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Concept explainers
(a)
Interpretation: The building blocks which represents the fatty acid residues in the given block diagram of sphingoglycolipid have to be predicted.
Concept introduction: The lipid that is composed of one fatty acid and one carbohydrate unit which is joined to a sphingosine molecule is known as sphingoglycolipid. The fatty acid of sphingoglycolipid is also joined to the sphingosine molecule with the help of amide linkage. The carbohydrate unit joined to a sphingosine molecule with the help of glycosidic linkage.
(b)
Interpretation: The building blocks which represents the carbohydrate residues in the given block diagram of sphingoglycolipid have to be predicted.
Concept introduction: The lipid that is composed of one fatty acid and one carbohydrate unit which is joined to a sphingosine molecule is known as sphingoglycolipid. The fatty acid of sphingoglycolipid is also joined to the sphingosine molecule with the help of amide linkage. The carbohydrate unit joined to a sphingosine molecule with the help of glycosidic linkage.
(c)
Interpretation: The linkage which represents the amide linkage in the given block diagram of sphingoglycolipid has to be predicted.
Concept introduction: The lipid that is composed of one fatty acid and one carbohydrate unit which is joined to a sphingosine molecule is known as sphingoglycolipid. The fatty acid of sphingoglycolipid is also joined to the sphingosine molecule with the help of amide linkage. The carbohydrate unit joined to a sphingosine molecule with the help of glycosidic linkage.
(d)
Interpretation: The linkage which involves a monosaccharide in the given block diagram of sphingoglycolipid has to be predicted.
Concept introduction: The lipid that is composed of one fatty acid and one carbohydrate unit which is joined to a sphingosine molecule is known as sphingoglycolipid. The fatty acid of sphingoglycolipid is also joined to the sphingosine molecule with the help of amide linkage. The carbohydrate unit joined to a sphingosine molecule with the help of glycosidic linkage.
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Chapter 19 Solutions
General, Organic, and Biological Chemistry Seventh Edition
- Curved arrows are used to illustrate the flow of electrons. Follow the curved arrows and draw the structure of the missing reactants, intermediates, or products in the following mechanism. Include all lone pairs. Ignore stereochemistry. Ignore inorganic byproducts. H Br2 (1 equiv) H- Select to Draw Starting Alkene Draw Major Product I I H2O 四: ⑦.. Q Draw Major Charged Intermediate Iarrow_forwardNH (aq)+CNO (aq) → CO(NH2)2(s) Experiment [NH4] (M) [CNO] (M) Initial rate (M/s) 1 0.014 0.02 0.002 23 0.028 0.02 0.008 0.014 0.01 0.001 Calculate the rate contant for this reaction using the data provided in the table.arrow_forward2CIO2 + 20H-1 CIO31 + CIO2 + H2O Experiment [CIO2], M [OH-1], M 1 0.0500 0.100 23 2 0.100 0.100 3 0.100 0.0500 Initial Rate, M/s 0.0575 0.230 0.115 ... Given this date, calculate the overall order of this reaction.arrow_forward
- 2 3 .(be)_[Ɔ+(be)_OI ← (b²)_IƆO+ (be)_I Experiment [1-] M 0.005 [OCI-] 0.005 Initial Rate M/min 0.000275 0.0025 0.005 0.000138 0.0025 0.0025 0.000069 4 0.0025 0.0025 0.000140 Calculate the rate constant of this reaction using the table data.arrow_forward1 2 3 4 I(aq) +OCl(aq) → IO¯¯(aq) + Cl¯(aq) Experiment [I-] M 0.005 [OCI-] 0.005 Initial Rate M/min 0.000275 0.0025 0.005 0.000138 0.0025 0.0025 Calculate the overall order of this reaction using the table data. 0.0025 0.000069 0.0025 0.000140arrow_forwardH2O2(aq) +3 I¯(aq) +2 H+(aq) → 13(aq) +2 H₂O(l)· ••• Experiment [H2 O2]o (M) [I]o (M) [H+]。 (M) Initial rate (M/s) 1 0.15 0.15 0.05 0.00012 234 0.15 0.3 0.05 0.00024 0.3 0.15 0.05 0.00024 0.15 0.15 0.1 0.00048 Calculate the overall order of this reaction using the table data.arrow_forward
- The U. S. Environmental Protection Agency (EPA) sets limits on healthful levels of air pollutants. The maximum level that the EPA considers safe for lead air pollution is 1.5 μg/m³ Part A If your lungs were filled with air containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lung volume of 5.40 L.) ΜΕ ΑΣΦ = 2.35 1013 ? atoms ! Check your rounding. Your final answer should be rounded to 2 significant figures in the last step. No credit lost. Try again.arrow_forwardY= - 0.039 (14.01) + 0.7949arrow_forwardSuppose 1.76 g of magnesium acetate (Mg (CH3CO2)2) are dissolved in 140. mL of water. Find the composition of the resulting electrolyte solution. In particular, list the chemical symbols (including any charge) of each dissolved ion in the table below. List only one ion per row. mEq Then, calculate the concentration of each ion in dwrite the concentration in the second column of each row. Be sure you round your answers to the L correct number of significant digits. ion Add Row mEq L x 5arrow_forward
- A pdf file of your hand drawn, stepwise mechanisms for the reactions. For each reaction in the assignment, you must write each mechanism three times (there are 10 reactions, so 30 mechanisms). (A) do the work on a tablet and save as a pdf., it is expected to write each mechanism out and NOT copy and paste the mechanism after writing it just once. Everything should be drawn out stepwise and every bond that is formed and broken in the process of the reaction, and is expected to see all relevant lone pair electrons and curved arrows.arrow_forwardNonearrow_forwardNonearrow_forward
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