Answer: the statements in 1 and 2 are true of IR spectroscopic region.
1. In general, the IR FUNDAMENTAL region has a longer wavelength region than the region we call the ultraviolet (uv) region.
2. We can sense some of the frequencies of the FUNDAMENTAL region of the IR as heat
Explanation:
IR has energy value between 10^-5eV - 10^-2eVwhile
UV has energy value of 4eV - 300eV
IR has low photon energy and cannot alter atoms and molecules while UV has sufficient energy to iodize atoms therefore UV has a higher energy band.
Infrared light falls just outside the visible spectrum, beyond the edge of what we can see as red.
Answer:
(S)-3-methoxy-3-methylbutan-2-ol
Explanation:
In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile (
) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.
See figure 1
I hope it helps!
Answer:
Explanation:
Of course you could do the separation chemically. Dissolve the salt up in water, pass thru a filter, wash the iron filings with ethanol, which would encourage the salt to precipitate from solution.
I do hope I helped you! :)
Answer:
1.8 × 10⁻⁴ mol M/s
Explanation:
Step 1: Write the balanced reaction
2 Br⁻ ⇒ Br₂
Step 2: Establish the appropriate molar ratio
The molar ratio of Br⁻ to Br₂ is 2:1.
Step 3: Calculate the rate of appearance of Br₂
The rate of disappearance of Br⁻ at some moment in time was determined to be 3.5 × 10⁻⁴ M/s. The rate of appearance of Br₂ is:
3.5 × 10⁻⁴ mol Br⁻/L.s × (1 mol Br₂/2 mol Br⁻) = 1.8 × 10⁻⁴ mol Br₂/L.s
Answer:
126.8, Iodine
Explanation:
- mass ×abundance/100
- (126.9045×80.45/100)+(126.0015×17.23/100)+(128.2230×2.23/100)
- 102.1+21.7+3=126.8
<em>IODINE</em><em> </em><em>has</em><em> </em><em>an</em><em> </em><em>atomic</em><em> </em><em>mass</em><em> </em><em>of</em><em> </em>126.8 or 126.9