12 moles of water H₂O are produced from the combustion of pentane.
Explanation:
We have the following combustion of pentane (C₅H₁₂):
C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O
Knowing the chemical reaction we devise the following reasoning:
if 1 moles of pentane C₅H₁₂ produces 6 moles of water H₂O
then 2 moles of pentane C₅H₁₂ produces X moles of water H₂O
X = (2 × 6) / 1 = 12 moles of water H₂O
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combustion of organic compounds
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Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have 
as nucleophile. Also, this compound is also in excess. So, we will have as solvent 
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).
Being frozen, staying a liquid, becoming ice, and becoming a gas (steam)
There are 19.5 g Na in 71.4 g NaHCO₃
Calculate the <em>molecular mass of NaHCO₃</em>.
1 Na = 1 × 22.99 u = 22.99 u
1 H = 1 × 1.008 u = 1.008 u
1 C = 1 × 12.01 u = 12.01 u
3 O = 3 × 16.00 u = <u>48.00 u
</u>
TOTAL = 84.008 u
So, there are 22.99 g of Na in 84.008 g NaHCO₃.
∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na
