It's a hazard symbol for irritant.
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:
518 mL
Explanation:
We can solve this using Boyle's Law Formula
P1V1 = P2V2
where p1 = initial pressure, p2 = final pressure, v1 = initial volume and v2 = final volume
here , the initial pressure is 1 atm and the initial volume is 725mL
we are given the final pressure 1.4 and we need to find the final volume
so we have p1v1 = p2v2
==> plug in p1 = 1 , v1 = 725 mL and p2 = 1.4
(1)(725) = (1.4)v2
==> multiply 1 and 725
725 = (1.4)(v2)
==> divide both sides by 1.4
v2 = 518
N2 would have a volume of 518mL at 1.4atm
Alkali Metals (Group 1) elements experience an increase in the vigour of their reaction in water as they go down the group (as the atomic number increase). As such the most reactive Alkali Metal would be
FRANCIUM, which is at the base of Group One.
Quite frankly, you do not want Francium to react with water- that's a huge explosion on your hand.
Identify<span> each </span>bond<span> as either </span>polar<span> or nonpolar.</span>