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lutik1710 [3]
2 years ago
15

A chemist is given a 3.00M solution of KBr and needs to measure out 0.733 moles of this solution. How many mL of the 3.00M KBr s

olution would the chemist need
Chemistry
1 answer:
Charra [1.4K]2 years ago
7 0

In order to measure 0.733 moles of KBr from a 3.00 M solution, the chemist needs 244 mL of solution.

<h3>What is molarity?</h3>

Molarity (M) is a unit of concentration of solutions, and it is defined as the moles of a solute per liters of a solution.

  • Step 1: Calculate the liters of solution required.

A chemist has a 3.00 M KBr solution and wants to measure 0.733 moles of KBr. The required volume is:

0.733 mol × (1 L/3.00 mol) = 0.244 L

  • Step 2: Convert 0.244 L to mL.

We will use the conversion factor 1 L = 1000 mL.

0.244 L × (1000 mL/1 L) = 244 mL

In order to measure 0.733 moles of KBr from a 3.00 M solution, the chemist needs 244 mL of solution.

Learn more about molarity here: brainly.com/question/9118107

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What is the standard electrode potential for a galvanic cell constructed in the appropriate way from these two half-cells?
____ [38]

E

θ

Cell

=

+

2.115

l

V

Cathode

Mg

2

+

/

Mg

Anode

Ni

2

+

/

Ni

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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l

e

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(

s

)

−

E

θ

=

−

2.372

l

V

Ni

2

+

(

a

q

)

+

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l

e

−

→

Ni

(

s

)

−

E

θ

=

−

0.257

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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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2 years ago
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AlladinOne [14]
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Answer:

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