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2 years ago

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The flame produced by the burner of a gas (propane) grill is a pale blue color when enough air mixes with the propane (C3H8) to

burn it completely. For every gram of propane that flows through the burner, what volume of air is needed to burn it completely? Assume that the temperature of the burner is 195.0 Celcius, the pressure is 1.1 atmosphere, and the mole fraction of O2 in air is 0.21.

1 answer:

adelina 88 [10]2 years ago

6 0

Answer:

At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane

Explanation:

The combustion stoichiometry is as follows:

C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:

MW 44 5x32 4x18 3x44

So each gram of propane is 1/44 = 0.02272 mol propane

and will need 5 x 0.02272 = 0.1136 mol oxygen

At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.

At the low pressure in the burner we can use the Ideal Gas Law

PV=nRT, or V = nRT/P

P = 1.1 x 101325 Pa = 111457 Pa

T = 195°C + 273 = 468 K

R = 8.314

and we calculated n = number of moles air = 0.54 mol

So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.

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Answer is: 0,133 mol/ l· atm.
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n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
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Answer:

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Explanation:

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Hello there!

To determine the fraction of the hydrogen atom's mass that is in the nucleus, we have to keep in mind that a Hydrogen atom has 1 proton and 1 electron. Protons are in the nucleus while electrons are in electron shells surrounding the nucleus. The mass of the nucleus will be equal to the mass of 1 proton and we can express the fraction as follows:

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So, the fraction of the hydrogen atom's mass that is in the nucleus is 0,9995. That means that almost all the mass of this atom is at the nucleus.

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2 years ago

Which of the following statements is true about the following reaction?

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The correct reaction equation is:

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Answer:

b) 1 mole of water is produced for every mole of carbon dioxide produced.

Explanation: <u>CONVERT EVERYTHING TO MOLES OR VOLUME, THEN COMPARE IT WITH THE COMPOUND'S STOICHIOMETRY IN CHEMICAL EQUATION.</u>

a) <u>22.4 L of $CO_{2}$ gas</u> is produced only when <u> $\frac{22.4}{3}$ L of $C_{6}H_{8}O_{7}$ </u> is reacted with 22.4 L of $NaHCO_{3}$ . So it is wrong.

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d) 54 g of water or 3 moles of $H_{2}O$ (<em>Molecular Weight of water is 18 g</em>) is produced when 3 moles of $NaHCO_{3}$ is used but in this option only one mole of $NaHCO_{3}$ is given. So it is wrong.

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dmitriy555 [2]

Answer: Reaction 1 is non spontaneous.

Explanation:

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T = temperature in Kelvin

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