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gogolik [260]
3 years ago
15

4.2 L of a gas has a pressure of 560mmHg at 72⁰F. What is the temperature at760mmHg if the volume is still 4.2 L? (5 pts)

Chemistry
1 answer:
IgorLugansk [536]3 years ago
8 0

Answer:

The temperature at 760mmHg if the volume is still 4.2 L is 400.862 K

Explanation:

Gay Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. This law can be expressed mathematically as follows:

\frac{P}{T} =k

Having an initial state 1 and a final state 2, the following is true:

\frac{P1}{T1} =\frac{P2}{T2}

In this case:

  • P1= 560 mmHg
  • T1= 72 F= 295.372 K (being 32 F= 273.15 K)
  • P2= 760 mmHg
  • T2= ?

Replacing:

\frac{560 mmHg}{295.372 K} =\frac{760 mmHg}{T2}

Solving:

T2= 760 mmHg*\frac{295.372 K}{560 mmHg}

T2= 400.862 K

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____ [38]

E

θ

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=

+

2.115

l

V

Cathode

Mg

2

+

/

Mg

Anode

Ni

2

+

/

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Explanation:

Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]

Mg

2

+

(

a

q

)

+

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s

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−

E

θ

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−

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l

V

Ni

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+

(

a

q

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+

2

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(

s

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−

E

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The standard reduction potential

E

θ

resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (

298

l

K

,

1.00

l

kPa

) is defined as

0

l

V

for reference. [2]

A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.

Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher

E

θ

and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower

E

θ

will experience oxidation and act the anode.

E

θ

(

Ni

2

+

/

Ni

)

>

E

θ

(

Mg

2

+

/

Mg

)

Therefore in this galvanic cell, the

Ni

2

+

/

Ni

half-cell will experience reduction and act as the cathode and the

Mg

2

+

/

Mg

the anode.

The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:

E

θ

cell

=

E

θ

(

Cathode

)

−

E

θ

(

Anode

)

E

θ

cell

=

−

0.257

−

(

−

2.372

)

E

θ

cell

=

+

2.115

Indicating that connecting the two cells will generate a potential difference of

+

2.115

l

V

across the two cells.

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