Answer: Rubbing alcohol molecules have a polar and nonpolar part, which means they are able to form hydrogen bonds with water and therefore able to mix with it.
Explanation:
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).
Answer:
Kc for this reaction is 0.43
Explanation:
This is the equilibrium:
N₂(g) + 2H₂O(g) → 2NO(g) +2H₂(g)
And we have all the concentration at equilibrium:
N₂: 0.25M
H₂ : 1.3M
NO: 0.33M
H₂: 1.2M
They are ok, because they are in MOLARITY. (mol/L)
Let's make the expression for Kc
Kc = ( [NO]² . [H₂]² ) / ([N₂] . [H₂O]²)
Kc = (0.33² . 1.2²) / (0.25 . 1.2²)
Kc = 0.4356
In two significant digits. 0.43
Copper chloride ...............
= 
= 9.95 mL
x 100 
x 100 = 0.5 % error