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rosijanka [135]
2 years ago
15

What mass of carbon dioxide (co2) can be produced from 86.17 grams of c6h14 and excess oxygen?

Chemistry
1 answer:
charle [14.2K]2 years ago
3 0
2C6H14 + 13O2 ---> 6CO2 +14H2O

M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol

86.17 g C6H14 is 1 mole.

                             2C6H14 + 13O2 ---> 6CO2 +14H2O
from reaction        2 mol                         6 mol
from the problem  1 mol                         3 mol

M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Answer : 132.0 g CO2


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There are 3 parts in this question:
1) To find the initial Boyle's constant k_{i}
2) To find the final Boyle's constant k_{f}
3) To verify whether gas is obeying Boyle's law or not

Given data:
The initial volume of the cylinder(in litres) = V_{i} = 12.0 L
The initial pressure(in atmospheric pressure) = P_{i} = 4.0 atm

The final pressure(in atmospheric pressure) = P_{f} = 8.0 atm
The final volume of the cylinder(in litres) = V_{f} = 6.0 L

First you need to know what Boyle's law is:
<span>Boyle's law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
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The Mathematical form of Boyle's law is:
P =  \frac{k}{V}

Where,
P = Pressure
V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
1. 
The initial volume of the cylinder(in litres) = V_{i} = 12.0 L
The initial pressure(in atmospheric pressure) = P_{i} = 4.0 atm
The Boyle's constant = k_{i} = ?

According to the Boyle's law,

P_{i} = \frac{k_{i}}{V_{i}}

=> k_{i} =  P_{i}V_{i}
Plug-in the values in the above equation, you would get:
k_{i} = 4.0 * 12.0 = 48

Ans-1) k_{i} = 48

2.
The final pressure(in atmospheric pressure) = P_{f} = 8.0 atm
The final volume of the cylinder(in litres) = V_{f} = 6.0 L
The Boyle's constant = k_{f} = ?

According to the Boyle's law,

P_{f} = \frac{k_{f}}{V_{f}}

=> k_{f} =  P_{f}V_{f}
Plug-in the values in the above equation, you would get:
k_{f} = 8.0 * 6.0 = 48

Ans-2) k_{f} = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

k_{i} = k_{f}

Since,
k_{i} = 48
k_{f} = 48

Therefore,
48=48.

Ans-3) Hence proved: The gas IS obeying the Boyle's law.

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3 years ago
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Answer:

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Explanation:

The half-cell reduction potentials are

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To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.

The anode is the electrode at which oxidation occurs.

The equation for the oxidation half-reaction is

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The answer to your question would be _B High Temp
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2) Molarity = M = number of moles of solute / liters of solution

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<em>Answer:</em>

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<em>Explanation:</em>

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<em>Density:   </em>

Density is the ratio of mass and volume as follow

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<em>Summary:</em>

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  • Purple liquid has low density, so it will be at top.
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