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Levart [38]
2 years ago
6

How many moles of xenon gas are necessary to exert a 0.64atm pressure in an 8.5L container at 300.0K?

Chemistry
2 answers:
navik [9.2K]2 years ago
5 0

0.026 moles is the right answer.

Sindrei [870]2 years ago
3 0

Answer:

0.026 moles

Explanation:

You might be interested in
An ideal gas originally at 0.85 atm and 66°C was allowed to expand until its final volume, pressure, and temperature were 94 mL,
svet-max [94.6K]

Answer : The initial volume was, 71.2 mL

Explanation :

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

P_1=0.85atm\\V_1=?\\T_1=66^oC=[66+273]K=339K\\P_2=0.60atm\\V_2=94mL\\T_2=43^oC=[43+273]K=316K

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

\frac{0.85atm\times V_1}{339K}=\frac{0.60atm\times 94mL}{316K}

V_1=71.2mL

Therefore, the initial volume was, 71.2 mL

4 0
3 years ago
The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at
horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol  

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0
3 years ago
What is the name of the average weather in an area over a long period of time?
Nezavi [6.7K]

Climate is the weather that occur over a long period in a particular place.

3 0
2 years ago
How should you remove a beaker from a hot plate after heating it?.
rewona [7]

Answer:

Explanation:

you will have to grabe  a towle or a meten and take it off

3 0
2 years ago
A sample of water at 20 degrees c contains what bonds
vladimir2022 [97]
Covalent and hydrogen bonds

3 0
3 years ago
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