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valina [46]
3 years ago
6

Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

a renewable resource and ethanol-blended fuels, like gasohol, appear to burn more efficiently in combustion engines. The heat of combustion of ethanol is 326.7 kcal/mol. The heat of combustion of octane, C8H18, is 1.308×103 kcal/mol. How much energy is released during the complete combustion of 406 grams of octane ? kcal Assuming the same efficiency, would 406 grams of ethanol provide more, less, or the same amount of energy as 406 grams of octane?
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Explanation:

Given,

The heat of combustion of octane = 1.308×10³ kcal/mol

It means that combustion of 1 mole of octane produces 1.308×10³ kcal of energy

Given mass = 406 g

Molar mass of octane = 114.23 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{406\ g}{114.23\ g/mol}

Moles of octane = 3.5542 moles

Combustion of 3.5542 moles produces 3.5542×1.308×10³ kcal of energy

<u>Energy released from 406 grams of octane = 4.649×10³ kcal</u>

Also given,

The heat of combustion of ethanol = 326.7 kcal/mol

It means that combustion of 1 mole of octane produces 326.7 kcal of energy

Given mass = 406 g

Molar mass of octane = 46.07 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{406\ g}{46.07\ g/mol}

Moles of octane = 8.8127 moles

Combustion of 8.8127 moles produces 8.8127×326.7 kcal of energy

<u>Energy released from 406 grams of ethanol= 2879.11 kcal = 2.879×10³ kcal</u>

<u>It is less.</u>

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A solution of aqueous ammonium sulfate has a mass of 482 grams. Calculate the mass of ammonium sulfate in solution if the mass p
umka2103 [35]

Answer:

The answer to your question is the mass of solute = 53.5 g

Explanation:

Data

mass of solution = 482 g

mass of solute = ?

mass percent = 11.1 %

Mass percent is a unit of concentration. It measures the mass of the solute divided by the total mass of the solution

Process

1.- Write the formula

        Mass percent = mass of solute / mass of solution x 100

-Solve for mass of solute

          mass of solute = Mass percent x mass of solution / 100

2.- Substitution

          mass of solute = 11.1 x 482 / 100

3.- Simplification

          mass of solute = 5350.2 / 100

4.- Result

          mass of solute = 53.5g

6 0
3 years ago
9. What are the advantages of using an indicator to inform pH measurements? What are the advantages of using a pH meter?
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Answer:

The advantages of using an indicator to inform pH measurements:

It gives a mathematically result of the pH, in addition, it gives the precise pH of solvent, and it also gives an idea of the straight of the solution also.

Now, the advantage of using a pH meter:

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Use the image above and describe the
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Answer:

If you see in the image above, there is an unbalance force applied while playing tug of war. Since it is 1 vs 2, there is a greater net force in the right side then the left side. If it was 2 vs 2 or 1 vs 1, then they are appling balance force. You can also see in the picture that the arrows are pointing outwards (--->) rather then inwards (<---) because you are pulling the rope not pushing the rope. If you add one person on the left side, then the newtons which is 20N will become to 35N and will be balanced, but since there in only 1 person, there is less force on the left side, the newtons gets subtracted having only 20N. Since you are pulling the rope, the friction is opposite (<---). Since you are pulling the rope, you are using Kinetic force and the rope stays in potential force since it stays constant.

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4 0
3 years ago
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dlinn [17]
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6 0
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Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r
Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

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We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

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1 N≡N = 1(945 kJ/mol)     3 H-H = 3 (432 kJ/mol)       6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑  H bonds broken  - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ)   + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

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