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Concept explainers
a.
Interpretation:
The IUPAC name of given compounds has to be drawn.
Concept Introduction:
The system which is used to name an organic compound is known as IUPAC Nomenclature. IUPAC are known as International Union of Pure and Applied Chemistry.
There are some rules followed for writing IUPAC Nomenclature for
- The longest chain in the compound is known as parent name which represent the number of carbon atom present in the continuous carbon chain. The root name for alkane compound is –ane. The single bonds are formed in alkane compounds.
- The suffix group denotes the
functional group present in a molecule. The prefix group indicates the identity, location and number of substituents attached to the carbon compound. - Then name and number the substituents present in compounds. Then use prefix di-to represent two groups, tri- refers to three groups and so on. The numbers can be separated by using commas and the letters from numbers can be separated by using dashes.
b.
Interpretation:
The one constitutional isomer of given compound has to be drawn.
Concept Introduction:
Isomers:
The organic compound contains same molecular formula and different arrangement of atoms is termed as isomers. The isomers have different physical and chemical properties.
Constitutional Isomer:
Constitutional isomer has same molecular formula but they have different connectivity to each other. It is also known as structural isomers.
c.
Interpretation:
The solubilty of given compounds in water has to be predicted.
d.
Interpretation:
The solubilty of given compounds in organic solvent has to be predicted.
e
Interpretation:
The skeletal structure of the given compound has to be drawn.
Concept Introduction:
Skeletal Structures:
The skeletal structure is the diagrammatic representation of organic compounds. It is used for compounds containing rings and chain of compounds. The bonds of atoms bonded together to form a skeletal structure. There is no need to mention the carbon and hydrogen element. There is no need to mention the lone pairs of electrons present in heteroatoms. The heteroatom has to be drawn and the hydrogen is bonded directly to them.
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Chapter 10 Solutions
Principles of General, Organic, Biological Chemistry
- Please answer the questions in the photos and please revise any wrong answers. Thank youarrow_forward(Please be sure that 7 carbons are available in the structure )Based on the 1H NMR, 13C NMR, DEPT 135 NMR and DEPT 90 NMR, provide a reasoning step and arrive at the final structure of an unknown organic compound containing 7 carbons. Dept 135 shows peak to be positive at 128.62 and 13.63 Dept 135 shows peak to be negative at 130.28, 64.32, 30.62 and 19.10.arrow_forward-lease help me answer the questions in the photo.arrow_forward
- For the reaction below, the concentrations at equilibrium are [SO₂] = 0.50 M, [0] = 0.45 M, and [SO3] = 1.7 M. What is the value of the equilibrium constant, K? 2SO2(g) + O2(g) 2SO3(g) Report your answer using two significant figures. Provide your answer below:arrow_forwardI need help with this question. Step by step solution, please!arrow_forwardZn(OH)2(s) Zn(OH)+ Ksp = 3 X 10-16 B₁ = 1 x 104 Zn(OH)2(aq) B₂ = 2 x 1010 Zn(OH)3 ẞ3-8 x 1013 Zn(OH) B4-3 x 1015arrow_forward
- Help me understand this by showing step by step solution.arrow_forwardscratch paper, and the integrated rate table provided in class. our scratch work for this test. Content attribution 3/40 FEEDBACK QUESTION 3 - 4 POINTS Complete the equation that relates the rate of consumption of H+ and the rate of formation of Br2 for the given reaction. 5Br (aq) + BrO3 (aq) + 6H (aq) →3Br2(aq) + 3H2O(l) • Your answers should be whole numbers or fractions without any decimal places. Provide your answer below: Search 尚 5 fn 40 * 00 99+ 2 9 144 a [arrow_forward(a) Write down the structure of EDTA molecule and show the complex structure with Pb2+ . (b) When do you need to perform back titration? (c) Ni2+ can be analyzed by a back titration using standard Zn2+ at pH 5.5 with xylenol orange indicator. A solution containing 25.00 mL of Ni2+ in dilute HCl is treated with 25.00 mL of 0.05283 M Na2EDTA. The solution is neutralized with NaOH, and the pH is adjusted to 5.5 with acetate buffer. The solution turns yellow when a few drops of indicator are added. Titration with 0.02299 M Zn2+ requires 17.61 mL to reach the red end point. What is the molarity of Ni2+ in the unknown?arrow_forward
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