{"id":3356,"date":"2023-10-05T07:55:53","date_gmt":"2023-10-05T07:55:53","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3356"},"modified":"2023-10-05T08:03:51","modified_gmt":"2023-10-05T08:03:51","slug":"week-3-ss2-third-term-chemistry-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-3-ss2-third-term-chemistry-notes\/","title":{"rendered":"Week 3 &#8211; SS2 Third Term Chemistry Notes"},"content":{"rendered":"<p>\u00a0<strong>WEEK THREE<br \/>\n<\/strong><strong>TOPIC: MASS\/VOLUME RELATIONSHIP<br \/>\n<\/strong><strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>Mole and Molar Quantities\n<\/li>\n<li>Relative Atomic Mass and Relative Molecular Mass.\n<\/li>\n<li>\n<div>Calculations involving Mass and Volume.\n<\/div>\n<p>\u00a0<strong>MOLE AND MOLAR QUANTITIES<br \/>\n<\/strong><\/li>\n<\/ul>\n<p><strong>THE MOLE<br \/>\n<\/strong>A mole is a number of particles of a substance which may be atoms, ions, molecules or electrons. This number of particles is approximately 6.02 x 10<sup>23<\/sup> in magnitude and is known as Avogadro&#8217;s number of particles.<\/p>\n<p>\u00a0<strong>The mole is defined<\/strong> as the amount of a substance which contains as many elementary units as there are atoms in 12g of Carbon-12.<\/p>\n<p>\u00a0<strong>RELATIVE ATOMIC MASS<br \/>\n<\/strong>The relative atomic mass of an element is the number of time the average mass of one atom of that element is heavier than one twelfth the mass of one atom of Carbon-12. It indicates the mass of an atom of an element. For e.g, the relative atomic mass of hydrogen, oxygen, carbon, sodium and calcium are 1, 16, 12, 23, and 40 respectively.<\/p>\n<p>\u00a0The atomic mass of an element contains the same number of atoms which is 6.02 x 10<sup>23<\/sup>atoms; 1 mole of hydrogen having atomic mass of 2.0g contains 6.02 x 10<sup>23<\/sup> atoms.<\/p>\n<p>\u00a0<br \/>\n\u00a0<strong>RELATIVE MOLECULAR MASS<br \/>\n<\/strong>The relative molecular mass of an element or compound is the number of times the average mass of one molecule of it is heavier than one-twelfth the mass of one atom of Carbon-12<br \/>\nIt is the sum of the relative atomic masses of all atoms in one molecule of that substance. It is also called the formula mass. The formula mass refers not only to the relative mass of a molecule but also that of an ion or radical.<\/p>\n<p>\u00a0<strong>CALCULATION<\/strong><br \/>\n\tCalculate the relative molecular mass of:<\/p>\n<ol>\n<li>Magnesium chloride\n<\/li>\n<li>Sodium hydroxide\n<\/li>\n<li>Calcium trioxocarbonate\n<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0[Mg=24, Cl=35.5, Na=23, O=16, H=1, Ca=40,C=12]<\/p>\n<p>\u00a0Solution:<\/p>\n<ol>\n<li>MgCl<sub> 2<\/sub> = 24 + 35.5&#215;2 = 24 + 71 = 95gmol<sup>-1<\/sup>\n\t\t<\/li>\n<li>NaOH = 23 + 16 + 1 = 40gmol<sup>-1<\/sup>\n\t\t<\/li>\n<li>\n<div>CaCO<sub>3<\/sub> = 40 + 12 +16&#215;3 = 100gmol<sup>-1<\/sup>\n\t\t\t<\/div>\n<p>\u00a0<\/li>\n<\/ol>\n<p><strong>EVALUATION<\/strong><\/p>\n<ol>\n<li>What is relative molecular mass of a compound?\n<\/li>\n<\/ol>\n<ol>\n<li>\n<div>Calculate the relative molecular mass of (a) NaNO<sub>3<\/sub> (b) CuSO<sub>4<\/sub>.5H<sub>2<\/sub>O\n<\/div>\n<p>\u00a0<\/li>\n<\/ol>\n<p><strong>MOLAR VOLUME OF GASES<br \/>\n<\/strong>The volume occupied by 1 mole of a gas at standard conditions of temperature and pressure (s.t.p) is 22.4 dm<sup>3<\/sup>. Thus 1 mole of oxygen gas of molar mass 32.0gmol<sup>-1<\/sup> occupies a volume of 22.4dm<sup>3<\/sup> at s.t.p and 1 mole of helium gas of molar mass 4.0gmol<sup>-1<\/sup> occupies a volume of 22.4 dm<sup>3<\/sup> at s.t.p.<br \/>\nNote: When the conditions of temperature and pressure are altered, the molar volume will also change. Also, standard temperature = 273K and standard pressure = 760mmHg.<\/p>\n<p>\u00a0<strong>RELATIONSHIP BETWEEN QUANTITIES<br \/>\n<\/strong>Molar mass = mass          (g)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0i.e. M = m   gmol<sup>-1<\/sup><br \/>\n\t                     Amount (moles)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0n<br \/>\nNote: Amount = Number of moles<\/p>\n<p>\u00a0Molar volume of gas = volume ( cm<sup>3<\/sup> or dm<sup>3<\/sup>)    i.e. Vm = v  dm<sup>3<\/sup>mol<sup>-1<\/sup><br \/>\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Amount (mole)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0n<br \/>\nAmount = Reacting mass (g)<br \/>\n\t                Molar mass (gmol<sup>-1<\/sup>)<br \/>\nAlso, Amount of substance = Number of particles<br \/>\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Avogadro&#8217;s constant<br \/>\nBut, Avogadro&#8217;s constant = 6.02 x 10<sup>23<\/sup><br \/>\n\tCombining the two expressions:<br \/>\nReacting mass = Number of particles<br \/>\n\t Molar mass         6.02 x 10<sup>23<\/sup><\/p>\n<p>\u00a0<strong>CALCULATIONS<\/strong><\/p>\n<ol>\n<li>\n<div>What is the mass of 2.7 mole of aluminium (Al=27)?\n<\/div>\n<p>\u00a0\u00a0\u00a0\u00a0Solution:<br \/>\nAmount = Reacting mass<br \/>\n\t\t\t                Molar mass<br \/>\nReacting mass = Amount x Molar mass<br \/>\n                       = 2.7mole x 27 gmol<sup>-1<\/sup>   = 72.9g.<\/p>\n<p>\u00a0<\/li>\n<li>\n<div>What is the number of oxygen atoms in 32g of the gas? (O=16, N<sub>A<\/sub> = 6.02 x 10<sup>23<\/sup>)\n<\/div>\n<p>Solution:<br \/>\nReacting mass = Number of atoms<br \/>\n Molar mass\u00a0\u00a0\u00a0\u00a06.02 x 10<sup>23<br \/>\n<\/sup>Number of atoms = Reacting mass x 6.02 x 10<sup>23<\/sup><sup><br \/>\n\t\t\t\t<\/sup><sup>\u00a0\u00a0\u00a0\u00a0<\/sup>Molar mass<br \/>\nMolar mass of O<sub>2 <\/sub>= 16&#215;2 =32gmol<sup>-1<br \/>\n<\/sup>Number of atoms = 32g x 6.02 x 10<sup>23<\/sup><br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a032gmol<sup>-1<\/sup><br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0= 6.02 x 10<sup>23<\/sup><br \/>\n\t\t\tThe number of oxygen atoms is 6.02 x 10<sup>23<\/sup><\/p>\n<p>\u00a0<strong>EVALUATION<\/strong>\n\t\t\t<\/li>\n<li>Define the molar volume of a gas\n<\/li>\n<li>\n<div>How many molecules are contained in 1.12dm<sup>3<\/sup> of hydrogen gas at s.t.p?\n<\/div>\n<p>\u00a0<strong>STOICHIOMETRY OF REACTION<br \/>\n<\/strong><\/li>\n<\/ol>\n<p>The calculation of the amounts (generally measured in moles or grams) of reactants and products involved in a chemical reaction is known as stoichiometry of reaction. In other words, the mole ratio in which reactants combine and products are formed gives the stoichiometry of the reactions.<\/p>\n<p>\u00a0From the stoichiometry of a given balanced chemical equation, the mass or volume of the reactant needed for the reaction or products formed can be calculated.<\/p>\n<p>\u00a0<strong>CALCULATION OF MASSES OF REACTANTS AND PRODUCTS<br \/>\n<\/strong><\/p>\n<ol>\n<li>Calculate the mass of solid product obtained when 16.8g of NaHCO<sub>3<\/sub>was heated strongly until there was no further change.\n<\/li>\n<\/ol>\n<p>\u00a0Solution:<br \/>\nThe equation for the reaction is:<br \/>\n2NaHCO<sub>3(s)<\/sub> \u2192 Na<sub>2<\/sub>CO<sub>3(s)<\/sub> + H<sub>2<\/sub>O<sub>(g)<\/sub> CO<sub>2(g)<\/sub><br \/>\n\tMolar mass of NaHCO<sub>3<\/sub> = 23 + 12 + 16&#215;3 = 84gmol<sup>-1<\/sup><br \/>\n\tMolar mass of Na<sub>2<\/sub>CO<sub>3<\/sub> = 23&#215;2 +12+16&#215;3 = 106gmol<sup>-1<\/sup><\/p>\n<p>\u00a0From the equation:<br \/>\n2 moles NaHCO<sub>3<\/sub> produces 1 mole Na<sub>2<\/sub>CO<sub>3<\/sub><br \/>\n\t2x84g NaHCO<sub>3<\/sub> produces 106g Na<sub>2<\/sub>CO<sub>3<\/sub><br \/>\n\t16.8g NaHCO<sub>3<\/sub> will produce Xg Na<sub>2<\/sub>CO<sub>3<\/sub><br \/>\n\tXg Na<sub>2<\/sub>CO<sub>3<\/sub> = 106g x 16.8g\u00a0\u00a0\u00a0\u00a0=10.6g<br \/>\n                         2x84g<br \/>\nMass of solid product obtained = 10.6g<\/p>\n<ol>\n<li>Calculate the number of moles of CaCl<sub>2<\/sub> that can be obtained from 25g of limestone [CaCO<sub>3<\/sub>] in the presence of excess acid.\n<\/li>\n<\/ol>\n<p>\u00a0Solution:<br \/>\nThe equation for the reaction is:<br \/>\nCaCO<sub>3(s) <\/sub>+ 2HCl \u2192 CaCl<sub>2(s)<\/sub> + H<sub>2<\/sub>0<sub>(l)<\/sub> + CO<sub>2(g)<\/sub><\/p>\n<p>\u00a0Number of moles = Reacting mass<br \/>\n\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Molar mass<br \/>\nMolar mass of CaCO<sub>3<\/sub> = 40 + 12 + 16&#215;3 = 100gmol<sup>-1<\/sup><br \/>\n\tNumber of moles of CaCO<sub>3<\/sub> = 25g           = 0.25 mole<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0100gmol<sup>-1<\/sup><br \/>\n\tFrom the equation of reaction,<br \/>\n1 mole CaCO<sub>3<\/sub> yields 1 mole CaCl<sub>2<\/sub><br \/>\n\tTherefore, 0.25 mole CaCO<sub>3<\/sub> yielded 0.25 mole CaCl<sub>2.<\/sub><\/p>\n<p>\u00a0<strong>EVALUATION<\/strong><\/p>\n<ol>\n<li>\n<div>What does the term &#8216;Stoichiometry of reaction&#8217; mean?\n<\/div>\n<p>Ethane [C<sub>2<\/sub>H<sub>6<\/sub>] burns completely in oxygen. What amount in moles of CO<sub>2<\/sub>will be produced when 6.0g of ethane are completely burnt in oxygen?<\/p>\n<p>\u00a0<strong>CALCULATION OF VOLUME OF REACTING GASES<br \/>\n<\/strong><\/li>\n<li>\n<div>In an experiment, 10cm<sup>3<\/sup> of ethene [C<sub>2<\/sub>H<sub>4<\/sub>] was burnt in 50cm<sup>3<\/sup> of oxygen.\n<\/div>\n<ol>\n<li>Which gas was supplied in excess? Calculate the volume of the excess gas remaining at the end of the reaction.\n<\/li>\n<li>Calculate the volume of CO<sub>2 <\/sub>gas produced\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>\u00a0Solution:<br \/>\nThe equation for the reaction is:<br \/>\nC<sub>2<\/sub>H<sub>4(g) <\/sub>+ 3O<sub>2(g)<\/sub> \u2192 2CO<sub>2(g)<\/sub> + 2H<sub>2<\/sub>O<sub>(g)<br \/>\n<\/sub><\/p>\n<ol>\n<li>\n<div>From the equation,\n<\/div>\n<p>1 mole of ethene reacts with 3mole of oxygen<br \/>\n1 volume of ethene reacts with 3 volumes of oxygen<br \/>\n10cm<sup>3<\/sup> of ethene will react with 30cm<sup>3<\/sup> of oxygen<br \/>\nSince 50cm<sup>3<\/sup> of oxygen was supplied, oxygen was in excess<br \/>\nHence volume of the excess gas = initial volume \u2013 volume used up = 50-30 = 20cm<sup>3<\/sup>\n\t\t\t<\/li>\n<li>\n<div>1 volume of ethene produces 2 volumes of CO<sub>2<\/sub>\n\t\t\t<\/div>\n<p>10 cm3 of ethene will produce 20cm3 of CO<sub>2 <\/sub><br \/>\n\t\t\tTherefore, 20cm<sup>3<\/sup> of CO<sub>2<\/sub> was produced <\/p>\n<p>\u00a0<\/li>\n<\/ol>\n<ol>\n<li>\n<div>20cm<sup>3<\/sup> of CO was mixed and sparked with 200cm<sup>3<\/sup> of air containing 21% of O<sub>2<\/sub>. If all the volumes are measured at s.t.p, calculate the total volume of the resulting gases.\n<\/div>\n<p>\u00a0<\/li>\n<\/ol>\n<p>Solution<strong>:<br \/>\n<\/strong>In 200cm<sup>3<\/sup> of air,<br \/>\nVolume of O<sub>2<\/sub> = 21   x 200cm<sup>3<\/sup> = 42cm<sup>3<\/sup><br \/>\n\t                        100<br \/>\nVolume of N<sub>2<\/sub> and rare gases = 200-42 = 158cm<sup>3<br \/>\n<\/sup>The equation for the reaction is:<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02CO<sub>(g)<\/sub> + O<sub>2(g)<\/sub> \u2192 2CO<sub>2(g)<\/sub><br \/>\n\tVolume ratio                2       :   1     :     2<br \/>\nBefore sparking        20cm<sup>3<\/sup>   42cm<sup>3<\/sup><br \/>\n\tReacting volume        20cm<sup>3<\/sup>   10cm<sup>3<\/sup><br \/>\n\tAfter sparking                    32cm<sup>3<\/sup>   20cm<sup>3<\/sup><br \/>\n\tVolume of resulting gases = 32 + 20 + 158 = 210cm<sup>3<br \/>\n<\/sup><br \/>\n\u00a0<strong>GENERAL EVALUATION\/REVISION<br \/>\n<\/strong><\/p>\n<ol>\n<li>Find the volume of oxygen produced by 1 mole of KClO<sub>3<\/sub> at s.t.p in the following reaction: 2KClO<sub>3(s) <\/sub> \u2192 2KCl<sub>(s)<\/sub> + 30<sub>2(g)<br \/>\n<\/sub><\/li>\n<li>Define the term &#8216;Relative atomic mass&#8217;<sub><br \/>\n\t\t\t<\/sub><\/li>\n<li><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_0755_Week3SS2Th1.png\" alt=\"\"\/>Balance the following redox equations I<sup>&#8211;<\/sup>  +  MnO<sub>4<\/sub><sup>&#8211; <\/sup>                   IO<sub>3<\/sub><sup>&#8211;<\/sup>  +  MnO<sub>2 <\/sub>in basic medium\n<\/li>\n<li>Write the symbols of the following elements: mercury, silver, gold, lead, tin, antimony.\n<\/li>\n<li>Define valency.\n<\/li>\n<\/ol>\n<p>\u00a0<strong>READING ASSIGNMENT<br \/>\n<\/strong>New School Chemistry for Senior Secondary School byO. Y. Ababio, Pg 156-164<\/p>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT<br \/>\n<\/strong><strong>SECTION A: <\/strong>Write the correct option ONLY<\/p>\n<ol>\n<li>Amount of a substance is expressed in a. mole b. grams c. kilograms d. mass\n<\/li>\n<li>Determine the mass of CO<sub>2<\/sub> produced by burning 104g of ethyne [C<sub>2<\/sub>H<sub>2<\/sub>]a. 256g b.352g c. 416g d. 512g\n<\/li>\n<li>The mole ratio in which reactants combine and products are formed is known as a. rate of reaction b. stoichiometry of reaction C. equation of reaction d. chemical reaction\n<\/li>\n<li>The unit for relative molecular mass is a. mole b. gmol<sup>-1<\/sup> c. grams d. mass\n<\/li>\n<li>What mass of Pb(NO<sub>3<\/sub>)<sub>2<\/sub> would be required to 9g of PbCl<sub>2<\/sub> on the addition of excess NaCl solution? [Pb=207, Na=23, O=16, N=14] a. 10.7g b. 1.2g c. 6.4g d. 5.2g\n<\/li>\n<\/ol>\n<p>\u00a0<strong>SECTION B<br \/>\n<\/strong><\/p>\n<ol>\n<li>Calculate the number of molecules of CO<sub>2<\/sub> produced when 10g of CaCO<sub>3<\/sub> is treated with 100cm<sup>3<\/sup> of 0.20moldm<sup>-3<\/sup>HCl\n<\/li>\n<li>Calculate the volume of nitrogen that will be produced at s.t.p from the decomposition of 9.60g ammonium dioxonitrate (III), NH<sub>4<\/sub>NO<sub>2<\/sub>.\n<\/li>\n<\/ol>\n<p>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0WEEK THREE TOPIC: MASS\/VOLUME RELATIONSHIP CONTENT Mole and Molar Quantities Relative Atomic Mass and Relative&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,268],"tags":[],"class_list":["post-3356","post","type-post","status-publish","format-standard","hentry","category-posts","category-third-term-ss2-chemistry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3356","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3356"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3356\/revisions"}],"predecessor-version":[{"id":3357,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3356\/revisions\/3357"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3356"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3356"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3356"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}