Answer: balanced reaction equation
C8H18(g) + 25/2 O2(g) ---------> 8CO2(g) + 9H2O(l)
Explanation:
Part A- coefficients
1, 25/2,8,9
Part B
Oxygen is the limiting reactant
Part C
If 1 mole of octane produced 9 moles of water from the balanced reaction equation
0.28 moles of octane will produce 0.28×9= 2.52 moles of water
Part D
If 12.5 moles of oxygen reacts with 1 mole of octane
0.63 moles of oxygen will react with 0.63/12.5=0.0504moles of octane
Amount of octane left= 0.280-0.0504=0.2296 moles
Answer:
The correct answer is electrons.
Explanation:
The electrons refer to the negatively charged constituents, which are arranged in the orbits around the nucleus of an atom. It determines all the chemical and physical characteristics of an atom except the radioactivity and mass. It is a stable subatomic particle witnessed in all the atoms and functioning as the chief carrier of electricity in solids.
Answer:
carbon dioxide and water
Explanation:
Example: Combustion of Methane (CH₄(g))
CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(g)**
____________________
Note: The combustion of any hydrocarbon produces CO₂ & H₂O. That is,
Ethane (C₂H₆) + O₂ => CO₂(g) + H₂O(g)
Propane (C₃H₈) + O₂ => CO₂(g) + H₂O(g)
Butane (C₄H₁₀) + O₂ => CO₂(g) + H₂O(g)
The issue remaining is to balance the reaction equation. For these type equation balance Carbon 1st, then Hydrogen and finish with Oxygen. Balancing in this order leaves Oxygen which can be balanced using fractions. If problem requires lowest whole number ratios of elements, simply multiply entire equation by 2 to get standard equation*
______________________
*Standard Equation is defined as the smallest whole number ratios of elements. The 'standard equation' is significant in that it is assumed to be at STP conditions; i.e., 0⁰C (=273K) & 1.0 Atmosphere pressure.
- Ethane (C₂H₆) + 7/2O₂(g) => 2CO₂(g) + 3H₂O(g)
=> 2C₂H₆ + 7O₂(g) => 4CO₂(g) + 6H₂O(g) <= Standard Form of Rxn
- Propane (C₃H₈) + 5O₂(g) => 3CO₂(g) + 4H₂O(g) <= Standard Form of Rxn (no need to balance with the '2' multiple)
- Butane (C₄H₁₀) + 13/2O₂ => 4CO₂(g) + 5H₂O(g)
=> 2C₃H₈ + 13O₂(g) => 4CO₂(g) + 5H₂O(g) <= Standard Form of Rxn
______________________
**Also, note that water, H₂O(g), is listed as a gas. In some cases it will be listed as a liquid, H₂O(l).
Answer:
Total Kcal energy produced in the catabolism of mannoheptulose = 1184 Kcal
Explanation:
The molecular formula of mannoheptulose is C₇H₁₄O₇.
The structure is as shown in the attachment below.
Number of C-C bonds present in mannoheptulose = 6
Number of C-H bonds present in mannoheptulose = 8
Since the each C-C bond contains 76 Kcal of energy,
Amount of energy present in six C-C bonds = 6 * 76 = 456 Kcal
Also, since each C-H bond contains 91 Kcal of energy;
amount of energy present in eight C-H bonds = 8 * 91 = 728 Kcal
Total Kcal energy produced in the catabolism of mannoheptulose = 456 + 728 = 1184 Kcal
