Answer:
The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.
As we know that
<span>V1/T1 = V2/T2
V1 = 9.10 L
T1 = 471 K
V2 = 2.50 L
T2 = 2.5 x 471 / 9.10 = 129.3 K
T2 = 129.3 - 273 =
-143.6 deg Celsiu
hope it helps</span>
It can be more effective in liquids.
15 is the group that phosphorus is found in.
amino group
carboxyl group
R-group
single Hydrogen arom
