Answer:
One gallon of octane produces approximately 7000 L of carbon dioxide.
Note:
I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.
Step1. You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. Step 2. Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on the mole ratio between octane and carbon dioxide in the balanced equation. Step 3. Then use the ideal gas law to determine the volume in liters of carbon dioxide that can be formed.
Answer:
So the answer is every day
Explanation:
Plants respire at all times of the day and night because their cells need a constant energy source to stay alive.
Answer:
1-Pentene
Explanation:
If we look at all the options listed, we will notice that the rate of reaction of bromine with each one differs significantly.
For 1-pentene, addition of bromine across the double bond is a relatively fast process. It is usually used as a test for unsaturation. Bromine water is easily decolorized by alkenes.
Cyclohexane, heptane are alkanes. They can only react with chlorine in the presence of sunlight. This is a substitution reaction. It does not occur easily. A certain quantum of light is required for the reaction to occur.
For benzene, bromine can only react with it by electrophilic substitution in which the benzene ring is retained. A Lewis acid is often required for the reaction to occur and it doesn't occur easily.
Answer:
![W=5743.1077\ J](https://tex.z-dn.net/?f=W%3D5743.1077%5C%20J)
Explanation:
The expression for the work done is:
![W=RT \ln \left( \dfrac{P_1}{P_2} \right)](https://tex.z-dn.net/?f=W%3DRT%20%5Cln%20%5Cleft%28%20%5Cdfrac%7BP_1%7D%7BP_2%7D%20%5Cright%29)
Where,
W is the amount of work done by the gas
R is Gas constant having value = 8.314 J / K mol
T is the temperature
P₁ is the initial pressure
P₂ is the final pressure
Given that:
T = 300 K
P₁ = 10 bar
P₂ = 1 bar
Applying in the equation as:
![W=8.314\times 300 \ln \left( \dfrac{10}{1} \right)](https://tex.z-dn.net/?f=W%3D8.314%5Ctimes%20300%20%5Cln%20%5Cleft%28%20%5Cdfrac%7B10%7D%7B1%7D%20%5Cright%29)
![W=300\times \:8.314\ln \left(10\right)](https://tex.z-dn.net/?f=W%3D300%5Ctimes%20%5C%3A8.314%5Cln%20%5Cleft%2810%5Cright%29)
![W=2.30258\times \:2494.2](https://tex.z-dn.net/?f=W%3D2.30258%5Ctimes%20%5C%3A2494.2)
![W=5743.1077\ J](https://tex.z-dn.net/?f=W%3D5743.1077%5C%20J)
Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL