upload 2 - National University Professor Ben Saida Chapter...

  • Test Prep
  • Sarah99
  • 10
  • 100% (1) 1 out of 1 people found this document helpful

Info icon This preview shows pages 1–4. Sign up to view the full content.

National University Professor Ben Saida Chapter 9 1-Calculate the Number of moles in these quantities (a) 25.0 g KNO3 Nm =mass/Molar mass Nm= 25.0 g / 101.1g Answer = 0.247 mole KN03 (b) 56 milimole NaOH 56/1000 mmol Answer =0.056 mol NaOH (c) 5.4 x 10^2 g (NH4)2 C2O4 / 124.1g Answer = 4.4 mol (NH4)2 C2O4 (d) 16.8 ml H2SO4 solution (d=1.727g/ml, 80.0% H2SO4 by mass) must covert the solution from ml to g of H2SO4. = 16.8 ml x 1.727g x 0.800g H2SO4 / 98.09g Answer= 0.237 mol H2SO4 3-Calculate the number of grams in these quantities: (a) 2.55mol Fe (OH) 3 = 2.55 mol x 106.9g Answer = 273g Fe (OH) 3 (b) 125 kg CaCO3 = 125 x 1000g = 1.25 x 10^5 g CaCO3 (c) 10.5 mol NH3 10.5mol x 17.03 g = 179 g NH3 (d) 72 milimole HCl /1000 mmol x 36.46g = 2.6g HCl (e) 500.0 ml of liquid Br2 (d=3.119g/ml) 500 ml Br2 x 3.119g / ml = 1.560 x 10 ^3g Br2 (5) Which contains the larger number of molecules. 10.0g H2O or 10.0g H2O2 Water has a lower molar mass than hydrogen peroxide: 10.0 grams of water has a lower molar mass, it has more moles which means more molecules than 10.0g of H2O2.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

(7) (a) Balance the equation for the synthesis of sucrose: 12 CO2 + 11 H2O ----------- C12H22O11 + 12 O2 (Balanced) (b) Set the mole ratio: (a) 12 mol CO2 / 11 mol H2O (b) 11MOL H2O / 1 Mol C12H22O11 (c) 12 mol O2/12mol CO2 (d) 1 mol C12H22O11/ 12 mol CO2 (e) 11 mol H2O /12 mol O2 (f) 12 mol O2/ 1 mol C12H22O11 (9) Given the unbalance equation: CO2 +H2 ----------- CH4 +H2O (a) How many moles of H2O can be produced from 25 moles of CO2? 25 mol CO2 x 2 mol H2O/ 1 mol CO2 Answer = 50 mol H2O (b) How many moles of CH4 will be produced along with 12 moles of H2O? 12 mol H2O x 1mole CH4 / 2 mol of H2O Answer = 6 mol CH4 Chapter 10 1-How many total and how many valence electrons are in each of the following neutral atoms? (a) Element name Valence Electron Total Electron Li 1 3 Mg 2 12 Ca 2 20 F 7 9 11-Electron configuration for the diagrams given (a) O 1s2 2s2 2p4 (b) Ca 1s2 2s2 2p6 3s2 3p6 4s2 (c) Ar 1s2 2s2 2p6 3s2 3p6 (d) Br 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 (e) Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6 39- Classification of elements (a) K is a metal (b) Pu is a metal (c) S is a nonmetal (d) Sb is a metalloid 49- Which ones will have the same number of valence electron? Answer: Na +, F- & Ne have the same valence electron: 8 valence electron 64-What is the family name for:
Image of page 2
(a) Group 1 A =Alkali metals (b) Group 2A = Alkali Earth metals (c) Group 7A = Halogens Chapter 11 5- Indicate which element is more positive and which one is more negative (a) H is more + O is more - (b) Rb is more + Cl is more (c) H is more + N is more - (d) Pb is more + S is more – (e) P is more + F is more – (f) H is more + C is more – 7- Classify the bond (a) Covalent
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 4
This is the end of the preview. Sign up to access the rest of the document.
  • Spring '14
  • Mole, mol, 0.125M H2SO4, Valence Electron Total Electron

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern