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Interpretation: To match the category of an organic compound for the given biological molecule.
Concept introduction: A
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Answer to Problem 5STP
Monosaccharides are most closely identified with a biological molecule i.e., carbohydrates.
Explanation of Solution
Carbohydrates are biochemical substances that contain a polyhydroxy
The classification of carbohydrates is done on the basis of molecular sizes which are as follows:
- Monosaccharide
- Disaccharide
- Polysaccharide
Monosaccharides are simple molecules or monomers which form carbohydrates. Such as glucose and fructose.
Disaccharides are carbohydrates formed when two monosaccharides are bonded together. In a disaccharide, the bond between two monosaccharides is known as glycosidic linkage. Some examples of disaccharides are sucrose, maltose, and lactose.
When more than two monosaccharide units are joined together, it yields into polysaccharide. Such as starch and cellulose.
Monosaccharides are most closely identified with a biological molecule i.e., carbohydrates.
Chapter 24 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
- Don't used hand raiting and don't used Ai solutionarrow_forward2' P17E.6 The oxidation of NO to NO 2 2 NO(g) + O2(g) → 2NO2(g), proceeds by the following mechanism: NO + NO → N₂O₂ k₁ N2O2 NO NO K = N2O2 + O2 → NO2 + NO₂ Ко Verify that application of the steady-state approximation to the intermediate N2O2 results in the rate law d[NO₂] _ 2kk₁[NO][O₂] = dt k+k₁₂[O₂]arrow_forwardPLEASE ANSWER BOTH i) and ii) !!!!arrow_forward
- E17E.2(a) The following mechanism has been proposed for the decomposition of ozone in the atmosphere: 03 → 0₂+0 k₁ O₁₂+0 → 03 K →> 2 k₁ Show that if the third step is rate limiting, then the rate law for the decomposition of O3 is second-order in O3 and of order −1 in O̟.arrow_forward10.arrow_forwardDon't used Ai solution and don't used hand raitingarrow_forward
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