is full of air at 15° C and standard pressure . The volume of the bottle is 0.500 liter . How many moles o

f air are in a bottle ?
Chemistry
1 answer:
Yuliya22 [10]3 years ago
6 0

Answer:

There are 0.0211 moles of air in the bottle

Explanation:

<u>Step 1:</u> Data given

Temperature of air = 15°C = 273.15 + 15 = 288.15 Kelvin

Standard pressure = 1 atm

Volume of the bottle = 0.500 L

<u>Step 2</u>: ideal gas law

p*V = n*R*T

with p = the pressure of the gas = 1 atm

with V = the volume of the bottle = 0.500 L

with n = the number of moles

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

with T = the temperature of the gas = 15 °C = 288.15 Kelvin

In this case we have:

1atm * 0.500L = n * 0.08206  L*atm/ mol*K * 288.15 K

Number of moles n = (1 atm * 0.500 L)/(0.08206  L*atm/ mol*K * 288.15 K)

Number of moles = 0.0211 moles

There are 0.0211 moles of air in the bottle

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Answer:

The concentration the student should write down in her lab is 2.2 mol/L

Explanation:

Atomic mass of the elements are:

Na: 22.989 u

S: 32.065 u

O: 15.999 u

Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.

Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.

For mole(s) of Na2S2O3 = (mass taken)/(molar mass)

= (17.240 g)/(158.105 g/mol) = 0.1090 mole.

Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)

= 0.05029 L.

To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:

= (moles of sodium thiosulfate)/(volume of solution in L)

= (0.1090 mole)/(0.05029 L)

= 2.1674 mol/L

6 0
3 years ago
How does the energy of an electron close to the nucleus compare to the energy of electrons farther away from the nucleus?
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The electrons closer to the nucleus would have more energy than the electrons farther away from the nucleus.
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Which of the following correctly describes the state of matter of one of the samples? D is a liquid because it retains its shape
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D is a solid because it retains its shape
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An ice cube tray full of water is put into a freezer. Which energy change occurs in the particles in the water as it undergoes a
o-na [289]

Answer:

A

Explanation:

increase in energy - hotter and/or more reactive

decrease in energy - colder and/or less reactive

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3 0
2 years ago
A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm
Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys)  =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

<u>Step 1</u>: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

<u>(a) reversibly</u>

<u>Step 2:</u> Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

<u>Step 3:</u> Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

<u>Step 4:</u> Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)  

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

<u>(b) against a constant external pressure of 1.00 atm</u>

<u>Step 1</u>: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

   ⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

     ⇒ Vf = (1*0.08206 *300)/1

      ⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

<u>Step 2:</u> Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

<u>Step 3: </u>Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0
3 years ago
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