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andriy [413]
4 years ago
5

Does temperature have an effect on capillary action? If it does, how so?

Chemistry
1 answer:
dybincka [34]4 years ago
5 0

Answer:

Yes, Temperature have an effect on capillary action.

Explanation:

increase in the temperature of liquid results to decrease in density. Hence,  this results in the decrease in capillary rise. Also,increase in temperature leads to expansion of the material of capillary, this also decrease capillary rise.

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3 years ago
If the solubility of a gas in water is 5.0 g/L when the pressure of the gas above the water is 2.5 atm, what is the pressure of
OLga [1]

Answer:

0.5 atm.

Explanation:

  • Henry’s law states that the amount of a gas dissolved in a  solution is directly proportional to the  pressure of the gas above the solution
  • We should use Henry’s law: <em>P = KC,</em>

where, P is the partial pressure of the  gaseous solute above the solution.

k is a constant.

C is the concentration of the  dissolved gas.

<em>At two different pressures:</em>

<em>P₁C₂ = P₂C₁,</em>

P₁ = 2.5 atm, C₁ = 5.0 g/L.

P₂ = ??? atm, C₂ = 1.0 g/L.

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3 years ago
Calculate the percent ionization of nitrous acid in a solution that is 0.139 M in nitrous acid. The acid dissociation constant o
anzhelika [568]
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 <-> H(+) + NO2(-)

Next, create an ICE table

           HNO2   <-->  H+        NO2-
[]i        0.139M          0M       0M
Δ[]      -x                   +x         +x
[]f        0.139-x          x           x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x / .139 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable, 

assume 0.139-x ≈ 0.139

4.5x10^-4 = x^2/0.139

Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.

Then to find percent dissociation, you do final concentration/initial concentration.

0.007M/0.139M = .0503 or 

≈5.03% dissociation.
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3 years ago
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