Combustion is a reaction between a combustible substance and oxygen, to ultimately produce carbon dioxide and water. Reaction between carbon and oxygen would give,
C + O2 ------> CO2
Here, we have 86.5 grams of carbon dioxide, CO2, which is a product of combustion. Dividing this mass by the molar mass of CO2, which is 44 grams, we can determine the number of moles of CO2.
<u> 86.5 g CO </u> = 1.966 moles CO2
44 g CO2/ mole
Considering that CO2 is composed of 1 mole of carbon and 2 moles of oxygen, and that with complete combustion, 1 mole of carbon reacts to produces 1 mole of CO2, we can then determine the mass of the carbon in the hydrocarbon fuel.
1.966 moles CO2 x <u> 1 mole C </u> x <u> </u><u>12 g C </u> = 23.59 g C
1 mole CO2 1 mole C
We were given 25.0 grams of the fuel hydrocarbon. A hydrocarbon is a substance consisting of carbon and hydrogen. To determine the mass of the hydrogen in the fuel, we simply subtract 23.59 grams from 25.0 grams.
25.0 g - 23.59 g = 1.41 grams Hydrogen
To know the number of moles of hydrogen, we divide the mass of the hydrogen in the fuel by the molar mass of hydrogen, which is 1.01 g/mole. Thus, we have 1.396 mole hydrogen.
To determine the empirical formula, we divide the number of moles carbon by the number of moles hydrogen, and find a factor that would give whole number ratios for the carbon and hydrogen in the fuel,
Carbon: <u> 1.966 mol </u> = 1.408 x 5 (factor) = 7
1.396 mol
Hydrogen: <u> 1.396 mol </u> = 1.00 x 5 (factor) = 5
1.396 mol
Thus, the empirical formula is C7H5
1. A radical is a reactive intermediate with a single ____________ electron, formed by ____________ of a covalent bond.
1. A: Unpaired, and homolysis
2. Allylic radicals are stabilized by ____________ , making them ____________ stable than tertiary radicals.
2. A: Resonance, and more
3. A compound that contains an especially weak bond that serves as a source of radicals is called a radical ____________ .
3. A: Initiator
4. Treatment of cyclohexene with N-bromosuccinimide in the presence of light leads to ____________ by ____________ intermediates.
4. A: Allylic substitution by radical
Answer:
Mass released = 8.6 g
Explanation:
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Solution:
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
Answer:
nBACO3=m/M=9,83/197=0,05(mol) ->nHCl=0,05.2/1=0,1(mol) =>VHCl=n/CM=0,1/0,44=0,227(lít)
Explanation:
