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Juli2301 [7.4K]
3 years ago
9

Carbon is often found in long chains composed of carbon and

Chemistry
2 answers:
EastWind [94]3 years ago
6 0
C.) <span>Carbon is often found in long chains composed of carbon and "Hydrogen"

Hope this helps!</span>
vladimir2022 [97]3 years ago
4 0
C) hydrogen as it has a property of catenation to form hydrocarbons as orgaic compounds
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Mrs. Jacobson is pushing a fridge to the right with a force of 100 N. The force pushing it
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Answer:

A

Explanation:

Let's illustrate this; see the attachment.

We see that Mrs. Jacobson is pushing to the right with a force of 100 N and there is another opposite force pushing with a force of 15 N. Since these are in opposite directions, we can say that the force opposite to Mrs. Jacobson is pushing the fridge -15 N to the right (instead of 15 N to the left).

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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

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