Answer: Charles's law, Avogadro's law and Boyle's law.
Justification:
Boyle's law states that at constant temperature PV = constant
Charles law states that at constant pressure V/T = constant
Avogadro's law states that at constant pressure ant temperature, equal volume of gases contain equal number of moles: V/n = constant
Ideal gas law states PV/nT = constant => PV = nT*constant = PV = nTR
Static electricity. they have the same charge and same charges repel each other.
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
Answer:
C. ↔
Explanation:
Resonating structure -
These are the set of lewis structures of the same compound .
Where the structure helps to show the delocalization of the electrons over the structure .
The charge on all the resonating structure remains the same .
All the structures can be interconverted to each other , and are shown by the arrow ↔ .
Hence , from the given information of the question,
The correct option is C. ↔