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Levart [38]
2 years ago
6

How many moles of xenon gas are necessary to exert a 0.64atm pressure in an 8.5L container at 300.0K?

Chemistry
2 answers:
navik [9.2K]2 years ago
5 0

0.026 moles is the right answer.

Sindrei [870]2 years ago
3 0

Answer:

0.026 moles

Explanation:

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Sibling rivalry may occur when:
sweet [91]

Answer:

Explanation: true

7 0
3 years ago
a gas has a volume of 350 cubic centimeters at 740 mmHg. how many cubic centimeters will the gas occupy at a pressure of 900 mmH
Harrizon [31]

Answer: 287.8 cm3

Explanation:

Given that:

Initial volume of gas V1 = 350 cm3

Initial pressure of gas P1 = 740 mmHg

New volume V2 = ?

New pressure P2 = 900 mmHg

Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law

P1V1 = P2V2

740 mmHg x 350 cm3 = 900mmHg x V2

V2 = (740 mmHg x 350 cm3) /900mmHg

V2 = 259000 mmHg cm3 / 900mmHg

V2 = 287.8 cm3

Thus, the gas will occupy 287.8 cubic centimeters at the new pressure.

3 0
3 years ago
What's larger centimeter or meter?
hoa [83]

Answer:

A meter is 100 times larger then a centimeter

6 0
2 years ago
What is the purpose of the ice?
kompoz [17]

Answer:

Ice act and help in condensation of the vapor

Explanation:

The energy released when gaseous water vapor condenses to form liquid water droplets is called latent heat. Latent heat from condensation causes an increase in air temperature surrounding the water droplets. The warmer air rises, causing the water vapor to condense when it meets cooler air at a higher altitude

3 0
2 years ago
We might think of a porous material as being a composite wherein one of the phases is a pore phase. Estimate upper and lower lim
eduard

Answer:

The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are

Ku = 38.252 W/mK

K lower = 0.199 W/mK

Explanation:

As we know  

Ku = Vp * Kair + Vmagnesium * K metal  

Ku = 0.10 *0.02 + (1-0.25) * 51

Ku = 38.252 W/mK

The lower limit  

K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)

K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)

K lower = 0.199 W/mK

8 0
2 years ago
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