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gogolik [260]
3 years ago
6

Please help

Chemistry
2 answers:
lara [203]3 years ago
7 0

Explanation:

Ideal gas law states that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.

Mathematically,      PV = nRT

where,   P = pressure

             V = volume

             n = number of moles

             R = gas constant

            T = absolute temperature

A gas behaves ideally at high temperature and low pressure because when pressure is low and temperature is high then there will be more number of collisions between the gas particles. As a result, kinetic energy of molecules will increase hence, molecule's size will become less significant as compared to the empty space between them.

Therefore, conditions for which the ideal gas law will be most accurate are as follows.

  • high temperatures
  • low pressures

kotegsom [21]3 years ago
4 0
High temp and low pressure
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A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains
Taya2010 [7]

Answer:

[Cu²⁺] = 2.01x10⁻²⁶

Explanation:

The equilibrium of Cu(CN)₄²⁻ is:

Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻

And Kf is defined as:

Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴

As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:

[Cu²⁺] = 0

[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M

[Cu(CN)₄²⁻] = 2.2x10⁻³

Some [Cu²⁺] will be formed and equilibrium concentrations will be:

[Cu²⁺] = X

[CN⁻] = 0.3212M + 4X

[Cu(CN)₄²⁻] = 2.2x10⁻³ - X

<em>Where X is reaction coordinate</em>

<em />

Replacing in Kf equation:

1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴

1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0

Solving for X:

X = 2.01x10⁻²⁶

As

[Cu²⁺] = X

<h3>[Cu²⁺] = 2.01x10⁻²⁶</h3>

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