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asambeis [7]
4 years ago
8

What is indicated by Benedict's solution changing from blue to orange when heated?

Chemistry
2 answers:
Andrew [12]4 years ago
4 0

Answer:

D.the presence of reducing sugars is the correct answer.

Explanation:

Benedict's solution is used tests to identify the presence of reducing sugars.

If a solution carrying Benedict's reagent and reducing sugars are heated, then the copper II ions present in Benedict's solution are reduced to copper I ions and change in the color from blue to orange is due to the presence of reducing sugars.

Sidana [21]4 years ago
3 0

1.Benedict�s reagent (blue in color):

Benedict's reagent (also called Benedict's solution or Benedict's test) is a reagent named after an American chemist, Stanley Rossiter Benedict. It can be prepared from sodium carbonate, sodium citrate and copper(II) sulfate.

Benedict's reagent is used as a test for the presence of reducing sugars such as glucose, fructose, galactose, lactose and maltose.

In the presence of reducing sugars and heat in water benedict�s reagent will change color from blue to orange/green and sometimes as well orange red.

Benedict's reagent contains blue copper(II) sulfate (CuSO4) which is reduced (takes up lost electrons from the reducing sugar) to a, ORANGE/RED colored red copper(I) oxide (Cu2O) by aldehydes, also oxidizing them to carboxylic acids. The copper(I) oxide is insoluble in water and so precipitates. (wiki sourced).

Summary: benedicts reagent color changes from a blue, to then a green, then final orange red color when electrons from an �unknown� reducing sugar, in a warm water bath suspension.

CLUE: *Benedicts reagent changes color to a Orange/Red color if a reducing sugar is your unknown.

2. Biuret reagent (blue in color normally, unreacted): is made of potassium hydroxide (KOH) and copper (II) sulfate (CuSO4), together with potassium sodium tartrate.

The Biuret blue color (no heat needed) reagent turns Violet in the presence of proteins (also called by some polypeptide chains), and changes to Violet/Pink when combined with short-chain (not long) polypeptides (amin acids {monomers of a protein} linke dtogethe rin a chain by bonds identified as peptide bonds, hence �poly (many) peptides�.

The reagent, Biuret protein assay, is a colorimetric (color change) yest used to determine an unknown solution�s protein concentration.

3.Sudan IV (dark red/burgundy color) is a lysochrome (fat-soluble dye) diazo dye used for staining of lipids, triglycerides and lipoproteins on frozen paraffin sections.

In industry, Sudan IV is used to color nonpolar substances like oils, fats, waxes, greases, various hydrocarbon products, and acrylic emulsions. (wiki sourced). In other words lipids �pick up� the color of the Sudan IV dye.

Scenario:

In a lab room, one (you) is handed three small reagent containers, some beakers, some test tubes, safety gloves and goggles, and SIX unknown solutions in testubes, and a known control of distilled water, all to be tested, by you, with the reagents, for the presence of either MONOSACCHARIDES (reducing sugar varieties), PROTEINS (long or short chain polypeptides) and potential LIPIDS: oils fats and waxes

The six unknowns and distilled water are labeled one, two, three, four, five, six, and seven.

ALL are all clear in color.     Only one of the test tube containers contains pure distilled water.

TEST ONE:

When one uses Benedict�s reagent on samples from each test tube, in a hot water bath heat source: samples ONE and FOUR turn color from Benedict�s blue being added to them, in color, from a blue to green to red-orange color (a color change definitely happens).

Samples TWO, THREE, FIVE, SIX, and, SEVEN change from clear to the Benedict�s reagent blue color when the benedicts was added. In essence NO color change happens with these unknon samples, even in the warm water bath. ONLY ONE and FOUR are positive color changes.

TEST TWO:

When one uses the Biuret reagent on samples from test tubes TWO, and THREE, a Pink color forms, and in sample FIVE the chemical unknown changes to a Violet purple color.

IN Samples   ONE FOUR SIX and SEVEN stay the blue color of the Biuret reagent that was added to the initially clear samples. NO color change occurs in these

TEST THREE:

When one adds Sudan IV to all seven samples ONLY sample SIX reacts and the liquid of the unknown sample becomes a reddish / burgundy color.

Samples ONE, TWO, THREE, FOUR, FIVE, and SEVEN turn color to a dilute red color from the stain added to the clear samples, but not in the intensity that is sample unknown SIX.

What can you deduce for the preceding set of experiments as to what was in each sample regarding: Monosaccharide / reducing sugar, a short chain polypeptide, a long chain polypeptide, or a lipid.  Now you just have to do the samples .

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If the self ionization reaction is endothermic in nature then in this condition at the equilibrium if we increases the temperature then the formation of ions increases as a result of the shift of the reaction towards the right side.

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2 years ago
An unknown compound contains only C, H, and O. Combustion of 6.10 g of this compound produced 14.9 g of CO₂ and 6.10 g of H₂O. W
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Answer:

The answer to your question is   C₃H₃O

Explanation:

Data

Combustion of a compound C, H, O

mass = 6.10 g

mass CO2 = 14.9 g

mass of water = 6.10 g

Reaction

                     Cx Hy Oz   +   O2   ⇒     CO2   +   H2O

Process

1.- Calculate the moles of carbon

                          44 g of CO2   --------------  12 g of carbon

                           14.9 g of CO2 -------------   x

                            x = (14.9 x 12) / 44

                            x = 4.06 g

                          12 g of C    ------------------ 1 mol

                          4.06 g       ------------------- x

                          x = (4.06 x 1) / 12

                          x = 0.34 moles

2.- Calculate the moles of hydrogen

                           18 g of water -------------  1 g of hydrogen

                            6.10 g of water ----------   x

                            x = (6.10 x 1) / 18

                            x = 0.33 g

                           1 g of H  ----------------  1 mol of H

                           0.33 g     ----------------  x

                           x = (0.33 x 1) / 1

                           x = 0.33 moles of H

3.- Calculate the mass of oxygen

mass of Oxygen = 6.10 - 4.06 - 0.33

                            = 1.71 g

                          16 g of O ---------------  1 mol of O

                          1.71 g of O -------------   x

                          x = (1.71 x 1) / 16

                          x = 0.11 moles

4.- Divide by the lowest number of moles

Carbon   = 0.34 / 0.11  = 3                          

Hydrogen = 0.33 / 0.11 = 3

Oxygen = 0.11 /0.11 = 1

5.- Write the empirical formula

                                    C₃H₃O

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