The balanced equation says that 2 moles C₂H₆ and 7 moles O₂ react together, i.e. in a ratio of 7:2 or 3.5 moles of O₂ to C₂H₆.
With molar masses 30.07 g/mol (C₂H₆) and 31.998 g/mol (O₂), the given quantities amount to
(19 g C₂H₆) × (1/30.07 mol/g) ≈ 0.63 mol C₂H₆
(115 g O₂) × (1/31.998 mol/g) ≈ 3.59 mol O₂
Now, 0.63/2 ≈ 0.32, and for every 0.32 mol C₂H₆ consumed, the reaction requires 7×0.32 ≈ 2.2 mol O₂. Then in order to consume all of the C₂H₆, the reaction would need 2×2.2 ≈ 4.4 mol O₂, which we don't have.
In other words, we have too much C₂H₆ and not enough O₂, so O₂ is the limiting reactant.
The correct answer is resource partitioning.
The concept of resource partitioning is applicable in the branch of ecology. It signifies towards the procedure by which natural selection mediates competing species into distinct patterns of different niches or resource utilization.
When the species differentiate a niche to prevent competition for food resources, it is known as resource partitioning. At certain times, the competition is among the species, known as interspecific competition, and at sometimes it is among the individuals of the similar species, that is, intraspecific competition.
C3H8 + 5O2 → 3CO2 + 4H2O .
This reaction is balanced correctly because number of atoms on reactant side and product side are same.
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Answer:</h3>
A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent.
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Explanation:</h3>
- A solution is made by dissolving a solute in a solvent.
- For example dissolving a salt in a solvent such as water results to a solution.
- Solution may either be saturated or unsaturated.
- Unsaturated solution is a solution that can dissolve more solute upon addition because it has not reached saturation.
- A saturated solution on the other hand is a solution that has maximum solute and the concentration of solute is maximum and thus the solvent can not dissolve any more solute.
- Therefore, a saturated solutions contain maximum concentration of a solute dissolved in the solute.
Explanation:
<em><u>in fact , we can use newtons second law of motion (see the SPT: Force topic) to calculate the acceleration in each of these cases</u></em>
<em><u>in fact , we can use newtons second law of motion (see the SPT: Force topic) to calculate the acceleration in each of these caseshope it helps you like me plz</u></em>
