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KengaRu
2 months ago
10

A certain chemical reaction releases 362.kj of heat energy per mole of reactant consumed. Suppose some moles of the reactant are

put into a calorimeter (a device for measuring heat flow). It takes 4.87j of heat energy to raise the temperature of this calorimeter by 5.5 C . Now the reaction is run until all the reactant is gone, and the temperature of the calorimeter is found to rise by . How would you calculate the number of moles of reactant that were consumed?
Set the math up. But don't do any of it. Just leave your answer as a math expression.

Also, be sure your answer includes all the correct unit symbols.
Chemistry
1 answer:
lions [2.9K]2 months ago
6 0

Answer:

The temperature increase of the calorimeter, which is missing in the problem, is necessary for the calculation.

Explanation:

Since the temperature rise (X) is unspecified, we'll express the calculation in terms of X, and demonstrate with an example value.

1) Calorimeter details:

  • Temperature increase: X °C
  • Heat capacity ratio: 4.87 J / 5.5 °C (given)
  • Energy absorbed by calorimeter at X °C rise:

                (4.87 J / 5.5 °C) × X

2) Reaction data:

  • Heat released: 362 kJ per mole of reactant
  • Number of moles consumed: n
  • Total energy from reaction:

     362 kJ/mol × 1000 J/kJ × n = 362,000 n J

3) Using energy conservation, assuming no heat loss to surroundings, the energy from the reaction equals the energy absorbed by the calorimeter:

  • 362,000 n = (4.87 J / 5.5 °C) × X

  • Solving for n gives:

  • n = [(4.87 / 5.5) × X] / 362,000

     n = 0.000002446 × X

This means for each degree Celsius rise in calorimeter temperature, 0.000002446 moles of reactant were consumed.

Example:

If the calorimeter temperature increases by 100 °C, then:

  • n = 0.000002446 × 100 = 0.0002446 mol

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14 days ago
A 2.950×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in w
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Answer:

The glycerol solution has a molality of 2.960×10^-2 mol/kg.

Explanation:

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7 0
1 month ago
32.7 grams of water vapor takes up how many liters at standard temperature and pressure (273 K and 100 kPa)?
alisha [2963]
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1 month ago
A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr
KiRa [2933]

Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

\frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

Now we can substitute the known variables:

\frac{0.6}{298.2}   = \frac{P_{2} }{345.4}

P₂ = 0.7atm

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1 month ago
Marie and Calvin dissolved 10 grams of KNO3 in 100 grams of water at 25oC. Next they added 5 grams more. Calvin told Marie that
Alekssandra [3086]
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8 0
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