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3 years ago

Why are standard electrode potentials usually measured at 1.0 M electrolyte concentration?

Chemistry

1 answer:

vovikov84 [41]3 years ago

7 0

Answer:

To avoid variation due to concentration

Explanation:

Hope this helps :)

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Which is the best revision of this sentence?

taurus [48]

Give more points, dont even feel like reading this for 5 pointsssss

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3 years ago

A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment,

nikitadnepr [17]

Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

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3 years ago

5,6-dimethyl-2-heptyne <br><br> Due in 1 hr pls help me

mina [271]

Answer:

Please see the attached pictures.

Explanation:

☆ To ensure that each carbon has 4 bonds, fill the other bonds with Hs.

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3 years ago

What is the difference between a quantity and a unit?

Irina-Kira [14]

$Question$

What is the difference between a quantity and a unit?

Answer:

Physical quantities are a characteristic or property of an object that can be measured or calculated from other measurements. Units are standards for expressing and comparing the measurement of physical quantities. All units can be expressed as combinations of four fundamental units.

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3 years ago

In lab (write this down in your lab protocol), you will be given a stock solution that has a glucose concentration of 60 mg/dL.

Wittaler [7]

Answer:

1. The dilution factor for the serial dilution = 2

2. V2 = 1 mL

3. V1 = 0.5 mL

Explanation:

1. Dilution factor is the ratio of the initial concentration to the final concentration.

Dilution factor = initial concentration / final concentration

First dilution: initial concentration = 60 mg/dL

final concentration = 30 mg/dL

Dilution factor = 60 mg/dL / 30 mg/dL = 2

Second dilution: initial concentration = 30 mg/dL

final concentration = 15 mg/dL

Dilution factor = 30 mg/dL / 15 mg/dL = 2

Therefore, the dilution factor for the serial dilution = 2

2. From the dilution formula, C1V1 = C2V2; V2 = final volume to be prepared.

Since 1 mL of the various glucose solutions are to be prepared, the final concentration, V2 = 1 mL

3. From the dilution formula, C1V1 = C2V2; V1 = initial concentration of the solution to be prepared.

C1/C2 = V2/V1

Since the dilution factor, C1/C2 is 2, V2/V1 = 2

V1 = V2/2

V1 = 1 mL / 2

V1 = 0.5 mL

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3 years ago