Q.
Show with diagrams why hydrochloric acid is used in the test for non-reducing sugars (e.g. sucrose)
A.
Sucrose is a combination of two simple sugars, fructose and glucose. The two simple sugars are combined in a way that gives a negative test for glucose or other reducing sugars. Hydrochloric acid will cause sucrose to break down into the two simple sugars from which it is composed. Study the diagrams below:
Glucose (or other reducing sugar) -----> Positive Benedict test
Sucrose (or other non-reducing sugar) ----> Negative Benedict test
Sucrose + HCl -------> glucose + fructose ----> Positive Benedict test
You can improve on the above diagrams if you spend enough time to understand them.
1/1/08
Related Reaction Rates
Q.
The average rate of disappearence of ozone in the reaction 2O3(g) --> 3O2(g) is found to be 3.0 x 10^-3 atm/sec over a certain interval of time. What is the rate of appearance of O2 during this interval?
A.
The mole ratio of O2(oxygen) to O3(ozone) is 3/2, the ratio of the two coefficients in the balanced chemical equation. Here, the partial pressure ratio, P(O2)/P(O3), of the two gases is also 3/2. The Ideal Gas Law written in subscript form is:
P1V1/P2V2 = n1T1/n2T2
If the temperatures and volumes are the same, this simplifies to:
P1/P2 = n1/n2 (pressure ratio = mole ratio)
or
P(O2)/P(O3) = n(O2)/n(O3) = 3/2
or
P(O2) = P(O3)*[ n(O2)/n(O3)]
or
P(O2) = P(O3)*[3/2]
(3.0 x 10^-3 atm.O3/sec)[(3 atm.O2/s)/(2 atm.O3/s)]=4.5x10^-3 atm.O2/s
=======
The average rate of disappearence of ozone in the reaction 2O3(g) --> 3O2(g) is found to be 3.0 x 10^-3 atm/sec over a certain interval of time. What is the rate of appearance of O2 during this interval?
A.
The mole ratio of O2(oxygen) to O3(ozone) is 3/2, the ratio of the two coefficients in the balanced chemical equation. Here, the partial pressure ratio, P(O2)/P(O3), of the two gases is also 3/2. The Ideal Gas Law written in subscript form is:
P1V1/P2V2 = n1T1/n2T2
If the temperatures and volumes are the same, this simplifies to:
P1/P2 = n1/n2 (pressure ratio = mole ratio)
or
P(O2)/P(O3) = n(O2)/n(O3) = 3/2
or
P(O2) = P(O3)*[ n(O2)/n(O3)]
or
P(O2) = P(O3)*[3/2]
(3.0 x 10^-3 atm.O3/sec)[(3 atm.O2/s)/(2 atm.O3/s)]=4.5x10^-3 atm.O2/s
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