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Drupady [299]
3 years ago
14

Consider three identical flasks filled with different gases. Flask A: CO at 760 torr and 08C Flask B: N2 at 250 torr and 08C Fla

sk C: H2 at 100 torr and 08C
a. In which flask will the molecules have the greatest average kinetic energy?
b. In which flask will the molecules have the greatest root mean square velocity?
c. Which flask will have the greatest number of colli- sions per second with the walls of the container?
Chemistry
1 answer:
jonny [76]3 years ago
7 0

Answer:

a, b, c Flask B

Explanation:

The  root mean square velocity  of a gas is  given by

V = √ ( 3RT/M)

where m is the molar mass, T is the temperature of the gas and R is the gas constant.

Therefore we have that the the velocity of the gas molecules is  inversely proportional to the square root of the molar mass . That is the heavier the molecules the slower they will move which makes sense and the lower its kinetic energy.

The greater the velocity the more collisions the gas will  have against the walls of the container.

The molar mass of CO and N2 are 28 g/mol while H2 is 2 g/mol. Hence hydrogen the lightest  will have a greater kinetic enrgy, the greatest root mean square velocity and the greatest number of collisions with the wall of the container.

Notice that the pressure given for each is of the gases is different being the greatest for CO but pressure does not have an effect on the kinetic enery of the molecules.

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 2.1 grams of ethane i
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Answer:

= 3.78 g H₂O

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Limiting Reactant:

A quick way to determine limiting reactant is to divide moles of reactant by its respective coefficient in the balanced molecular equation. The smaller value is the limiting reactant.

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Since the smaller value is associated with ethane, then ethane is the limiting reactant and the problem is worked from the 0.07 moles of ethane in an excess of O₂.

From the equation stoichiometry ...

2 moles C₂H₆  in an excess of O₂ => 6 moles H₂O

then 0.07 mole C₂H₆  in an excess of O₂ => 6/2(0.07 moles H₂O = 0.21 mole

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