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Concept explainers
Interpretation:
Carbon dioxide is nonpolar and sulfur dioxide is polar has to be explained by using Lewis structures of these molecules.
Concept introduction:
Lewis structures are also known as Lewis dot structures which represent the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
Covalent bonds are formed by sharing of electrons between atoms.
The polarity of a bond is arises due to the difference in electronegativity of atoms present in that bonding.
The direction of net polarity is the resultant vector of the vectors drawn for dipole moment of each bonds.
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Chapter 4 Solutions
Fundamentals of General, Organic, and Biological Chemistry, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (8th Edition)
- Draw the structures of the following acids:(a) 2-Ethyl-3-hydroxyhexanoic acid (b) m-Nitrobenzoic acidarrow_forwardIdentify the acid on the left and its conjugate base on the right in the following equations:(a) HOCl + H2O ↔ H3O+ + OCl-(b) HONH2 + H2O ↔ HONH3+ + OH-(c) NH4+ + H2O ↔ NH3 + H3O+(d) 2HCO3-2 ↔ H2CO3 + CO3-2 (e) PO4-3 + H2PO4- ↔ 2HPO4-2arrow_forwardConsider the following acids and their ionization constant, determine which conjugate base is HCOOH Ka = 1.7 x 10-4 (b) HCN Ka = 4.9 x 10-10arrow_forward
- Which of the following define the stereochemistry of alanine (as per the structure shown)? Note: Functional groups arranged horizontally are facing towards the front, and the functional groups arranged vertically are facing towards the back. СООН + H₂N to OS- Od- CH, OR-arrow_forwardIDENTIFY THE FUNCTIONAL GROUP PRESENT IN THESE COMPOUNDS. A = ? B = ? C = ?arrow_forwardUse Frost Circles to complete the molecular orbital diagram for cyclooctatetrane. Label the bonding, non bonding, and anti bonding MO’s. If the molecule is planar, would it be aromatic, antiaromatic, or nonaromatic? If the molecule is nonplanar, would it be aromatic, antiaromatic, or nonaromatic?arrow_forward
- Compound P was discovered by a scientist. Compound P is a dipeptide, optically active and has the molecular formula C„H14N2O3. Compound P is formed when compound Q and compound R joined together by condensation reaction. While, monomers S and T are formed by modifying the compounds Q and R. Polymer U is formed by the condensation reaction of monomers S and T. Draw the possible structural formulae of compounds P, Q, R, S, T and U. Label the peptide bond(s) for compound P. Draw the possible structural formulae for repeating unit of polymer U. Please state the number of functional groups present in compound P.arrow_forwardHow many peptide bonds were produced by creating the following How many H2O particle were produced by creating the following structure if the starting reagent is glucose? * structure if the starting reagent is glucose? * CH,OH CH,OH CH,OH CH;OH CH;OH CH;OH OH он он OH 250 ÓH OH OH ÓH OH OH 300 А.) 125 B.) 250 А.) 30 c.) 25 B.) 301 D.) 500 C.) 302 E.) O D.) 300 E.) 150 F.) Oarrow_forwardAn incomplete structure of a porphyrin ring is shown below. The structure is missing three pi bonds and does not show the non-zero formal charges on two atoms. Complete the structure of the porphyrin ring with the missing pi bonds and formal charges. 0 0 +t DC 12⁰ H C N O S F P Cl Br Iarrow_forward
- Histidine has three ionizable groups. On the titration curve below, show: a) 2 molar equivalents of OH b) pka of the third ionizable group Histidine Titration 12.0 10.0 8.0 4 pH 6.0 4.0 2.0 0.0 Equivalents OHarrow_forwardb) Determine which ionization state corresponds to aspartic acid (shown below) at pH 1, 7, and 10 by selecting the most suitable answer from the statements i-iv. The pK of the amine, carboxylic acid, and side chain of aspartic acid are 9.6, 1.88 and 3.65, respectively. ОН ÓH ÑH2 i) the amine is charged; the carboxylic acids are neutral. ii) the amine and carboxylic acids are all charged. iii) the amine is neutral, and the carboxylic acids are charged. iv) the amine and the carboxylic acids are all neutral. pH Statement (i, ii, iii, or iv) 1 7 10 %3Darrow_forwardDraw all possible carboxylic acids with the formula C5H10O2.arrow_forward
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