A locating agent is needed in an experiment to separate amino acids by chromatography as it helps to analyze colorless substances on paper.
<h3>What is chromatography?</h3>
Chromatography is an analytical method frequently used to separate a chemical mixture into its components, allowing for in-depth analysis of each component.
There are many different types of chromatography including liquid, gas, ion-exchange, and affinity chromatography, but they all use the same basic principle.
<h3>What is the use of a locating agent in Chromatography?</h3>
In paper chromatography, colorless compounds are examined using a locating agent.
It is a chemical that combines with colorless substances to produce colorful compounds that are easy to see for analysis. Ninhydrin spray is an example of a locating agent of this type.
Thus, a locating agent is used in an experiment to separate amino acids by chromatography as it helps in determining colorless substances on paper.
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1 milliliters in one milligram.
Answer:
Hello There!!
Explanation:
1.C nH 2n
2.Family
3.carbonyl group
4.Catalyst
5.Nickel
hope this helps,have a great day!!
~Pinky~
The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.
A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)
NH₃ + O₂ → NO₂ + H₂O
The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Calculate the number of moles from the reactant
- Ammonia gas
Molar mass N = 14 gr/mol
Molar mass H = 1 gr/mol
Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
mass = 28.5 grams
n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol - Oxygen gas
Molar mass O = 16 gr/mol
Molar mass O₂ = 16 × 2 = 32 gr/mol
mass = 83.4 grams
n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol - n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.
According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.
- Theoretically the number moles of NO₂
n O₂ : n NO₂ = 7 : 4
2.61 : n NO₂ = 7 : 4
n NO₂ = 4 x 2.61 : 7 = 1.49 mol - The actual number of moles NO₂
Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol
The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.
P = (1.35 ÷ 1.49) × 100%
P = 0.9052 × 100%
P = 90.52%
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