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Pepsi [2]
3 years ago
15

Hydrocarbons, compounds containing only carbon and hydrogen, are important in fuels. The heat of combustion of cyclopentane, C5H

10, is 786.6 kcal/mol. Write a balanced equation for the complete combustion of cyclopentane. How much energy is released during the complete combustion of 416 grams of cyclopentane
Chemistry
1 answer:
pishuonlain [190]3 years ago
6 0

Answer:

4664.54  kcal of energy is produced on the complete combustion of 416 grams of cyclopentane

Explanation:

Combustion is the process by which hydrocarbons react with oxygen to heat to produce heat energy.

The heat energy produced when one mole of an hydrocarbon is completely burnt or combusted in air is called the heat of combustion

The balanced equation for the complete combustion of cyclopentane (C5H10) is given below:

             2C₅H₁₀ + 15O₂ → 10CO₂ + 10H₂O

                                                             ΔH combustion = 786.6 kcal/mol

What the above equation means is that each time one mole of cyclopentane is completely burnt, 786.6kcal of energy is released.

Hence we need to calculate the number of moles of cyclopentane in 416grams; therefore

No of moles cyclopentane = mass of cyclopentane / molar mass of cycopentane  

                                             = 416 g / 70.1 g/mol

                                             = 5.9343moles

                                             ≅ 5.93 moles of cyclopentane in 416grams

From the question, since the  cyclopentane is completely combusted, it implies that the energy released would be

    1 mole of cyclopentane     = 786.6 kcal of energy

   5.93 moles of cyclopentane would produce    

                                                = 5.93 moles x 786.6kcal/mol

                                                = 4664.54  kcal of energy

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alexira [117]

Answer:

B

Explanation:

5 0
3 years ago
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Two (2) moles of an ideal gas are in a container at 200 kpa and a temperature of 300 k. The volume occupied by the gas in the co
k0ka [10]

The volume occupied by the gas in the container is 1 m³

Boyles law applies

P₁ V₁ = P₂ V₂

Where P₁ = 200kpa

P₂ = 300kpa

if its initial volume is 1.5

then,

P₁ V₁ = P₂ V₂

200 × 1.5 = 300 × V₂

V₂ = 200 × 1.5 / 300

    = 1 m³

Hence  the volume occupied by the gas container is 1 m³

Learn more about the Boyles law on

brainly.com/question/13759555

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4 0
2 years ago
If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit
Blizzard [7]

<u>Answer:</u>

<u>For a:</u> The volume of the box is 217.5 mL

<u>For b:</u> The volume of the box is 0.2175 L

<u>Explanation:</u>

The box is a type of cuboid.

To calculate the volume of cuboid, we use the equation:

V=lbh

where,

V = volume of cuboid

l = length of cuboid = 10.00 cm

b = breadth of cuboid = 7.25 cm

h = height of cuboid = 3.00 cm

Putting values in above equation, we get:

V=10.00\times 7.25\times 3.00=217.5cm^3

  • <u>For a:</u>

To convert the volume of cuboid into milliliters, we use the conversion factor:

1mL=1cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1mL}{1cm^3})\\\\\Rightarrow 217.5mL

Hence, the volume of the box is 217.5 mL

  • <u>For b:</u>

To convert the volume of cuboid into liters, we use the conversion factor:

1L=1000cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1L}{1000cm^3})\\\\\Rightarrow 0.2175L

Hence, the volume of the box is 0.2175 L

3 0
4 years ago
Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
GrogVix [38]

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

8 0
3 years ago
Two 10g blocks, one of copper and one of iron, were heated from 300 K to 400K (a temperature difference of 100 K).
vova2212 [387]

Energy absorbed by Iron block E (iron) = 460.5 J

Energy absorbed by Copper block E (Copper) = 376.8 J

<u>Explanation:</u>

To find the heat absorbed, we can use the formula as,

q = m c ΔT

Here, Mass = m = 10 g = 0.01 kg

ΔT = change in temperature = 400 - 300 = 100 K = 100 - 273 = -173 °C

c = specific heat capacity

c for iron = 460.5 J/kg K

c for copper = 376.8 J/kg K

Plugin the values in the above equation, we will get,

q (iron) = 0.01 kg × 460.5 J/kg K × 100 K

            = 460.5 J

q (copper) = 0.01 kg ×  376.8 J/kg K × 100 K

                 = 376.8 J

3 0
3 years ago
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