<h3>
Answer:</h3>
8CO₂
<h3>
Explanation:</h3>
We are given;
- Butane is a hydrocarbon in the homologous series known as alkane.
We are required to determine the other product produced in the combustion of butane apart from water.
- We know that the complete combustion of alkane yields carbon dioxide and water.
- Therefore, combustion of butane will yield carbon dioxide and water.
- The balanced equation for the complete combustion of butane will be;
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
A. 0.25 g/mL
Because 6/24 is 0.25
Answer: Organisms breakdown __into smaller particles___ in order to produce__new compounds__?
Explanation:
branliest pls :)
I think the correct answer would be the third option. The correct name for the hydrocarbon described above would be 2-heptyne. It has a chemical formula written as CH3 - CH2 - CH2 - CH2 - C ≡ C - CH3. Counting the number of carbons, we have 7 carbon atoms so we use the prefix hepta-. Since it has a triple bond then it is an alkyne. So, it would be named as heptyne. The triple bond is located on the second carbon atom so we write 2 before the name to indicate the location of the triple bond. The name of the compound would be 2-heptyne.
Answer:
71.5g
Explanation:
The reaction equation is given as:
C + O₂ → CO₂
Mass of C = 42g
Mass of O₂ = 52g
Unknown:
Mass of CO₂ produced = ?
Solution
Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.
The extent of the reaction is controlled by this reactant.
Find the number of moles of the given species;
Number of moles = 
Number of moles of C = 
= 3.5mol
Number of moles of O₂ = 
= 1.63mol
Now;
From the balanced reaction equation;
1 mole of C reacted with 1 mole of O₂
We see that C is in excess and O₂ is the limiting reactant.
1 mole of O₂ will produce 1 mole of CO₂
So; 1.63mole of O₂ will produce 1.63 mole of CO₂
Mass of CO₂ = number of moles x molar mass
Molar mass of CO₂ = 44g/mol
Mass of CO₂ = 1.63 x 44 = 71.5g
