10. Identify the products in the combustion of methane.
1 answer:
Answer:
a. CO2 and H20
Explanation:
Chemically, this combustion process consists of a reaction between methane and oxygen in the air. When this reaction takes place, the result is carbon dioxide (CO2), water (H2O), and a great deal of energy. The following reaction represents the combustion of methane:
CH4[g] + 2 O2[g] -> CO2[g] + 2 H2O[g] + energy
One molecule of methane, (the [g] referred to above means it is gaseous form), combined with two oxygen molecules, react to form a carbon dioxide molecule, and two water molecules usually given off as steam or water vapor during the reaction and energy.
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Answer:
the concentration of in the original solution= 0.0088 M
the concentration of in the original solution = 0.058 M
Explanation:
Given that:
The volume of the sample containing Cd2+ and Mn2+ = 50.0 mL; &
was treated with 64.0 mL of 0.0600 M EDTA
Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+
i.e the strength of the Ca2+ = 0.0310 M
Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+
To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :
Volume of newly freed EDTA =
=
= 7.3367 mL
concentration of =
=
= 0.0088 M
Thus the concentration of in the original solution = 0.0088 M
Volume of excess unreacted EDTA =
=
= 8.318 mL
Volume of EDTA required for sample containing and
= (64.0 - 8.318) mL
= 55.682 mL
Volume of EDTA required for = Volume of EDTA required for
sample containing and
-- Volume of newly freed EDTA
Volume of EDTA required for = 55.682 - 7.3367
= 48.3453 mL
Concentration of =
Concentration of =
Concentration of in the original solution= 0.058 M
Thus the concentration of = 0.058 M