Benedict's solution is commonly used to test for the presence of reducing sugars, such as glucose and fructose. In this test, Benedict's solution is mixed with the substance to be tested and heated. If a reducing sugar is present, it will undergo oxidation and reduce the copper(II) ions in Benedict's solution to copper(I) oxide, which precipitates as a red or orange precipitate.
To determine which of the two substances can be oxidized with Benedict's solution, we need to know the nature of the functional group present in each substance. Without this information, it is difficult to determine the substance's reactivity with Benedict's solution.
However, if we assume that both substances are monosaccharides, such as glucose and fructose, then they both contain an aldehyde functional group (CHO). In this case, both substances can be oxidized by Benedict's solution. The aldehyde group is oxidized to a carboxylic acid, resulting in the reduction of copper(II) ions to copper(I) oxide.
The balanced equation for the oxidation reaction of a monosaccharide with Benedict's solution can be represented as follows:
C₆H₁₂O₆ (monosaccharide) + 2Cu₂+ (Benedict's solution) + 5OH- (Benedict's solution) → Cu₂O (copper(I) oxide, precipitate) + C₆H₁₂O₇ (carboxylic acid) + H₂O
It is important to note that without specific information about the substances involved, this is a generalized explanation assuming they are monosaccharides. The reactivity with Benedict's solution may vary depending on the functional groups present in the actual substances.
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a) perform a linear search by hand for the array [20,−20,10,0,15], loching for 0 , and showing each iteration one line at a time b) perform a binary search by hand fo the array [20,0,10,15,20], looking for 0 , and showing each iteration one line at a time c) perform a bubble surt by hand for the array [20,−20,10,0,15], shouing each iteration one line at a time d) perform a selection sort by hand for the array [20,−20,10,0,15], showing eah iteration one line at a time
In the linear search, the array [20, -20, 10, 0, 15] is iterated sequentially until the element 0 is found, The binary search for the array [20, 0, 10, 15, 20] finds the element 0 by dividing the search space in half at each iteration, The bubble sort iteratively swaps adjacent elements until the array [20, -20, 10, 0, 15] is sorted in ascending order and The selection sort swaps the smallest unsorted element with the first unsorted element, resulting in the sorted array [20, -20, 10, 0, 15].
The array is now sorted: [-20, 0, 10, 15, 20]
a) Linear Search for 0 in the array [20, -20, 10, 0, 15]:
Iteration 1: Compare 20 with 0. Not a match.
Iteration 2: Compare -20 with 0. Not a match.
Iteration 3: Compare 10 with 0. Not a match.
Iteration 4: Compare 0 with 0. Match found! Exit the search.
b) Binary Search for 0 in the sorted array [0, 10, 15, 20, 20]:
Iteration 1: Compare middle element 15 with 0. 0 is smaller, so search the left half.
Iteration 2: Compare middle element 10 with 0. 0 is smaller, so search the left half.
Iteration 3: Compare middle element 0 with 0. Match found! Exit the search.
c) Bubble Sort for the array [20, -20, 10, 0, 15]:
Iteration 1: Compare 20 and -20. Swap them: [-20, 20, 10, 0, 15]
Iteration 2: Compare 20 and 10. No swap needed: [-20, 10, 20, 0, 15]
Iteration 3: Compare 20 and 0. Swap them: [-20, 10, 0, 20, 15]
Iteration 4: Compare 20 and 15. No swap needed: [-20, 10, 0, 15, 20]
The array is now sorted: [-20, 10, 0, 15, 20]
d) Selection Sort for the array [20, -20, 10, 0, 15]:
Iteration 1: Find the minimum element, -20, and swap it with the first element: [-20, 20, 10, 0, 15]
Iteration 2: Find the minimum element, 0, and swap it with the second element: [-20, 0, 10, 20, 15]
Iteration 3: Find the minimum element, 10, and swap it with the third element: [-20, 0, 10, 20, 15]
Iteration 4: Find the minimum element, 15, and swap it with the fourth element: [-20, 0, 10, 15, 20]
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