Answer:
Sure because I need more friends
1. Democritus characterized atoms as indestructible and indivisible.
Answer:
carbon dioxide and water
Explanation:
Example: Combustion of Methane (CH₄(g))
CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(g)**
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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*
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*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
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**Also, note that water, H₂O(g), is listed as a gas. In some cases it will be listed as a liquid, H₂O(l).
The temperature increase is from 19.5 to 100 degrees centigrade or 80.5 degrees centigrade. The calorie increase is 2.50 x 1000 x 0.238902957619 or a total of 597.25 calories. 597.25/80.5 = 7.419 calories per degree centigrade. 7.419/135 grams = 0.0549 calories/gram/degree centigrade. The conversion from kilo joules involves multiplying the calories per joule x 1000 to get the number of calories in one kilo joule and then by the 2.5.
A whiteout occurs when strong updrafts and downdrafts combine
