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Serggg [28]
3 years ago
5

According to the ideal gas law, a 9.585 mol sample of krypton gas in a 0.8032 L container at 504.9 K should exert a pressure of

494.4 atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For Kr gas, a = 2.318 L2atm/mol2 and b = 3.978×10-2 L/mol. %
Chemistry
1 answer:
natka813 [3]3 years ago
6 0

Answer:

1.17%

Explanation:

The pressure using vander waals' equation = 610.955atm

The difference

  • 610.955-494.4 =116.555atm

The % difference = 1.17%

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Answer:

P O 2 = 5.21 atm P C O 2 = 4.79 atm

Explanation:

Hope it helps!

6 0
3 years ago
Determine whether each of the molecules below is polar or nonpolar. Tetrahedral C C l 4 CClX4 Nonpolar Tetrahedral C H 3 O H CHX
liraira [26]

Answer:

CCl4 - Nonpolar

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NH3 - polar

CS2 - Nonpolar

Explanation:

One important thing that we should know is that polarity has to do with the presence of a resultant dipole moment in a molecule.

Dipole moment is a vector quantity, This means that its direction is also taken into account when discussing the dipole moment of molecules.

Hence, symmetrical molecules such as CS2 and CCl4 are non-polar even though they have polar bonds because their dipoles cancel out(zero resultant dipole moment).

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3 years ago
The 10x SDS gel electrophoresis buffer contains 250mM Tris HCl, 1.92M Glycine, and 1% (w/v) SDS. Buffers are always used at 1x c
olchik [2.2K]

Answer:

25 mM Tris HCl and 0.1% w/v SDS

Explanation:

A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:

(250 mM Tris HCl)/10 = 25 mM Tris HCl

(1.92 M glycine)/10 = 0.192 M glycine

(1% w/v SDS)/10 = 0.1% w/v SDS

Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.

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