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4 years ago

If 8.00 mol of an ideal gas at stp were confined to a cube, what would be the length in cm of an edge of this cube?

1 answer:

6 0

To solve this problem it is fundamentally, just look for the volume of the gas and convert it to cm3. At STP 1 mole = 22.4 liters. 8.00 moles x 22.4 liters/mole = 179.2 liters = 179,200 cm^3 Then. get the cube root of 179,200 cm^3. This would be equal to 56.38 cm and thus would be the length of the edge of this cube.

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50 POINTS! PLEASE HELP!Solid sodium nitrate reacts with sulfuric acid to produce sodium sulfate and nitric acid. You have 36 gra

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3 years ago

A substance absorbed 895 J and caused its temperature to increase by 10 ºC. If the substance has a mass of 27.9 g, what is the s

Irina18 [472]

Answer:

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3 years ago

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kipiarov [429]

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2 years ago

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What is the freezing point of a solution prepared from 45.0 g ethylene glycol (C2H6O2) and 85.0 g H2O? Kf of water is 1.86°C/m.

Andrews [41]

Answer:

$T_{sol}=-15.9\°C$

Explanation:

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In this case, we can analyze the colligative property of solutions - freezing point depression - for the formed solution when ethylene glycol mixes with water. Thus, since water freezes at 0 °C, we can compute the freezing point of the solution as shown below:

$T_{sol}=T_{water}-i*m*Kf$

Whereas the van't Hoff factor for this solute is 1 as it is nonionizing and the molality is:

$m=\frac{mol_{solute}}{kg\ of\ water}=\frac{45.0g*\frac{1mol}{62g} }{85.0g*\frac{1kg}{1000g} } =8.54m$

Thus, we obtain:

$T_{sol}=0\°C+(-8.54m*1.86\frac{\°C}{m} )\\\\T_{sol}=-15.9\°C$

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3 years ago

Can someone PLEASE help me with this im struggling bad

Art [367]

Answer:

4.5+2.34= 6.84
4.5-5 =-0.5
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3.4×2.32 = 7.888
7.77/2.3= 3.37

7. 1200×23.4=28080

9. = 78.512

10. =341.199

11= 7.45

12 =65.0023

13.=3400210.34

Explanation:

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3 years ago