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According to the rules on significant figures, the product

General Chemistry: Principles and Modern Applications | 10th Edition | ISBN: 9780132064521 | Authors: Ralph Petrucci ISBN: 9780132064521 175

Solution for problem 67 Chapter 1

General Chemistry: Principles and Modern Applications | 10th Edition

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General Chemistry: Principles and Modern Applications | 10th Edition | ISBN: 9780132064521 | Authors: Ralph Petrucci

General Chemistry: Principles and Modern Applications | 10th Edition

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Problem 67

According to the rules on significant figures, the product of the measured quantities 99.9 m and 1.008 m should be expressed to three significant figures Yet, in this case, it would be more appropriate to express the result to four significant figures 100.7 m2. Explain why. 101 m2.

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Week 4: Day 10: VSEPR (Valence Shell Electron Pair Repulsion). Electronic Shape: ● Linear (2 sites). ● Trigonal Planar (3 sites). ● Tetrahedral (4 sites). Molecular Shape: ● Linear (2 sites). ● Trigonal Planar (3 sites). ● Tetrahedral (4 sites). ● Bent. Linear​: ● 2 atom y­x­y ● 1 atom x­y E Shape And Molecular Shape Are The Same Most Of The Time. E Shape Of Trigonal Planar: ● Three atoms bonded to the central atom. ● Molecular shape is also trigonal planar. Two Atoms Bonded W/ Trigonal Planar Shape: E Shape Tetrahedral: ● Four atoms bonded to the central atom. ● Molecular shape is also tetrahedral. 3 Atoms Bonded + 1 Lone Pair: Bent Molecule: 2 atoms bonded to a central atom with a tetrahedral e shape gives you a bent molecule. Example 1: CO ​ C = 4 valence electrons. O​2= 12 valence electrons. 4 + 12 = 16 ­ 4 ​because each line is worth 2 valence electrons) ​ = 12 remaining valence electrons. The lewis structure that is labeled the “best lewis structure” is the best because each element contains 8 valence electrons each (an octet). It is satisfied. KNOW: H needs 2 valence electrons to be satisfied. Be needs 4 valence electrons to be satisfied. B needs 6 valence electrons to be satisfied. All 3 need less than 8 valence electrons to be satisfied. E Shape: Linear. Molecular Shape: Linear. Polarity: Non­Polar. Example 2: SO S = 6 valence electrons. O​2= 12 valence electrons. 6 + 12 = 18 ­ 4 ​ecause each line is worth 2 valence electrons) ​ = 14 remaining valence electrons. E Shape: Trigonal Planar. Molecular Shape: Bent. Polarity: Polar. Example 3: NH3 N = 5 valence electrons. H​3​= 3 valence electrons. 5 + 3 = 8 ­ 6 (​because each line is worth 2 valence electrons​ ) = 2 remaining valence electrons. E Shape: Tetrahedral. Molecular Shape: Trigonal. Polarity: Polar. Example 4: NH​4​1+ N = 5 valence electrons. H​​= 4 valence electrons. 5 + 4 = 9 ­ 1ake one out because it has a positive one charge)​ = 8 ­ 8because each line is worth 2 valence electrons) = 0 E Shape: Tetrahedral. Molecular Shape: Tetrahedral. Polarity: Polar (Charged Species Are Polar). Example 5: SO 3​^2­ S = 6 valence electrons. O​​= 18 valence electrons. 6 + 18 = 24 + ​ dd two on because it has a negative two char = 26 ­ 6because each line is worth 2 valence electr​ = 20 remaining valence electrons. E Shape: Tetrahedral. Molecular Shape: Trigonal. Polarity: Polar (Charged Species Are Polar). Day 11: * Ionic = no tri’s or di’s. Strong Acids (Put Into H2O → 100% Dissociate ((1 Proton And 1 Ion)) = H+): H2SO​4 HCl HBr HI HClO​4 HNO ​ General Rules For Naming: 1. Hx Hydro_____ic acid. 2. “ate” forms “ic” acids. → SO ​ sulfate. → H ​SO​​sulfuric acid. 3. “ite” forms “ous” acids. → H ​SO​ sulfurous acid. Ex: H3​O​4 ^3­ = Phosphoric Acid (phosphate). HNO​2 ^1­ = Nitrous Acid (nitrous acid). 1 carbon = meth 2 carbon = eth 3 carbon = prop 4 carbon = but 5 carbon = pent 6 carbon = hex 7 carbon = hept 8 carbon = oct 9 carbon = non 10 carbon = dex Functional Groups Determine Ending: 1­ethanol: 1­butanol: * Know Acetic Acid Structure, name, etc. C2H​3​​^1­ = Acetate Ion * IONS ARE ALWAYS POLAR * (CH​CO 2 ^1­) Test Example 1​: Draw Lewis Structure For The Nitrate Ion. NO​3 ^1­ : How Many Valence Electrons Do You Have N = 5 valence electrons. O3 = 18 valence electrons. 5 + 18 = 23 + 1 (because it has a negative 1 charge) = 24 ­ 6 (because each line is worth 2 valence electrons) = 18 remaining valence electrons. Put in all 18 remaining valence electrons. Each element must be satisfied with 8 valence electrons EACH! Before turning the 2 valence electrons into a shared electron, O is satisfied (8), but N is not (6)...in order to satisfy N, you turn the 2 valence electrons into a shared electron, which then satisfies both N and O. Test Example 2​ : Boron Trichloride lewis structure has ___ sites of e­ density, e­ shape, molecular shape, polar/nonpolar 24 valence electrons! Day 12: Sodium Phosphate → will 100% dissociate in water which means it’ll break apart into ions. 1+ 3­ Na 3P O4 → 3Na(aq) + 1PO ​(aq) Whatever you start with (Na​ 3) you will bring the 3 to the other side always!! * 1 Mole = 6.022x10^23 * g = gas l = liquids aq = aqueous s = solids Limiting Reagents: A + 2B → C + D Example 1: 18 Na+ + 8 Cl­ → How many NaCl 1 Na+ + 1 Cl­ → 1 NaCl. 18 Na+ x1 NaCl = 18 NaCl. 1 Na+ 8 Cl­ 1 NaCl =8 NaCl. 1 Cl­ The answer is always the smallest one. Example 2: 36 Na ^1+ + 24 P4 ^3­ → How many Na​PO4​ 3 Na + 1 P​^3­ → 1 Na3PO​. 36 Na ^1+ 1 Na​PO = 12 NaPO​4. 3 Na ^1+ 24 PO​ ^3­ 1 Na​PO4 = 24 N3PO . ​^3­ The answer is always the smallest one. How to make 45 Iron(II) Phosphorus: How many Iron(II) How many Phosphates 45 Fe(PO ​​ → __ Fe ^2+ + __ 4​3­ Fe​(PO​​2→ 3 Fe ^2+ + 2 P ^3­ 3 Fe ^2+ + 2 P4^3­ → 1 F(PO ​2 Iro: 45 F3(PO4​ x3 Fe ^2+= 135 Fe ^2+. 13(PO4​ Phos​: 45 3​PO4) x​ PO ^3​= 90 PO ^3­ 3(POF)​ 135 ­ 90 = 45 (PO​)2.

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Chapter 1, Problem 67 is Solved
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Textbook: General Chemistry: Principles and Modern Applications
Edition: 10
Author: Ralph Petrucci
ISBN: 9780132064521

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According to the rules on significant figures, the product