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Natalija [7]
3 years ago
7

When determining the standard reduction potential of a substance by using a standard hydrogen electrode as a reference, the stan

dard reduction potential will always be equal to:______.
ta. he reduction potential for the standard hydrogen electrode.
b. one-half the cell potential.
c. the overall cell potential.
d. impossible to predict.
Chemistry
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

the overall cell potential

Explanation:

We must bear in mind that the standard hydrogen electrode is a reference electrode whose electrode potential has been arbitrarily set at 0 V.

The standard hydrogen electrode consists of hydrogen ion solution and hydrogen gas together with a platinum electrode.

The overall cell potential is the reduction potential of the substance being determined using the standard hydrogen electrode as a reference electrode since its electrode potential is set at zero volts.

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Some of the negative impact of disruption in water cycle are adverse environment, with affected farming, hydroelectricity, and deforestation.

The water cycle has been operated in the environment for the conversion of water in the different states. The water cycle has been operated with the conversion of water from the water body to the clouds by evaporation.

The evaporated water has resulted in clouds, that have been condensed and precipitated in the form of rain. The water cycle helps in maintaining the water balance in the ecosystem. It made the water available for use as well.

The drought has been the condition of the scarcity of water. In this condition, there has been a disruption of the water cycle. Some of the negative impacts resulting from the disruption of the water cycle are:

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Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

\Delta T = change in temperature = 1.643^oC

Now put all the given value in the above formula, we get:

26380J/g=0.235g\times c\times 1.643^oC

c=68323.38J/^oC

Thus, the specific heat of calorimeter is 68323.38J/^oC

Now we have to calculate the heat of combustion of caffeine.

Formula used :

Q=c\times \Delta T

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = 68323.38J/^oC

\Delta T = change in temperature = 1.584^oC

Now put all the given value in the above formula, we get:

Q=68323.38J/^oC\times 1.584^oC

Q=108224.23J=108.2kJ

Now we have to calculate the moles of caffeine.

\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

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