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olga_2 [115]
3 years ago
13

. A large quantity of very dilute aqueous HCl solution is neutralized by the addition of the stoichiometric amount of a 10-mol-%

aqueous NaOH solution. Estimate the heat effect per mole of HCl neutralized if the tank is maintained at 25°C and 1(atm) and the neutralization reaction goes to completion. ∙ For NaOH(9H2O), Δ˜H = 45.26 kJ·mol−1 ∙ For NaCl, ~ limn→[infinity]˜ΔH = 3.88 kJ·mol−1
Chemistry
1 answer:
ad-work [718]3 years ago
5 0

Answer:

Make the question more clear for me

Explanation:

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If it takes 22974 J to heat 96.5 g of a substance from 13.46°C to 73.86 °C, what is the specific heat of the substance?​
yulyashka [42]

The specific heat capacity, C, of the substance is 3.94 \frac{J}{g-^oC}. To raise one gram of a substance by one degree Celsius, a specific amount of heat must be absorbed. This is called the specific heat of the substance. The higher the specific heat of a substance, the more energy is needed to increase its temperature and vice versa.

<u>Further Explanation: </u>

The specific heat capacity, C, of a substance is related to the amount of substance and the temperature change using the following equation

q \ = \ mC(T_{f}-T_{i})

where

q = amount of heat absorbed (usually in J)

m = amount of substance (in g)

C = specific heat capacity (in \frac{J}{g-^oC})

T(f) = final temperature

T(i) = initial temperature.

To answer the problem, we sort the given information:

q = 22974 J (5 significan figures)

m = 96.5 g (3 significant figures)

C = ?

T(f) = 73.86 °C (4 significant figures)

T(i) = 13.46 °C (4 significant figures)

Then, to find C, rearrange the equation:

C \ = \frac{q}{m(T_{f} \ - \ T_{i})}

Plug in the values to the equation above to get the value of C.

C \ = \ \frac{22974 \ J}{96.5 \ g \ (73.86 \ \ -13.46 \°C)} \\C \ = \ \frac{22974 \ J}{96.5 \ g \ (60.4\°C)} \\C \ = \ \frac{22974 \ J}{5828.6 \ g -\°C} \\C \ = \ 3.941598 \ \frac{J}{g -\°C} \\

Since the least number of significant figures in the given is 3, then the final answer must have 3 significant figures, too. Therefore,

\boxed {C = 3.94 \ \frac{J}{ g -\°C}}

<u></u>

<u>Learn More: </u>

1. Learn more about heat capacity brainly.com/question/8828503

2. Learn more about calorimetry brainly.com/question/8168263

3. Learn more about heat of reaction brainly.com/question/10122365

<u>Keywords:</u> heat capacity, specific heat, calorimetry

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