In the complete combustion of 1.60 moles of benzene, C6H6, 12 moles of oxygen, O2, is consumed.
Combustion is defined as the process of burning something. In chemistry, combustion refers to the chemical process between a fuel and an oxidant, usually oxygen to produce heat and light in the form of flame.
In a complete combustion, oxygen is sufficient to react with any hydrocarbons to produce carbon dioxide and water.
Balancing the combustion reaction of benzene, we have:
2C6H6 + 15 O2 = 12CO2 + 6H2O
Based on the balanced combustion reaction above, 2 moles of benzene requires 15 moles of oxygen to have a complete combustion.
If we have 1.60 moles C6H6,
moles O2 = mole ratio x mole of benzene
moles O2 = (15 moles O2/2 moles C6H6) x 1.60 moles C6H6
moles O2 = 12
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Answer: D
Explanation:
Chlorine is in group 7 or (VII) in Roman numerals, which means it has 7 balance electrons. It only needs one electron to become stable, hence it is next to the noble gases
Actually, it could indeed be the independent variable.
Answer: 24.1%, under below assumptions.
Justification:
The question is quite ambiguous, because one of the data is not clearly stated. It says that the mixture consists on two compounds:
- sodium bicarbonate, and
- ammonium bicarbonate
.After, it says that it is 75.9 % bicarbonate, but it does not specify which bicarbonate, it might be the sodium bicarbonate or the ammonium bicarbonate. It is apparent that you omitted that information by error.
Given that later, the question is <span>what the mass percent of sodium bicarbonate is in the mixture, it is supposed that the 75.9% content is of ammonium bicarbonate.
With that said, you can calculate the mass percent of sodium bicarbonate, because there are only two compounds and so you know that both add up the 100% of the mixture.
In formulas:
100% = %m/m sodium bicarbonate + %m/m ammonium bicarbonate = 100%
=> % m/m sodium bicarbonate = 100% - % m/s ammonium bicarbonate
=> % sodium bicarbonate = 100% - 75.9% = 24.1%
Answer: 24.1%
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