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ikadub [295]
3 years ago
15

Stoichiometry problems which give a known amount of one reactant and solve for an amount of product frequently state that there

is "an excess" or "plenty" of the reactant(s) not given. (For example, in the equation 2Na + 2H2O → 2NaOH + H2, how many grams of sodium hydroxide are produced from 3.0 mol of sodium with an excess of water?) Why is it necessary to know that there is an excess of these reactant(s)?
Chemistry
1 answer:
gtnhenbr [62]3 years ago
5 0
Dumb. Can. Cabbage. Hn. D ikkk
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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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the answer is c

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\\ \sf\longmapsto E=hv

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