Answer:
a) The functional group that will be evident in the IR spectrum is the OH group.
b) OH group appears between 3200-3600 cm⁻¹
c) An important impurity that have the same functional group is water.
Explanation:
Eugenol is a chemical substance that consist in a benzene that have in 1 an alcohol, in position 2 a methyl ether and in position 4 an 1-propene bonded by the terminal alkyl carbon.
a) Having this in mind, the functional group that will be evident in the IR spectrum is the OH group.
b) This OH group appears between 3200-3600 cm⁻¹
c) An important impurity that have the same functional group is water. When you have water in your sample a big signal will appear in this zone and it is possible that overlapes the OH signal of eugenol.
I hope it helps!
Liquid? maybe, its really inbetween if you get what i mean
Answer:
we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
Explanation:
when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
The speed of the polar spot depends largely on the level of polarity, an increase in the polarity will see both spots of Neat hexane run when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate
Here is a site my buddie has to help you. Well co-owner..
https://www.quora.com/Why-is-fresly-prepared-FeSO4-required-for-the-ring-test
Answer:
392.97 litres
Explanation:
From the equation of reaction, we can see that 1 mole of methane yielded 1 mole of carbon iv oxide. Hence, 15.9 moles of methane will yield 15.9 moles of carbon iv oxide.
At s.t.p one mole of a gas occupies a volume of 22.4L ,hence 15.9 moles of a gas will occupy a volume of 22.4 × 15.9 which equals
356.16L.
Now, we can use the general gas equation to get the volume produced at the values given.
We have the following values;
V1 = 356.16L P1= 1 atm ( standard pressure) T1 = 273K ( standard temperature) V2 = ? T2 = 23.7 + 273 = 296.7K P2 = 0.985 atm
The general form of the general gas equation is given as :
(P1V1)T1 = (P2V2)/T2
After substituting the values , we get V2 to be 392.97Litres
