31.12.2021

what is the final pressure for C3H8

. 4

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Biology
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P Answered by Specialist

D. In a low-pressure system, air sinks, creating fair weather. In a high-pressure system, air rises, creating storms.

Explanation :A low-pressure system usually brings clouds and precipitation, while a high-pressure system brings cool dry air and clear skies. A low-pressure system usually brings warm dry air and clear skies, while a high-pressure system usually brings warm moist air and precipitation. In low pressure system air rises while in a high pressure system air sinks

Physics
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P Answered by PhD

The pressure at a depth 01 25 m is 2.5 x 10^5 Pa. This is almost 2.5 times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.

Explanation:

The depth is h = 25 m

The density of water p = 1000 kg/m^3

Acceleration due to gravity g = 9.81 m/s^3.

Pressure due to a depth is gotten from

P = pgh

P = 1000 x 9.81 x 25 =245250 Pa

==> 2.45 x 10^5 Pa

Approximately 2.5 x 10^5 Pa

Atmospheric pressure = 1.01325 x 10^5 Pa

Dividing the pressure at the bottom of the pond by the atmospheric pressure gives a value of about 2.5,which means that the pressure is 2.5 times the atmospheric pressure.

Biology
Step-by-step answer
P Answered by Specialist

D. In a low-pressure system, air sinks, creating fair weather. In a high-pressure system, air rises, creating storms.

Explanation :A low-pressure system usually brings clouds and precipitation, while a high-pressure system brings cool dry air and clear skies. A low-pressure system usually brings warm dry air and clear skies, while a high-pressure system usually brings warm moist air and precipitation. In low pressure system air rises while in a high pressure system air sinks

Physics
Step-by-step answer
P Answered by PhD

The pressure at a depth 01 25 m is 2.5 x 10^5 Pa. This is almost 2.5 times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.

Explanation:

The depth is h = 25 m

The density of water p = 1000 kg/m^3

Acceleration due to gravity g = 9.81 m/s^3.

Pressure due to a depth is gotten from

P = pgh

P = 1000 x 9.81 x 25 =245250 Pa

==> 2.45 x 10^5 Pa

Approximately 2.5 x 10^5 Pa

Atmospheric pressure = 1.01325 x 10^5 Pa

Dividing the pressure at the bottom of the pond by the atmospheric pressure gives a value of about 2.5,which means that the pressure is 2.5 times the atmospheric pressure.

Chemistry
Step-by-step answer
P Answered by Specialist

Explanation:

A) Pressure at which HPLC procedure is running = P = 2.04\times 10^8 Pa

1 Torr = 133.322 Pascal

P=2.04\times 10^8 Pa=\frac{2.04\times 10^8 }{133.322 } Torr=1,530.13 Torr

The running pressure in Torr is 1,530.13.

B)Initial temperature if the gas in balloon = T_1=37^oC=310.15 K

Initial volume of the gas in the balloon = V_1=1.12\times 10^3 L

Final temperature if the gas in balloon = T_2=58^oC=331.15 K

Final volume of the gas in the balloon = V_2=?

Using Charles law:

\frac{V_1}{T_1}=\frac{V_2}{T_2} (constant pressure)

V_2=\frac{V_1\times T_2}{T_1}=\frac{1.12\times 10^3 L\times 331.15 K}{310.15 K}=1.196\times 10^3 L

1.196\times 10^3 L is the new volume of the gas.

C) Initial temperature if the gas in balloon = T_1=20^oC=293.15 K

Initial volume of the gas in the balloon = V_1=5.00 L

Initial pressure of the gas in the balloon = P_1=760 mmHg

Final temperature if the gas in balloon = T_2=-50^oC=223.15 K

Final volume of the gas in the balloon = V_2=?

Final pressure of the gas in the balloon = P_2=76 mmHg

Using combine gas law:

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

V_2=\frac{P_1V_1\times T_2}{T_1\times P_2}

=\frac{760 mmHg\times 5.00 L\times 23.15 K}{293.15 K\times 76 mmHg}

V_2=38.06 L

38.06 liters is the new volume of the balloon.

D) Initial temperature if the gas in container= T_1=20^oC=293.15 K

Initial volume of the gas in the container = V_1=4.60 L

Initial pressure of the gas in the container= P_1=365 mmHg

Final temperature if the gas in container= T_2=30^oC=303.15 K

Final volume of the gas in the container= V_2=2.40 L

Final pressure of the gas in the container= P_2=?

Using combine gas law:

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

P_2=\frac{P_1V_1\times T_2}{T_1\times V_2}

=\frac{365 mmHg\times 4.60 L\times 303.15 K}{293.15 K\times 2.4 L}

P_2=723.45 mmHg

723.45 mmHg is the new pressure inside the container.

Chemistry
Step-by-step answer
P Answered by Specialist

Explanation:

A) Pressure at which HPLC procedure is running = P = 2.04\times 10^8 Pa

1 Torr = 133.322 Pascal

P=2.04\times 10^8 Pa=\frac{2.04\times 10^8 }{133.322 } Torr=1,530.13 Torr

The running pressure in Torr is 1,530.13.

B)Initial temperature if the gas in balloon = T_1=37^oC=310.15 K

Initial volume of the gas in the balloon = V_1=1.12\times 10^3 L

Final temperature if the gas in balloon = T_2=58^oC=331.15 K

Final volume of the gas in the balloon = V_2=?

Using Charles law:

\frac{V_1}{T_1}=\frac{V_2}{T_2} (constant pressure)

V_2=\frac{V_1\times T_2}{T_1}=\frac{1.12\times 10^3 L\times 331.15 K}{310.15 K}=1.196\times 10^3 L

1.196\times 10^3 L is the new volume of the gas.

C) Initial temperature if the gas in balloon = T_1=20^oC=293.15 K

Initial volume of the gas in the balloon = V_1=5.00 L

Initial pressure of the gas in the balloon = P_1=760 mmHg

Final temperature if the gas in balloon = T_2=-50^oC=223.15 K

Final volume of the gas in the balloon = V_2=?

Final pressure of the gas in the balloon = P_2=76 mmHg

Using combine gas law:

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

V_2=\frac{P_1V_1\times T_2}{T_1\times P_2}

=\frac{760 mmHg\times 5.00 L\times 23.15 K}{293.15 K\times 76 mmHg}

V_2=38.06 L

38.06 liters is the new volume of the balloon.

D) Initial temperature if the gas in container= T_1=20^oC=293.15 K

Initial volume of the gas in the container = V_1=4.60 L

Initial pressure of the gas in the container= P_1=365 mmHg

Final temperature if the gas in container= T_2=30^oC=303.15 K

Final volume of the gas in the container= V_2=2.40 L

Final pressure of the gas in the container= P_2=?

Using combine gas law:

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

P_2=\frac{P_1V_1\times T_2}{T_1\times V_2}

=\frac{365 mmHg\times 4.60 L\times 303.15 K}{293.15 K\times 2.4 L}

P_2=723.45 mmHg

723.45 mmHg is the new pressure inside the container.

Geography
Step-by-step answer
P Answered by PhD
Actually,  A. is incorrect, I just took the test right now and I got 80 %(i passed but only 1 wrong). on it and I looked it over, and the incorrect answer I got was the last one witch was this one. So, I looked at it to see what was the correct one, and the correct answer is B. Cool air near surface forms high-pressure areas, warm air forms low pressure areas. I hope this helps :D :)
Biology
Step-by-step answer
P Answered by Specialist
B, because if you think about it, you're climbing a mountain and the mountain is at a very high elevation right? it is very cold and the pressure of the air is high. now when you go down, it gets warmer and the pressure isn't as bad.

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