Calculating Density of an Irregularly Shaped Metal Object using Archimedes Principle - Pro, Study notes of Chemistry

Download Calculating Density of an Irregularly Shaped Metal Object using Archimedes Principle - Pro and more Study notes Chemistry in PDF only on Docsity! He Ne Ar Kr Xe Rn The Periodic Table of the Elements CrMn Fe Co Ni Mo W Tc Re Ru Os Rh Ir Pd Pt Most Probable Oxidation State +1 +2 +3 +4 +3 +_4 - 3 - 2 - 1 0 H Li Na K Rb Cs Fr Sc Y Be Mg Ca Sr Ba Ra La Ac B Al Ga In Tl Ti Rf Hf Zr C Si Ge Sn Pb F Cl Br I At O S Se Te Po N P As Sb Bi Zn Cd Hg + 2+1 Cu Ag Au +5 V Nb Ta Ce Th Pr Nd PmSmEu Gd Tb Dy Ho Er TmYb Lu Pa U Np Pu AmCmBk Cf Es FmMd No Lr +3 +3 Db Sg Bh Hs Mt Ds Martin &. Silberberg, Chemiatry: The Molecular Nature of Matter and Change. 24 Edition. Capyright @ The McGraw-Hill Companies, Inc. dll rights reserved. Table 1.4 Common SI-English Equivalent Quantities £ (I'll give you a similar table on the test.) c 5 English English to co si SI Equivalents Equivalents Sl Equivalent Length 1 kilometer (km) (1000 (10°) meters 0.6214 mile (mi) 1 mile = 1.609 km 1 meter (m) 100 (107) centimeters 1.094 yards (yd) 1 yard =0.9144m 1000 millimeters (mm) 39.37 inches (in) 1 foot (ft) = 0.3046 m 1 centimeter (cm) 0.01 (107) meter 0.3937 inch 1 inch = 2.54 cm (exactly) Volume 1 cubic 1,000,000 (10°) 35.31 cubic 1 cubic foot = meter (m°) cubic centimeters feet (ft9) 0.02832 m? 1 cubic 1000 cubic 0.2642 gallon 1 gallon = 3.785 dm® decimeter (dm*) centimeters (gal) 1.057 quarts (qt) 1 quart = 0.9464 dm? 1 cubic 0.001 dm? 0.03381 fluid 1 quart = 946.4 cm? centimeter (cm*) ounce 1 fluid ounce = 29.57 cm? Mass 1 kilogram (kg) 1000 grams. 2.205 pounds 1 pound = 0.4536 kg (Ib) 1 gram (g) 1000 milligrams (mg) 0.03527 1 ounce = 28.35 g ounce (oz) Notes: mass of e- tiny relative to p+, n. p+, n have same mass (almost). e-, p+ have same charge, opposite sign. 1+ + Also: Know something about the famous people (what they did which led to the picture of an atom and the properties of its particles: Rutherford, Mulliken, Dalton. Atomic Definitions I: Symbols, Isotopes, Numbers X A Z X = Atomic symbol of the element, or element symbol Z = The Atomic Number, the Number of Protons in the Nucleus (All atoms of the same element have the same no. of protons.) A = The Mass number; A = Z + N N = The Number of Neutrons in the Nucleus Isotopes = atoms of an element with the same number of protons, but different numbers of Neutrons in the Nucleus The Nuclear Symbol of the Atom, or Isotope Modern Reassessment of the Atomic Theory 1. All matter is composed of atoms. Although atoms are composed of smaller particles (electrons, protons, and neutrons), the atom is the smallest body that retains the unique identity of the element. 2. Atoms of one element cannot be converted into atoms of another element in a chemical reaction. Elements can only be converted into other elements in nuclear reactions in which protons are changed. 3. All atoms of an element have the same number of protons and electrons, which determines the chemical behavior of the element. Isotopes of an element differ in the number of neutrons, and thus in mass number, but not in chemical behavior (much). A sample of the element is treated as though its atoms have an average mass. 4. Compounds are formed by the chemical combination of two or more elements in specific ratios, as originally stated by Dalton. Copyright © Houghton Mifflin Company. All rights reserved. Demos: cations with multiple oxidation states: V5+ yellow V4+ blue; Mn ions Copyright © Houghton Mifflin Company. All rights reserved. Rules for naming binary ionic compounds Always list cation (+) name first, anion (-) name second. Monatomic anion’s name = first part of element name + “-ide” If the ion is not listed in the Type II table, assume it’s Type I. TABLE 2.5 Common Polyatomic lons Ton Name Name NH,* NO,~ ammonium nitrite nitrate sulfite sulfate hydrogen sulfate (bisulfate is a widely used common name) hydroxide cyanide phosphate hydrogen phosphate dihydrogen phosphate carbonate hydrogen carbonate (bicarbonate is a widely used common name) hypochlorite chlorite chlorate perchlorate acetate permanganate dichromate chromate peroxide MOLE – just a number, like a dozen • The Mole is based upon the definition: The amount of substance that contains as many elementary particles (atoms, molecules, ions, or other?) as there are atoms in exactly 12 grams of carbon -12 (i.e., 12C). 1 Mole = 6.022045 x 1023 particles (atoms, molecules, ions, electrons, apples, hairs, etc…) = NA particles = Avogadro’s number ~100 million x 100 million x 100 million Calculating the “Average” Atomic Mass of an Element 24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu 26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu 24.309268 amu With Significant Digits = 24.3 amu Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%). Martin &. Silberberg, Chemistry: The Molecular Nature of Matter and Change. 2° Edition. Copyright © The McGraw-Hill Companios, Inc. All rights reserved. iz Summary of Mass Terminology* Term Definition Unit Isotopic mass Mass of an isotope of an element amu Atomic mass Average of the masses of the amu (also called atomic naturally occurring isotopes weight) of an element weighted according to their abundance Molecular (or formula) Sum of the atomic masses of amu mass (also called the atoms (or ions) ina molecular weight) molecule (or formula unit) Molar mass (AC) Mass of 1 mole of chemical entities g/mol (also called gram- (atoms, ions, molecules, formula molecular weight) units) All terms based on the '*C standard: 1 atomic mass unit = + 5 Mass of one 126 atom. Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound Mass % of X Mass fraction of X Mass (g) of X in one mole of compound Multiply by M (g / mol of X) Divide by mass (g) of one mole of compound Multiply by 100 % Steps to Determine Empirical Formulas Mass (g) of Element in sample Moles of Element Preliminary Formula Empirical Formula ÷ M (g/mol ) for that element Use no. of moles as subscripts. Change to integer subscripts: ÷ smallest, conv. to whole #. Determining a Chemical Formula from Combustion Analysis - I Problem: Erthrose (M = 120 g/mol) is an important chemical compound used often as a starting material in chemical synthesis, and contains Carbon, Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded: 1.027 g CO2 and 0.4194 g H2O. Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula. Martin &. Silberberg, Chemistry: The Molecular Nature of Mattar and Change. 2°4 Edition. Copyright © The McGraw-Hill Companion, Inc. All rights reserved. Summary of the Mass-Mole-Number Relationships in a Chemical Reaction A A _ M (g/mol) of : M (g/mol) of _ compound A _ compound B y y ’ molar ratio from y mo|) Dalanced equation a a i - Avogadro's Avogadro's __ number __ humber \. (molecules/mol) \, (molecules/mol) Sample Problem: Calculating Reactants and Products in a Chemical Reaction - I Problem: Given the following chemical reaction between Aluminum Sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles of water are required for the reaction? b) What mass of H2S & Al(OH)3 would be formed? Al2S3 (s) + 6 H2O(l) 2 Al(OH)3 (s) + 3 H2S(g) Plan: Calculate moles of Aluminum Sulfide using its molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it’s molecular weight. Solution: a) molar mass of Aluminum Sulfide = 150.17 g / mol moles Al2S3 = = 0.4382 moles Al2S365.80 g Al2S3150.17 g Al2S3/ mol Al2S3 Calculating the Amounts of Reactants and Products in a Reaction Sequence - I Problem: Calcium Phosphate could be prepared in the following reaction sequence: 4 P4 (s) + 10 KClO3 (s) 4 P4O10 (s) + 10 KCl (s) P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq) 2 H3PO4 (aq) + 3 Ca(OH)2 (aq) 6 H2O(aq) + Ca3(PO4)2 (s) Given: 15.5 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What mass of Calcium Phosphate could be formed? Plan: (1) Calculate moles of P4. (2) Use molar ratios to get moles of Ca3(PO4)2. (3) Convert the moles of product back into mass by using the molar mass of Calcium Phosphate. DEMO: Match head reaction- Somorjai POW!!