Huh what type of question is that?!
Answer:
The product is cyclohexanol
Explanation:
Firstly,
A ketone undergo a borohydride reduction reaction to form an alcohol as below,
R-CO-R' ⇒ R-CO(OH)-R'
- IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹
- From 1H-NMR, the product has 10 hydrogen atoms, the MS suggest that the formula is C₅H₁₀O (M = 86). With this formula, the alcohol is monosaturated. Since, the substance already underwent reduction reaction, the only way to suggest a monosaturated compound is a cyclic alcohol. So the compound is cyclopentanol.
- Check with other spectroscopic properties,
- 3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-<u>C</u>H₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-<u>C</u>H₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-<u>C</u>H(CH₂-)-OH).
1.56 δ (4H, triplet) - (-C<u>H</u>₂-CH₂-CH-OH) ; triplet as coupling with 2 H,
1.78 δ (4H, multiplet) - (-CH₂-C<u>H</u>₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH
3.24 δ (1H, quintet); - (-CH₂-CH₂-C<u>H</u>(CH₂-)-OH), coupling with4 H of 2 group of CH₂
3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-O<u>H</u>), hydrogen of alcohol group, not tend to coupling with other hydrogen
the compound is B.
Water is a compound because it is made up of water molecules.
N=m/M
n=118/58.93=2
Answer: 2 moles
To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-<span>1 * 298.15K)
</span>n= 1.636 moles
The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= <span>3.38 L</span>