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Vikentia [17]
3 years ago
13

Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 18.0 atm. Calculate the work, w , if the gas exp

ands against a constant external pressure of 1.00 atm to a final volume of 18.0 L.
Chemistry
1 answer:
Dominik [7]3 years ago
6 0

Answer:

-32821.2 J (negative sign implies that work is done by the system)

Explanation:

The expression for the calculation of work done is shown below as:

w=-P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

From the question,  

\Delta V = 18.0 - 1.00 L = 18.0 L

P = 18.0 atm

w=-18.0\times18.0\ atmL

Also, 1 atmL = 101.3 J

So,  

w=-18.0\times 18.0\times 101.3\ J=-32821.2\ J (negative sign implies that work is done by the system)

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Predict what the Co, levels will be when you are "old". How many plants/trees need to be planted now to keep carbon dioxide in t
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Answer:

A lot of them.

Explanation:

It would take hundreds of thousands of trees to clear all of the emmisions.

5 0
3 years ago
URGENT!
irga5000 [103]

Answer:

GAY-LUSSAC'S LAW.

Explanation:

P1 = 765 torr

T1 = 23°C = 296K

P2 = 560. torr

T2 = ?

(765 torr)/(296K) = (560. torr)/ T2

T2 = 226 K = -57°C

7 0
2 years ago
To understand the relation between the strength of an acid or a base and its pKa and pKb values. The degree to which a weak acid
elena-14-01-66 [18.8K]

Answer:

pKa = 3.675

Explanation:

  • pKa = - Log Ka

∴ <em>C</em> X-281 = 0.079 M

∴ pH = 2.40

let X-281 a weak acid ( HA ):

∴ HA ↔ H+ + A-

⇒ Ka = [H+] * [A-] / [HA]

mass balance:

⇒<em> C</em> HA = 0.079 M = [HA] + [A-]

⇒ [HA] = 0.079 - [A-]

charge balance:

⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water

⇒ [H+] = [A-]

∴ pH = - log [H+] = 2.40

⇒ [H+] = 3.981 E-3 M

replacing in Ka:

⇒ Ka = [H+]² / ( 0.079 - [H+] )

⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )

⇒ Ka = 2.113 E-4

⇒ pKa = - Log ( 2.113 E-4 )

⇒ pKa = 3.675

8 0
3 years ago
A student weighed an empty graduated cylinder. It weighed 35.86 g. She then carefully added water to the graduated cylinder unti
Firdavs [7]

Question:

A student weighed an empty graduated cylinder. It weighed 35.86 g. She then carefully added water to the graduated cylinder until it reached the 7.5 mL mark. When she weighed the graduated cylinder again, this time with the 7.5 mL of water in it, it weighed 43.18 g. What was this student's experimental density of water?

Answer:

0.976 g/mL

Explanation:

Weight of empty cylinder = 35.86g

Volume of water = 7.5mL

Weight of cylinder + water = 43.18g

Experimental density = ?

Density of water = Mass of water / volume of water

Mass of water = (Weight of cylinder + water) - Weight of empty cylinder

Mass of water = 43.18 - 35.86 = 7.32g

Density = 7.32 / 7.5 = 0.976 g/mL

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3 years ago
Where is the earth curst thinkest
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The crust of the Earth is thickest beneath the continents. 
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