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bagirrra123 [75]
3 years ago
8

Calculate the amount of heat (in kJ) required to raise the temperature of a 79.0 g sample of ethanol from 298.0 K to 385.0 K. Th

e specific heat capacity of ethanol is 2.42 J/g°C. Calculate the amount of heat (in kJ) required to raise the temperature of a 79.0 g sample of ethanol from 298.0 K to 385.0 K. The specific heat capacity of ethanol is 2.42 J/g°C. 16.6 kJ 28.4 kJ 12.9 kJ 73.6 kJ 57.0 kJ
Chemistry
1 answer:
irina [24]3 years ago
4 0

Answer:

The amount of heat required to raise the temperature of the sample from 298 to 385 Kelvin, is 16.6 kJ

Explanation:

<u>Step 1: </u>Given data

A 79.0 g sample of ethanol raises from 298 K to 385 K

The specific heat of ethanol is 2.42J/g°C

<u>Step 2:</u> Calculate the heat transfer

Q = m*Cp*ΔT

with m = the mass of the ethanol sample (in grams)

⇒ mass = 79 grams

with Cp = the specific heat capacity of ethanol (in J/g°C)

⇒ Cp = 2.42 J/g°C

with ΔT = the change of temperature (T2-T1)

⇒ ΔT = 385 K - 298K = 112 °C - 25 °C = 87

Q = 79 grams * 2.42 J/g°C * 87 = 16632.66 j = 16.6 kJ

The amount of heat required to raise the temperature of the sample from 298 to 385 Kelvin, is 16.6 kJ

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The average kinetic energy of a gas when the particles of the gas collide against each other at a constant temperature and volume will not changes.

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The product of half the mass of each gas molecule with the square of root mean square speed is the average kinetic energy of the molecules.

And this average kinetic energy is deirectly proportional to the absolute temperature of the system but in the given question temperature and volume of the gas is constant. So that there is no change in average kinetic energy of molecules when they collide.

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How much carbon dioxide will be formed if 12.5 grams of oxygen reacts with 7.2 grams of propane (C3H8 )? Balanced equation: C3H8
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10.3125 grams of carbon dioxide will be formed.

Explanation:

The balanced reaction is:

C₃H₈ + 5 O₂→ 3 CO₂ + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of the compounds:

  • C₃H₈: 44 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
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then by stoichiometry of the reaction, the following amounts of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole = 44 g
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  • CO₂: 3 moles* 44 g/mole= 132 g
  • H₂O: 4 moles* 18 g/mole= 72 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: If by reaction stoichiometry 44 grams of propane react with 160 grams of oxygen, 7.2 grams of propane react with how much mass of oxygen?

mass of oxygen=\frac{7.2 grams of propane*160 grams of oxygen}{44 grams of propane}

mass of oxygen= 26.18 grams

But 26.18 moles of O₂ are not available, 12.5 grams are available. Since you have less mass than you need to react with 7.2 grams of propane, oxygen O₂ will be the limiting reagent.

Then you can apply the following rule of three: if by stoichiometry of the reaction 160 grams of oxygen form 132 grams of carbon dioxide, 12.5 grams of oxygen will form how much mass of carbon dioxide?

mass of carbon dioxide=\frac{12.5 grams of oxygen*132 grams of carbon dioxide}{160 grams of oxygen}

mass of carbon dioxide= 10.3125 grams

<u><em>10.3125 grams of carbon dioxide will be formed.</em></u>

<u><em> </em></u>

<u><em></em></u>

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