Interpretation:
The
Concept introduction:
Chromatography is the technique used to identify the components of a mixture. The division of the distance covered by the component of a compound to the distance covered by solvent in the chromatography technique is known as the
Answer to Problem 1ASA
The
Explanation of Solution
The chromatograph for a mixture that is prepared by a student is shown below.
Figure 1
The modified chromatogram with proper divided parts is shown below.
Figure 2
According to the above shown chromatogram, four compounds A, B, C and D are present in the mixture.
The total distance travelled by the solvent is
The total distance travelled by the compound A is
The total distance travelled by the compound B is
The total distance travelled by the compound C is
The total distance travelled by the compound D is
The expression to represent the
To calculate the
To calculate the
To calculate the
To calculate the
Therefore, the
The
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Chapter 2 Solutions
EBK CHEMICAL PRINCIPLES IN THE LABORATO
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- Which one of the following statements explain why protecting groups are referred to as “a necessary evil in organic synthesis”? Question 12Select one or more: A. They increase the length and cost of the synthesis B. Every synthesis employs protecting groups C. Protecting group have no role to play in a synthesis D. They minimize the formation of side productsarrow_forwardWhich of the following attributes is a key advantage of the chiral auxiliary approach over the chiral pool approach in asymmetric synthesis? Question 10Select one: A. Chiral auxiliaries are cheaper than chiral pool substrates B. Chiral auxiliary can be recovered and recycled unlike chiral pool substrates. C. The use of chiral auxiliaries provide enantiopure products, while chiral pool reactions are only enantioselective D. The chiral auxiliaries are naturally occurring and do not require synthesisarrow_forwardIn the following molecule, indicate the hybridization and shape of the indicated atoms. CH3 CH3 H3C HO: CI:arrow_forward
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- Draw the major organic product of the Bronsted acid-base reaction. Include all lone pairs and charges as appropriate. Ignore any counterions. :0: NaOH Harrow_forward5. Calculate the total amount of heat transferred as 50 g of wat Specific heat H₂O (g) 2.00 J/g°C -10 °C. Specific heat H₂O (1) Specific heat H₂O (s) 4.18 J/g°C 2.11 J/g°C Heat of vaporization 2260 J/g Heat of fusion 334 J/g Melting point 0°C 6. Calculate the total amount of heat transferred as 25 g of water is heated from 50 °C to 100 °C as a gas. Boiling point 100 °Carrow_forwardCalculate the total amount of heat transferred as 50 g of Water -10°C. Calculate the total amount of heat transferred as 25 g of water is heated from 50°C to 100°C as a gas. \table[[Specific heat H₂O(g), 2.00°C Η 2 g 5. Calculate the total amount of heat transferred as 50 g of wat Specific heat H₂O (g) 2.00 J/g°C -10 °C. 4.18 J/g°C 2.11 J/g°C 2260 J/g 334 J/g Specific heat H₂O (1) Specific heat H₂O (s) Heat of vaporization Heat of fusion Melting point 6. Calculate the total amount of heat transferred as 25 g of water is heated from 50 °C to 100 °C as a gas. Boiling point 100 °C 0°Carrow_forward
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