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Helga [31]
3 years ago
10

A 25-liter sample of steam at 100°c and 1.0 atm is cooled to 25°c and expanded until the pressure is 19.71 mmhg. if no water con

denses, calculate the final volume of the water vapor. 770 l 240 l 1.3 x 10-3 l 1.0 l
Chemistry
2 answers:
qwelly [4]3 years ago
8 0
I will use [pV/T] in the state 1 = [pV/T] in the state 2.

State 1:
p = 1.0 atm
V = 25 liter
T = 100 + 273.15 = 373.15 K

State 2:

p = 19.71 mmHg * 1.atm / 760 mmHg = 0.0259atm
V= ?
T = 25 + 273.15 = 298.15 K

Application of the formula

1.0 atm * 25 liter / 373.15 k = 0.0259 atm * V / 298.15 K =>
V = [1.0atm * 25 liter / 373.15 K]*298.15K/0.0259atm = 771 liter

 
abruzzese [7]3 years ago
6 0

the correct answer is 770 L. I had this question this morning.

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The right answer is 2.

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B. A precipitate will form since Q > Ksp for calcium oxalate

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Ksp of CaC₂O₄ is:

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Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:

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In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.

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[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.

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Thus, right answer is:

<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>

<em></em>

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