Answer:
The laboratory demonstration consists of the following;
The compounds present in the combustion reaction = Methane, CH₄ and Oxygen, O₂
The chemical equation for the combustion reaction is given as follows;
CH₄ + 2O₂ → CO₂ + 2H₂O
Therefore;
A. The equation given as CH₄ + O → CO₂ is not correct because;
1) Oxygen gas exist as diatomic molecules, O₂, and given that the experiment involves the mixture of gases, the oxygen gas present which can exist as a separate compound, should be represented as O₂
2) The number of oxygen molecules in the reaction is two rather than one
3) The product also includes two molecules of water (vapor) H₂O
B. The correct equation for the reaction should be given as follows;
CH₄ + 2O₂ → CO₂ + 2H₂O
B i) The constituents of the equation is obtained by the knowledge of the fact that the combustion reaction of an organic substance such as methane in the presence of oxygen yields, carbon dioxide and water (vapor)
The equation showing the relative amounts the reacting compounds is by balancing the basic equation of the combustion of methane in the presence of oxygen
Explanation:
The state of matter is liquid.
Answer:
24g of NaOH are required
Explanation:
Molarity, M, is an unit of concentration widely used in chemistry defined as the ratio between moles of solute (In this case, NaOH), and volume of solution in liters.
We can find the moles of NaOH and its mass with the volume and desired concentration as follows:
<em>Moles NaOH:</em>
400.0mL = 0.400L * (1.50mol / L) = 0.600 moles NaOH
<em>Mass NaOH -Molar mass: 40.0g/mol-:</em>
0.600 moles * (40.0g / mol) =
<h3>24g of NaOH are required</h3>
Answer:
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Explanation:
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