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babunello [35]
2 years ago
9

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.
Chemistry
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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At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2
s344n2d4d5 [400]
Answer is:  0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K. 
Ideal gas law: p·V = n·R·T
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.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.

4 0
2 years ago
Olivia, a Latina student, is told that she can check only two books out of the library at a time, but Leann, a white student, is
Sliva [168]

Answer:

a

Explanation:

the others are rude, and rather support this, while a helps to support the ending of white privlige

4 0
2 years ago
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What is the fraction of the hydrogen atom's mass (11h) that is in the nucleus? the mass of proton is 1.007276 u, and the mass of
ololo11 [35]
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:

Mass Fraction= \frac{mass 1 Proton}{mass H atom} = \frac{1,007276 u}{1,007825}=0,9995

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.

Have a nice day!
3 0
2 years ago
Which of the following statements is true about the following reaction?
MAVERICK [17]

The correct reaction equation is:

3NaHCO_{3} (aq) + C_{6}H_{8}O_{7} (aq) \rightarrow 3CO_{2} (g) + 3H_{2}O (l) + Na_{3}C_{6}H_{5}O_{7} (aq)

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.

b) Since in the chemical equation the stoichiometric coefficient of CO_{2}  and H_{2}O are same so the number of moles or volume of each of them will be same whatever the amount of reactants taken. <u>Therefore it is correct option.</u>

c)  6.02\times 10^{23} molecules is equal 1 mole of Na_{3}C_{6}H_{5}O_{7} if produced then 3 moles of NaHCO_{3} is required, which is not given in the option. So it is wrong.

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.

8 0
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dmitriy555 [2]

Answer: Reaction 1 is non spontaneous.

Explanation:

According to Gibb's equation:

\Delta G=\Delta H-T\Delta S

\Delta G = Gibbs free energy  

\Delta H = enthalpy change

\Delta S = entropy change  

T = temperature in Kelvin

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\Delta G= -ve, reaction is spontaneous

\Delta G= 0, reaction is in equilibrium

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As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.

6 0
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