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Sonja [21]
4 years ago
6

ASAP HELP, PLEASE HELP ASAP

Chemistry
2 answers:
kondor19780726 [428]4 years ago
8 0

Answer: gravity affect the amount of both kinetic and potential energy

Explanation:

AnnZ [28]4 years ago
8 0

Answer:

i agree,  gravity affect the amount of both kinetic and potential energy.

Explanation:

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Octane (C8H18) is a component of gasoline. Complete combustion of octane yields H2O and CO2. Incomplete combustion produces H2O
mestny [16]

86.5\; \% of octane had been converted to carbon dioxide CO₂.

<h3>Explanation</h3>

Octane has a molar mass of

12.01 \times 8 + 1.008 \times 18 = 114.22 \; \text{g} \cdot \text{mol}^{-1}

1.000 gallon of this fuel would have a mass of 2.650 kilograms or 2.65 \times 10^{3} \; \text{g}, which corresponds to 2.65 \times 10^{3} / 114.22 = 23.2\; \text{mol} of octane.

Octane undergoes complete combustion to produce carbon dioxide and water by the following equation:

2\; \text{C}_8\text{H}_{18} + 25 \; \text{O}_2 \to 16 \; \text{CO}_2 + 18 \; \text{H}_2\text{O}

An incomplete combustion of octane that gives rise to carbon monoxide and water but no carbon dioxide would consume not as much oxygen:

2\; \text{C}_8\text{H}_{18} + 17 \; \text{O}_2 \to 16 \; \text{CO} + 18 \; \text{H}_2\text{O}

The mass of the product mixture is 11.53 - 2.65 = 8.88 \; \text{kg} heavier than that of the octane supplied. Thus 8.88 \; \text{kg} = 8.88 \times 10^{3} \; \text{g} of oxygen were consumed in the combustion. There are 277.5 \; \text{mol} of oxygen molecules in 8.88 \times 10^{3} \; \text{g} of oxygen.

Let the number of moles of octane that had undergone complete combustion as seen in the first equation be x (0 \le x \le 23.2). The number of moles of octane that had undergone incomplete combustion through the second equation would thus equal 23.2 - x.

25 moles of oxygen gas is consumed for every two moles of octane that had undergone complete combustion and 17 moles if the combustion is incomplete.

n(\text{O}_2, \; \text{Complete Combustion}) + n(\text{O}_2, \; \text{Incomplete Combustion} ) = n(\text{O}_2, \; \text{Consumed})\\

\frac{25}{2} \; x + \frac{17}{2} \; (23.2 - x) = 277.5\\4 \; x = 277.5 - \frac{17}{2} \times 23.2\\x = 20.1

Therefore 20.1 \; \text{mol} out of the 23.2 moles of octane had undergone complete combustion to produce carbon dioxide.

\%n(\text{Complete Combustion}) = 20.1 / 23.2 \times 100 \; \%= 86.5\; \%

7 0
3 years ago
What is the acknowledgment to ''Do all liquids evaporate at the same rate?'
natita [175]
<span>'Do all liquids evaporate at the same rate
 that would be false

</span>
4 0
3 years ago
Consider the synthesis of water as shown in Model 3. A container is filled with 10,0 g of H, and
kondor19780726 [428]

Answer:

Oxygen, O₂ is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2H₂ + O₂ —> 2H₂O

Next, we shall determine the masses of H₂ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 2 × 2 = 4 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂O from the balanced equation = 1 × 32 = 32 g

SUMMARY:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Therefore, 10 g of H₂ will react with

= (10 × 32)/4 = 80 g of O₂.

From the calculations made above, we can see that a higher mass (i.e 80 g) of O₂ than what was given (i.e 5 g) is required to react completely with 10 g of H₂. Therefore, O₂ is the limiting reactant.

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3 years ago
Przedstaw sposób tworzenia wiązania jonowego w w MgF2 i AlBr3.<br> Kto pomoże daje naj!!!
stiks02 [169]
Is this a question? Or just letters
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Which property would cesium most likely have?
Romashka-Z-Leto [24]
The answer is a because its not low ductile or gasst

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