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Lana71 [14]
3 years ago
11

A 10.0 g sample of propane, C3H8, was combusted in a constant-volume bomb calorimeter. The total heat capacity of the bomb calor

imeter and water was 8.0 kJ/°C. The molar heat of combustion of propane is -2 222 KJ/mol. If the starting temperature of the water was 20 °C, what will be the final temperature of the bomb calorimeter?
Chemistry
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

83ºC

Explanation:

A bomb calorimeter is an instrument used to measure the heat that release or absorb a particular reaction.

The reaction of combustion of propane is:

C₃H₈ +  5O₂ → 3 CO₂ + 4 H₂O ΔH = -2222kJ/mol

<em>1 mole of propane release 2222kJ</em>

10.0g of propane (Molar mass: 44.1g/mol).

10.0g ₓ (1mol/ 44.1g) = <em>0.227 moles of C₃H₈</em>

If 1 mole of propane release 2222kJ, 0.227moles will release (Release because molar heat is < 0):

0.227 moles of C₃H₈ ₓ (2222kJ / mol) = 504kJ.

Our calorimeter has a constant of 8.0kJ/ºC, that means if there are released 8.0kJ, the bomb calorimeter will increase its temperature in 1ºC. As there are released 504kJ:

504kJ ₓ (1ºC / 8.0kJ) = 63ºC will increase the temperature in the bomb calorimeter.

As initial temperature was 20ºC, final temperature will be:

<h2>83ºC</h2>
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The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}

Given: -\frac{d[HBr]}{dt}]=0.301

Putting in the values we get:

\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}

+\frac{1d[Br_2]}{dt}=0.151

Thus the rate of appearance of Br_2 is 0.151

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Density of water calculation using a 10 mL graduated cylinder
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Answer:

Average density of the liquid = 0.992 g/mL

Explanation:

Density = mass/volume

mass of liquid = (mass of liquid + mass of cylinder) - mass of cylinder

Trial 1: mass of liquid = 19.731 - 9.861 = 9.87

volume of liquid = 10 mL

density of liquid = 9.87 g / 10 mL = 0.987 g/mL

Trial 2: mass of liquid = 19.831 - 9.861 = 9.97

volume of liquid = 10 mL

density of liquid = 9.97 g / 10 mL = 0.997 g/mL

Trial 3: mass of liquid = 19.831 - 9.861 = 9.97

volume of liquid = 10 mL

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Trial 4: mass of liquid = 19.771 - 9.861 = 9.91

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density of liquid = 9.91 g / 10 mL = 0.991 g/mL

Trial 5: mass of liquid = 19.751 - 9.861 = 9.89

volume of liquid = 10 mL

density of liquid = 9.89 g / 10 mL = 0.989 g/mL

Average density = (0.987 + 0.997 + 0.997 + 0.991 + 0.989)/5 = 4.961/5

Average density of the liquid = 0.992 g/mL

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