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swat32
3 years ago
5

A supersaturated solution can be prepared by dissolving solute in solvent while adding ______. A supersaturated solution contain

s more ______ than can ordinarily be dissolved in the solvent at room temperature. A solution may remain supersaturated until _______ is initiated, often by adding solid to the solution or by allowing solvent to evaporate.
Chemistry
1 answer:
Delicious77 [7]3 years ago
4 0

Answer:

Heat, solutes and high temperature.

Explanation:

A supersaturated solution can be formed by dissolving solute more solute in solvent by increasing temperature of the solution. A supersaturated solution contains more quantity of solutes than can be dissolved in the solvent at room temperature. A solution may remain supersaturated until the solution has high temperature and when the temperature started lower, the extra dissolve solutes begin undissolved and remain suspended in the solution.

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At 400 K oxalic acid decomposes according to the reaction:H2C2O4(g)→CO2(g)+HCOOH(g)In three separate experiments, the intial pre
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Answer:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

Explanation:

For the reaction:

H₂C₂O₄(g) → CO₂(g) + HCOOH(g)

At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:

H₂C₂O₄(g) = P₀ - x

CO₂(g) = x

HCOOH(g) = x

P at t=20000 is:

P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x

For 1st point:

x = 92,8-65,8 = 27

Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8

2nd point:

x = 130-92,1 = 37,9

H₂C₂O₄(g): 92,1 - 37,9 = 54,2

3rd point:

x = 157-111 = 46

H₂C₂O₄(g): 111-46 = 65

Now, as the rate law is :

v = k P[H₂C₂O₄]

Based on integrated rate law, k is:

(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k

1st point:

k = 2,64x10⁻⁵

2nd point:

k = 2,65x10⁻⁵

3rd point:

k = 2,68x10⁻⁵

The averrage of this values is:

k = 2,66x10⁻⁵

That means law is:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

I hope it helps!

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at t= t_{1/2}, x=\frac{a}{2}

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