<h3>Answer:</h3>
89.6 L of O₂
<h3>Solution:</h3>
The balanced chemical equation is as,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,
44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂
So,
88 g CO₂ will be produced by = X L of O₂
Solving for X,
X = (88 g × 44.8 L) ÷ 44 g
X = 89.6 L of O₂
Only the Fe is unbalanced.
:)
There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.
Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
Answer:
Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.
Explanation:
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