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asambeis [7]
3 years ago
5

The gas in the piston is at constant temperature. A student increases the pressure on the piston from 2 atm to 3 atm. The observ

ation will be summarized in a row of the incomplete table below.



Row

Name

Observation

Constants

1

Boyle's law

?

?

2

Charles’s law

?

?

3

Gay-Lussac’s law

?

Volume, moles of gas

4

Combined gas law

?

?


Which row of the table and which observation would the student fill?

row 1, volume decreases when temperature decreases

row 1, volume decreases when pressure increases

row 3, volume decreases when temperature decreases

row 3, volume decreases when pressure increases
Chemistry
1 answer:
Vesna [10]3 years ago
7 0

Answer: row 1, the volume decreases when the pressure increased

Explanation:

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The amount of solute added.

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Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

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Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

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