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Andre45 [30]
3 years ago
5

One tablespoon of peanut butter has a mass of 15 g . It is combusted in a calorimeter whose heat capacity is 120 kJ/∘C . The tem

perature of the calorimeter rises from 22.4 ∘C to 25.5 ∘C. Find the food caloric content of peanut butter.
Chemistry
1 answer:
devlian [24]3 years ago
8 0

Answer:

5927.3488 calorie/g

Explanation:

For peanut butter:

Mass = 15 g

Initial temperature = 22.4 °C

Final temperature = 25.5 °C

Specific heat of calorimeter, C = 120 kJ/°C

Considering

Q=C\times (T_f-T_i)

Q=120\ kJ/^0C\times (25.5-22.4)\ ^0C

Q=372\ kJ

Also,

1 kJ = 239.006 calories

So, Heat = 372*239.006 calories = 88910.232 calories

15 g of the peanut butter gives 88910.232 calories

<u>So, the food caloric content of peanut butter = 88910.232 calories/ 15 g = 5927.3488 calorie/g</u>

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Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

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The balanced two-half reactions will be,

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Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

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Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

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F = Faraday constant = 96500 C

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T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

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E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

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4 0
3 years ago
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Explanation:

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