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sertanlavr [38]
3 years ago
9

Which would not be associated with stable atmospheric conditions

Chemistry
2 answers:
dlinn [17]3 years ago
7 0

The correct answer is - afternoon thunder showers.

When we say stable atmospheric conditions it means that the atmosphere should be relatively calm, without having any disturbances in it. A nice example for a stable atmospheric conditions is a clear sunny sky and a calm wind. On the other hand, we have the afternoon thunder showers. The afternoon thunder showers are the total opposite of stable atmospheric conditions. When they occur, there's lot of disturbances, air pressures are mixing, the winds are changing speeds, heavy rainfall occurs, the air pressure changes rapidly, there's electrical charges, all in all creating a real mess in the atmosphere.

Genrish500 [490]3 years ago
5 0

Answer:

afternoon thunder showers.

Explanation:

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A flask of volume 2.0 liters, provided with a stopcock, contains oxygen at 20 oC, 1.0 ATM (1.013X105 Pa). The system is heated t
leonid [27]

Answer:

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

2.6592 grams of oxygen remain in the flask.

Explanation:

Volume of the flask remains constant = V = 2.0 L

Initial pressure of the oxygen gas = P_1=1.0 atm

Initial temperature of the oxygen gas = T_1=20^oC =293.15 K

Final pressure of the oxygen gas = P_2=?

Final temperature of the oxygen gas = T_2=100^oC =373.15 K

Using Gay Lussac's law:

\frac{P_1}{T_1}=\frac{P_2}{T_2}

P_2=\frac{P_1\times T_2}{T_1}=\frac{1 atm\times 373.15 K}{293.15 K}=1.27 atm

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

Moles of oxygen gas = n

P_1V_1=nRT_1 (ideal gas equation)

n=\frac{P_1V_1}{RT_1}=\frac{1 atm\times 2.0 L}{0.0821 atm l/mol K\times 293.15 K}=0.08310 mol

Mass of 0.08310 moles of oxygen gas:

0.08310 mol × 32 g/mol = 2.6592 g

2.6592 grams of oxygen remain in the flask.

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