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3 years ago

Please help will mark Brainly

Chemistry

1 answer:

Murljashka [212]3 years ago

5 0

C: tt

Remember, individuals must have two alleles for a gene, so someone could be homozygous dominant (EE), homozygous recessive (ee), or heterozygous (Ee). In this example, the recessive trait is having attached earlobes and is only seen in someone with the genotype ee (homozygous recessive)

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C6H6 miscible or immiscible

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C6H6 is benzene, an organic compound. I would imagine it would be immiscible, since most organic compounds don’t dissolve in water very well.

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If 1.00 mol of argon is placed in a 0.500-L container at 27.0 degree C , what is the difference between the ideal pressure (as p

ElenaW [278]

Answer:

2.0 atm is the difference between the ideal pressure and the real pressure.

Explanation:

If 1.00 mole of argon is placed in a 0.500-L container at 27.0 °C

Moles of argon = n = 1.00 mol

Volume of the container,V = 0.500 L

Ideal pressure of the gas = P

Temperature of the gas,T = 27 °C = 300.15 K[/tex]

Using ideal gas equation:

$PV=nRT$

$P=\frac{1.00 mol\times 0.0821 L atm/mol K\times 300.15 K}{0.500 L}=49.28 atm$

Vander wall's of equation of gases:

The real pressure of the gas= $p_v$

For argon:

$a=1.345 L^2 atm/mol^2$

b=0.03219 L/mol.

$(p_v+(\frac{an^2}{V^2})(V-nb)=nRT$

$(p_v+(\frac{(1.345 L^2 atm/mol^2)\times (1.00 mol)^2}{(0.500 L)^2})(0.500 L-1.00 mol\times 0.03219L/mol)=1.00 mol\times 0.0821 L atm/mol K\times 300.15 K$

$p_v = 47.29 atm$

Difference : $p - p_v= 49.28 atm - 47.29 atm = 1.99 atm\approx 2.0 atm$

2.0 atm is the difference between the ideal pressure and the real pressure.

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