Amy is asked to determine the chemical composition of an unknown liquid sample. She determines that the sample is negative for protein, simple sugars, and lipids. However, someone else has used up all of the iodine reagent. Which scenario would allow Amy to test the unknown sample for the presence of starch

Amy can apply the Tyndall Beam effect ,by shining a beam of light through the sample solution.

Explanation:

The chemical test for starch is add iodine solution to the tested sample which is yellow/brown and observe the color change to blue/black color. However, the physical test to distinguish simple sugars from starch is the shine a beam of light through the sample solution. Because that sample solution contained no simple sugars, then the presence of starch in the solution will be indicated by dispersal of light because the large starch molecules are enough to affect the light.

In order to test the unknown sample for the presence of starch, Amy should B. Incubate the unknown sample with amylase and then test for simple sugars.

Amylase is an enzyme that breaks starch down to sugars so if Amy incubates the substance with amylase and the substance has starch, the starch will be broken down to sugars which can then be tested for.

If there was starch, a sugar test would be positive.

Other options are wrong because:

In conclusion therefore, the best way to test for starch would be to incubate with amylase and then test for simple sugars.

For more information, look at link.

Options for this question include:

A. Incubate the unknown sample with pepsin and then test for proteins.

B. Incubate the unknown sample with amylase and then test for simple sugars.

C. Add a known amount of starch to the unknown sample and then run the Benedict's test.

D. Incubate the unknown sample with catalase and then test for proteins.

E. Put a small amount of the unknown sample onto a fresh potato slice and observe for color changes.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.