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elena-s [515]
3 years ago
9

The aromatic hydrocarbon cymene (C10H14) is found in nearly 100 spices and fragrances, including coriander, anise, and thyme. Th

e complete combustion of 1.608 g of cymene in a bomb calorimeter (Calorimeter = 3.640 kJ/°C) produced an increase in temperature of 19.35°C. How much thermal energy is produced during the complete combustion of one mole of cymene?
Chemistry
1 answer:
andrezito [222]3 years ago
4 0

Answer : The thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole

Explanation :

First we have to calculate the heat released by the combustion.

q=c\times \Delta T

where,

q = heat released = ?

c = specific heat of calorimeter = 3.640kJ/^oC

\Delta T = change in temperature = 19.35^oC

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

q=3.640kJ/^oC\times (19.35)^oC

q=70.43kJ

Thus, the heat released by the combustion = 70.43 kJ

Now we have to calculate the molar enthalpy combustion.

\Delta H=-\frac{q}{n}

where,

\Delta H = molar enthalpy combustion = ?

q = heat released = 70.43 kJ

n = number of moles cymene = \frac{\text{Mass of cymene}}{\text{Molar mass of cymene}}=\frac{1.608g}{164.22g/mol}=0.009792mole

\Delta H=-\frac{70.43kJ}{0.009792mole}=-7193kJ/mole

Therefore, the thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole

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Answer:

0.005404 M

Explanation:

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Since you added an excess of sodium carbonate you warrantied that all the Pb^{+2} in the sample reacted with it. So we can say that the insoluble lead (II) carbonate PbCO_{3} contains all the Pb^{+2} ions in the original sample.  

The moles of PbCO_{3} are:

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Knowing this, 0.00054 is also the number of moles of Pb^{+2} in the original sample.  

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Answer:

Part 1

Where we have that the phase temperature at which fuel 1 changes to liquid due to its low temperature after the energy transfer is reached, fuel 1 changes to liquid

Where we have that the temperature of fuel 2 is still above its gas to liquid phase transition temperature, the fuel in fuel 2 will remain gaseous

Part 2

The kinetic energy of the individual molecules in fuel 1 is less than the intermolecular forces holding the molecules of fuel 1 in the liquid state such that fuel 1 molecules undergoes phase transformation from gas to liquid

The kinetic energy of the molecules in fuel 2 is higher than the fuel 2 liquid state intermolecular forces fuel 2 does not undergo phase transformation and remain a gas

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Why fuel 1 change phase but fuel 2 stayed the same can be explained by the combination of the following physical phenomena

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The energy transferred out, ΔQ, can be expressed as follows;

ΔQ = m·c·ΔT

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c = The specific heat capacity of the fuel

ΔT = The temperature change of the fuel

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For a given amount of transferred energy, when the specific heat capacity is high, the temperature change will be low and vice versa

Taking the specific heat capacity of fuel 1, c₁ as lower than the specific heat capacity of fuel 2, c₂₂, we have;

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