(a)
Interpretation:
To draw a condensed structure for an
Concept Introduction:
Condensed structure is accustomed to composing organic structure in a line of text. It demonstrates all molecules, however, overlooks the vertical bonds and most or all the horizontal single bonds. It utilizes brackets to demonstrate that polyatomic gatherings inside in a formula are joined to the closest non-hydrogen atom on the left side.
(b)
Interpretation:
To indicate all polar bonds present in
Concept Introduction:
The unequal sharing of valence electrons in a bond is called polar bond. Polar bond result when the bond formed between two atoms in which one atom is more electronegative than the other one. One example of polar bond is
In
(c)
Interpretation:
To determine the geometry around the N atom.
Concept Introduction:
The following table should be used while determining the shape around an atom.
Number of groups | Number of atoms | Number of lone pairs | Shape | Bond angle |
2 | 2 | 0 | Linear | |
3 | 3 | 0 | Trigonal planar | |
4 | 4 | 0 | Tetrahedral | |
4 | 3 | 1 | Trigonal pyramidal | |
4 | 2 | 2 | Bent |
(d)
Interpretation:
To determine the drawn compound as polar or nonpolar.
Concept Introduction:
Polar compound is that compound in which polar bonds are present. Bond formed due to unequal sharing of valence electrons is called polar bond. Polar bond result when the bond formed between two atoms in which one atom is more electronegative than the other one. One example of polar bond is

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Chapter 11 Solutions
GENERAL,ORGANIC, & BIOLOGICAL CHEM-ACCES
- 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
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
