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Sindrei [870]
3 years ago
8

An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 C:

Chemistry
2 answers:
yarga [219]3 years ago
8 0

Answer: 1.79 x 10^-5

Explanation: The equilibrium constant of a reaction can be calculated from the quotient of the concentrations of the products over the concentrations of the reactants, with each termed raised to their respective stoichometric coefficients.

For acetic acid, this equilibrium expression is:

Kc=\frac{[H+] [CH3COO-]}{[CH3COOH]}

Replacing the equilibrium concentrations given by the exercise into the expression above, the equilibrium constant, Kc will be obtained and it is found to be equal to 1.79 x 10^-5.

valentina_108 [34]3 years ago
6 0

Answer:

The Kc for the ionization of acetic acid is 1.79x10^-5 M

Explanation:

the reaction is as follows:

CH3COOH = CH3COO- + H+

the constant Kc is calculated with the following equation:

Kc = ([CH3COOH-] * [H+])/[CH3COOH]

Kc = (5.44x10^-4 * 5.44x10^-4)/(1.65x10^-2) = 1.79x10^-5 M

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Mashcka [7]

Answer:

CO

Explanation:

4 0
3 years ago
Calculate the molar solubility of copper(II) arsenate (Cu3(AsO4)2) in water. Use 7.6 x 10^-36 as the solubility product constant
Molodets [167]
Molar solubility<span> is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated. We calculate as follows:

</span>3Cu2+ + 2(AsO4)3-<span> = Cu3(AsO4)2
</span>
7.6 x 10^-36 = (3x^3)(2x^2)
x = 6.62 x 10^-8 M
8 0
3 years ago
Read 2 more answers
What do you call the things in an experiment that must be the same to make it fair​
Pavel [41]

Answer:

controlled variables

Explanation:

3 0
3 years ago
Read 2 more answers
BEST GETS BRAINLIEST!!
tatuchka [14]
Gee.  I'll have to guess at what's "commonly thought".

One thing is the scale.  Nobody has an accurate picture of the scale in
his head, because we never see a true-scale drawing.  THAT's because
it's almost impossible to draw one on paper.

Example:
Shrink the solar system and everything in it so that the Sun
is the size of a quarter (the 25¢ coin).
Then:
-- The Earth is in orbit around the sun, 8.6 feet from it. 
That's close enough that you might think you could find the
shrunken Earth.  Unfortunately, it's only 0.009 inch in diameter.

-- The shrunken Jupiter is a 'huge' gas giant almost 0.1 inch in diameter.
It's orbiting the sun, about 45 feet away from it.

-- The shrunken Uranus is another gas giant, about 0.035 inch in diameter.
It's orbiting the sun, about 165 feet away from it.

-- The nearest star outside of the solar system is 441 MILES away !
On the same shrunken scale !
And there's NOTHING between here and there !  

I think that's the biggest point to make about the REAL solar system ...
its utter emptiness.  With the sun reduced to something you can hold
in your hand, the planets are the size of grains of sand, with hundreds
of feet of nothingness between them.

Same for its mass:  The solar system is approximately nothing but a star.
That's it.  A star, with some dust and some gas around it, and here and there
in the neighborhood a microscopic pebble or a chip of mineral.  But mostly
it's nothing but a star ... if you went around and gathered up all that other
rubbish in the same bag and called it a part of the same solar system, the
sun would still have more than 99% of the total mass, and the bag would
hold less than 1% of it.

Book ... It's getting late, Hillary's fading, and that's all I can think of.
I hope this much is some help.
3 0
3 years ago
Which of these half-reactions represents reduction?
gogolik [260]

Answer: The half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element  takes place is called reduction-half reaction.

For example, the oxidation state of Cr in Cr_{2}O^{2-}_{7} is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}

Similarly, oxidation state of Mn in MnO^{-}_{4} is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}

Thus, we can conclude that half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}
3 0
3 years ago
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