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(a)
Interpretation: The ion formed by the given element is to be identified.
Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.
(a)
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Answer to Problem 1SP
The ion formed by the given element is an anion.
Explanation of Solution
The given element is selenium. The
(b)
Interpretation: The ion formed by the given element is to be identified.
Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.
(b)
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Answer to Problem 1SP
The ion formed by the given element is a cation.
Explanation of Solution
The given element is barium. The atomic number of Ba is 56. It belongs to group 2 in the periodic table which is alkaline earth metals. It has 2 valence electrons. In order to complete the octet, it can lose two electrons to form a cation with a +2 charge.
(c)
Interpretation: The ion formed by the given element is to be identified.
Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.
(c)
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Answer to Problem 1SP
The ion formed by the given element is an anion.
Explanation of Solution
The given element is phosphorus. The atomic number of phosphorus is 15. It belongs to group 15 in the periodic table which is the nitrogen family. It has 5 valence electrons. In order to complete the octet, it can gain three electrons to form an anion with a -3 charge.
(d)
Interpretation: The ion formed by the given element is to be identified.
Concept Introduction : There are two types of charged species, anion (negatively charged species) and cation (positively charged species). Anion is formed by the gaining of electron/s whereas a cation is formed by losing electron/s.
(d)
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Answer to Problem 1SP
The ion formed by the given element is an anion.
Explanation of Solution
The given element is iodine. The atomic number of iodine is 53. It belongs to group 17 which is the halogen family. It has 7 valence electrons. Thus, in order to complete the octet, it can gain one electron to form an anion with a -1 charge.
Chapter 9 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
- propose synthesisarrow_forwardExplanation O Conjugated Pi Systems Deducing the reactants of a Diels-Alder reaction Can the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? ? Δ If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. • If your answer is no, check the box under the drawing area instead. Click and drag to start drawing a structure. Xarrow_forwardDiels Alder Cycloaddition: Focus on regiochemistry (problems E-F) –> match + of thedienophile and - of the diene while also considering stereochemistry (endo).arrow_forward
- HELP! URGENT! PLEASE RESOND ASAP!arrow_forwardQuestion 4 Determine the rate order and rate constant for sucrose hydrolysis. Time (hours) [C6H12O6] 0 0.501 0.500 0.451 1.00 0.404 1.50 0.363 3.00 0.267 First-order, k = 0.210 hour 1 First-order, k = 0.0912 hour 1 O Second-order, k = 0.590 M1 hour 1 O Zero-order, k = 0.0770 M/hour O Zero-order, k = 0.4896 M/hour O Second-order, k = 1.93 M-1-hour 1 10 ptsarrow_forwardDetermine the rate order and rate constant for sucrose hydrolysis. Time (hours) [C6H12O6] 0 0.501 0.500 0.451 1.00 0.404 1.50 0.363 3.00 0.267arrow_forward
- Draw the products of the reaction shown below. Use wedge and dash bonds to indicate stereochemistry. Ignore inorganic byproducts. OSO4 (cat) (CH3)3COOH Select to Draw ઘarrow_forwardCalculate the reaction rate for selenious acid, H2SeO3, if 0.1150 M I-1 decreases to 0.0770 M in 12.0 minutes. H2SeO3(aq) + 6I-1(aq) + 4H+1(aq) ⟶ Se(s) + 2I3-1(aq) + 3H2O(l)arrow_forwardProblem 5-31 Which of the following objects are chiral? (a) A basketball (d) A golf club (b) A fork (c) A wine glass (e) A spiral staircase (f) A snowflake Problem 5-32 Which of the following compounds are chiral? Draw them, and label the chirality centers. (a) 2,4-Dimethylheptane (b) 5-Ethyl-3,3-dimethylheptane (c) cis-1,4-Dichlorocyclohexane Problem 5-33 Draw chiral molecules that meet the following descriptions: (a) A chloroalkane, C5H11Cl (c) An alkene, C6H12 (b) An alcohol, C6H140 (d) An alkane, C8H18 Problem 5-36 Erythronolide B is the biological precursor of erythromycin, a broad-spectrum antibiotic. How H3C CH3 many chirality centers does erythronolide B have? OH Identify them. H3C -CH3 OH Erythronolide B H3C. H3C. OH OH CH3arrow_forward
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