The percent yield of carbon dioxide will be 49.0 %.
<h3>Percent yield</h3>
First, let's look at the equation of the reaction:

The mole ratio of octane to oxygen is 2:25.
Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol
Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol
Thus, octane is limiting.
Mole ratio of octane to carbon dioxide = 2:16.
Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol
Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams
Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %
More on percent yield can be found here: brainly.com/question/17042787
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You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles
Explanation:
The three sub atomic particles are;
Neutrons, Electrons and Protons
Hydrogen - H
Neutrons: 0
Electrons: 1
Protons: 1
Deuterium - D
Neutrons: 1
Electrons: 1
Protons: 1
Tritium - T
Neutrons: 2
Electrons: 1
Protons: 1
Answer:
a H2CO3 b HCO3- and c H+ and HCO3-
Explanation:
As the pKa value of phenol is more than that of carbonic acid(H2CO3), the carbonic acid will have high Ka value than that of phenol.
The acid that contain high Ka value act as stong acid.From that point of view H2CO3 is a strong acid than phenol as the Ka value of carbonic acid is greater than that of phenol.
The conjugate base of H2CO3 is bicarbonate ion(HCO3-)
c The species that predorminates at equilibrium are H+ and HCO3-