Wiki User
∙ 9y ago??
Shafait Anwar
The o-toluidine method is specific to detecting glucose because o-toluidine reacts specifically with aldehyde functional groups on glucose molecules. Fructose does not have an aldehyde functional group, so it will not react with o-toluidine in the same way as glucose, making this method ineffective for detecting fructose in a solution.
Anonymous
The reason why Ortho-Toluidine cannot be recommended for estimation of fructose is because o-Toluidine is fairly specific for aldosugar and fructose is a ketosugar.
Anonymous
Because like glucose it does not have aldehyde group it contain ketose group so o toluidine cannot be used for detection of fructose
No, you cannot find yellow adulteration by chana dal using iodine solution. Iodine solution is typically used to detect the presence of starch in food products. Yellow coloring in chana dal may be due to artificial coloring agents or other adulterants that may require different testing methods for detection.
Cellulose cannot be ingested by humans because our bodies lack the necessary enzymes to break down its beta-linkages. Sucrose, maltose, and fructose are all types of sugars that can be metabolized by the human body for energy.
Benedict's solution changes colors (blue to green to yellow to orange to red) in the presence of "reducing" sugars, which are not normally present in saliva. An interesting experiment, however, is testing table sugar with Benedict's solution. Table sugar is a glucose sugar joined to a fructose sugar, so they cannot react with the Benedict's solution and no color change occurs. Put table sugar in your mouth for a few moments, and then test the saliva. Now the Benedict's solution will react! (The reason: saliva has an enzyme, amylase, which breaks the glucose and fructose apart so that they can react to the Benedict's.)
Monosaccharides are simple sugars that cannot be hydrolyzed into smaller units. They are the most basic unit of carbohydrates and include glucose, fructose, and galactose. These sugars are typically the building blocks for more complex carbohydrates like disaccharides and polysaccharides.
Sucrose cannot conduct electricity as a solid because it does not contain free ions to carry the charge. However, when it is dissolved in water to form a solution, sucrose is broken down into its constituent ions (glucose and fructose) and can conduct electricity due to the presence of these ions.
Lugol's solution is a reagent commonly used to detect the presence of starch. It cannot detect simple sugars because its active ingredient, iodine, does not react with simple sugars like glucose or fructose. Simple sugars do not contain the necessary chemical structure for the iodine in Lugol's solution to form a complex with, so they remain undetected in this test.
A filtrate is already in solution.
Glucose, fructose, and galactose are common examples of monosaccharides. They are simple sugars that cannot be broken down further into simpler sugars.
No, you cannot find yellow adulteration by chana dal using iodine solution. Iodine solution is typically used to detect the presence of starch in food products. Yellow coloring in chana dal may be due to artificial coloring agents or other adulterants that may require different testing methods for detection.
High fructose corn syrup, HFCS, is a blend of two simple sugars, fructose and glucose. Both sugars have the same chemical formula, C6H12O6, but have unique chemical structures. "Regular" fructose, found in fruit or sucrose ( table sugar) has the same chemical formula. Sucrose, however, is a disaccharide of fructose and glucose which means that the fructose and glucose are chemically linked. Therefore, the ratio of the fructose to glucose in sucrose is precisely 50:50, 1:1, and the %fructose cannot exceed 50%. This is not the case with HFCS. Depending on the needs of the manufacturers the %fructose can range from 42% to 90%. Dairy and baked products use 42%, Pepsi and Coke use 55% fructose, and low-cal products may contain as much as 90% fructose.
Cellulose cannot be ingested by humans because our bodies lack the necessary enzymes to break down its beta-linkages. Sucrose, maltose, and fructose are all types of sugars that can be metabolized by the human body for energy.
Nothing that follows can be in the solution set of 1 so all of it cannot.
Sucrose is made up of glucose and fructose. To convert sucrose into glucose, you can use an enzyme called invertase, which hydrolyzes sucrose into its constituent glucose and fructose molecules. Heating sucrose in water can also break it down into glucose and fructose.
Cellulose cannot be digested by humans.Cellulosecellulosehumans cannot digest cellulosecelluloseCelluloseCellulose (aka Fiber) can not be digested by humans because, we don't have the bacteria needed to break down cellulose. Sucrose, Maltose, and Fructose are all disaccharides (carbohydrates/sugars) and are all able to be broken down to glucose in the body.
Benedict's solution changes colors (blue to green to yellow to orange to red) in the presence of "reducing" sugars, which are not normally present in saliva. An interesting experiment, however, is testing table sugar with Benedict's solution. Table sugar is a glucose sugar joined to a fructose sugar, so they cannot react with the Benedict's solution and no color change occurs. Put table sugar in your mouth for a few moments, and then test the saliva. Now the Benedict's solution will react! (The reason: saliva has an enzyme, amylase, which breaks the glucose and fructose apart so that they can react to the Benedict's.)
Monosaccharides are simple sugars that cannot be hydrolyzed into smaller units. They are the most basic unit of carbohydrates and include glucose, fructose, and galactose. These sugars are typically the building blocks for more complex carbohydrates like disaccharides and polysaccharides.
No answer is possible. As soon as you have one valid line, all points that are not on that line cannot be part of the solution set. Therefore the solution set cannot be all real numbers.No answer is possible. As soon as you have one valid line, all points that are not on that line cannot be part of the solution set. Therefore the solution set cannot be all real numbers.No answer is possible. As soon as you have one valid line, all points that are not on that line cannot be part of the solution set. Therefore the solution set cannot be all real numbers.No answer is possible. As soon as you have one valid line, all points that are not on that line cannot be part of the solution set. Therefore the solution set cannot be all real numbers.