Answer:
The answer to your question is 98.9 %
Explanation:
Data
moles of methane = CH₄ = 2.0
excess air
Percent yield = ?
mass of CO₂ = 87 g
- Balanced chemical reaction
CH₄ + 2O₂ ⇒ CO₂ + 2H₂O
Reactants Elements Products
1 C 1
4 H 2
4 O 2
- Calculate the molar mass of CH₄
CH₄ = 12 + 4 = 16 g
- Convert the moles to mass
16 g of CH₄ ----------------- 1 mol
x ----------------- 2 moles
x = (2 x 16) / 1
x = 32 g of CH₄
-Calculate the theoretical formation of CO₂
16 g of CH₄ ----------------- 44 g of CO₂
32 g of CH₄ ---------------- x
x = (32 x 44) / 16
x = 88 g of CO₂
-Calculate the Percent yield
Percent yield = Actual yield/Theoretical yield x 100
Percent yield = 87/88 x 100
Percent yield = 98.9 %
Answer:
31 moles
Explanation:
The balanced combustion reaction of the wax, 
is shown below as:
As seen from the reaction,
1 mole of wax, on combustion produces 31 moles of carbon dioxide, 
<u>Hence, moles of when 1 mole of wax, is burnt = 31 moles</u>
Answer:
The correct answer to this is
e. all of the above
Explanation:
An hypothesis is a proposal to solve or answer a new or strange phenomenon that is not suitably explained by current scientific knowledge. The outcome of an hypothesis is not predetermined. An hypothesis can be termed a scientific hypothesis only when the statements of the hypothesis can be tested also known as the testability of the hypothesis. Hypothesis are normally focused on observations that are unexplained by current scientific theories
Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.
<h3>
What is the keto-enol means in tautomer?</h3>
They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.
- The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
- If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
- If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
- This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.
