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Tanzania [10]
3 years ago
12

Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomi

c gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are diff erent. Determine the ratio KEKrypton KEArgon of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom.

Chemistry
1 answer:
azamat3 years ago
7 0

Answer:

2.1

Explanation:

The solution is in the attached file below

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Which of the following BEST describes a way in which water moves in the water cycle?
____ [38]

Answer: Its A or D                                                        

wish i had an actual answer sorry..  

6 0
3 years ago
Read 2 more answers
Which corresponds to a pressure of 1.23 atm??
Vadim26 [7]

Atmospheric pressure<span>, sometimes also called barometric pressure, is the pressure exerted by the weight of air in the </span>atmosphere of Earth<span> (or that of another planet)</span>

1 atm is equivalent to = 101325 Pa

= 760 mmHg

= 760 torr

= 1.01325 bar

So 1.23 atm is equal to

= 124629.8 Pa

= 934.8 mmHg

= 934.8 torr

<span>= 1.2462 bar</span>

6 0
3 years ago
Liquids that do not mix are called
djyliett [7]

Answer:

Immiscible liquids

Explanation:

Immiscible comes from two words; 'im' indicating a negation or a contradiction; and 'miscere' meaning to mix.

Putting the words together, immiscible means not able to mix.

3 0
3 years ago
A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t
soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K  .  296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K  .  296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K  .  296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0
3 years ago
You measure salt water in a tank to have a density of 1.02 g/mL. A balloon weighs 2.0 g and you weights have a mass of 30.0 g ea
Elden [556K]

Weight of the balloon = 2.0 g

Six weights each of mass 30.0 g is added to the balloon.

Total mass of the balloon = 2.0 g + 6*30.0 g = 182 g

Density of salt water = 1.02 g/mL

Calculating the volume from mass and density:

182g*\frac{mL}{1.02g} =178mL

Converting the volume from mL to cubic cm:

178 mL * \frac{1cm^{3} }{1mL} =178cm^{3}

Assuming the balloon to be a sphere,

Volume of the sphere = \frac{4}{3}πr^{3}

178 cm^{3} = \frac{4}{3}(\frac{22}{7})r^{3}

r = 3.49 cm

Radius of the balloon = 3.49 cm

Diameter of the balloon = 2 r = 2*3.49 cm = 6.98cm


4 0
3 years ago
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