{"id":2298,"date":"2023-10-02T12:03:12","date_gmt":"2023-10-02T12:03:12","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2298"},"modified":"2023-10-02T12:15:14","modified_gmt":"2023-10-02T12:15:14","slug":"week-3-ss1-second-term-chemistry-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-3-ss1-second-term-chemistry-notes\/","title":{"rendered":"Week 3 &#8211; SS1 Second Term Chemistry Notes"},"content":{"rendered":"<p>\u00a0<br \/>\n\u00a0<strong>WEEK  3<br \/>\n<\/strong><br \/>\n\u00a0<strong>Balancing Chemical Equations<\/strong><\/p>\n<p>\u00a0\u00a0 Atoms are neither created nor destroyed during any chemical reaction. Chemical changes merely rearrange the atoms.<br \/>\n\t\t\tThe statement above is supported by: <\/p>\n<ul>\n<li>\n<div>Law of conservation of mass\/matter\n\t\t\t\t<\/div>\n<\/li>\n<li>\n<div>Law of definite proportions\n\t\t\t\t<\/div>\n<\/li>\n<li>\n<div>Law of multiple proportions\n\t\t\t\t<\/div>\n<\/li>\n<\/ul>\n<p>Chemical reactions are represented in a concise way by <strong>chemical equations<\/strong>.<br \/>\n2 H<sub>2<\/sub> + O<sub>2<\/sub> \u2192 2 H<sub>2<\/sub>O <\/p>\n<ul>\n<li>\n<div>The reacting substances, called <strong>reactants<\/strong>, are located on the left side of the arrow. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>The substances formed, called <strong>products<\/strong>, are located on the right side of the arrow. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>In a chemical equation, the + sign is read as &#8220;reacts with&#8221; and the arrow is read as &#8220;produces&#8221;. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Numbers in front of the formulas are <strong>coefficients<\/strong>, indicating the relative number molecules or ions of each kind involved in the reaction.\n<\/div>\n<\/li>\n<\/ul>\n<p>Coefficients of 1 are never written &#8211; they are understood. <\/p>\n<ul>\n<li>\n<div>Numbers to the lower right of chemical symbols in a formula are <strong>subscripts<\/strong>, indicating the specific number of atoms of the element found in the substance.\n<\/div>\n<\/li>\n<li>\n<div>Subscripts of 1 are never written &#8211; they are understood.\n<\/div>\n<\/li>\n<\/ul>\n<p><img decoding=\"async\" align=\"right\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1203_Week3SS1Se1.gif\" alt=\"\"\/>A chemical equation must have the same number of atoms of each element on both sides of the arrow. When this condition is met, the equation is said to be <strong>balanced<\/strong>.<br \/>\n\t\tTo count atoms, multiply the formula&#8217;s coefficient by each symbol&#8217;s subscript.<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0For example: 2Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub><\/p>\n<ul>\n<li>\n<div>For Al &#8211; coefficient of 2, times subscript of 2 = 4 aluminum atoms \u00a0\n<\/div>\n<\/li>\n<li>\n<div>For S &#8211; coefficient of 2, times subscript inside parenthesis of 1, times subscript outside parenthesis of 3 = 6 sulfur atoms \u00a0\n<\/div>\n<\/li>\n<li>\n<div>For O &#8211; coefficient of 2, times subscript inside parenthesis of 4, times subscript outside parenthesis of 3 = 24 oxygen atoms\n<\/div>\n<\/li>\n<\/ul>\n<p>\u00a0 The order in which the following steps are performed is important. While shortcuts are possible, (and you will learn about one), following these steps in order is the best way to be sure you are correct.<br \/>\n<strong>Balance equations &#8220;by inspection&#8221; with these steps:<\/strong><\/p>\n<ol>\n<li>\n<div>Check for diatomic molecules. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Balance the metals (not Hydrogen). \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Balance the nonmetals (not Oxygen). \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Balance oxygen. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Balance hydrogen. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>The equation should now be balanced, <strong>but<\/strong>recount all atoms to be sure. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Reduce coefficients (if needed). ALL coefficients must be reducable before you can reduce. An equation is not properly balanced if the coefficients are not written in their lowest whole-number ratio.\n<\/div>\n<\/li>\n<\/ol>\n<p>HINT: NEVER change subscripts to balance equations.<br \/>\n\t\tThe <strong>physical state<\/strong> of each substance in a reaction may be shown in an equation by placing the <\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0following symbols to the right of the formula: <\/p>\n<ul>\n<li>\n<div>(<em>g<\/em>) \u00a0 for gas\n<\/div>\n<\/li>\n<li>\n<div>(<em>l<\/em>) \u00a0 for liquid\n<\/div>\n<\/li>\n<li>\n<div>(<em>s<\/em>) \u00a0 for solid\n<\/div>\n<\/li>\n<li>\n<div>(<em>aq<\/em>) \u00a0 for aqueous (water) solution\n<\/div>\n<\/li>\n<\/ul>\n<p>\u00a0 <strong>Stoichiometry<\/strong> is the quantitative study of chemical changes.<br \/>\n\t\tThe most common type of stoichiometry calculation is a mass-mass problem. Generally, a mass-mass problem looks like this: &#8220;given this amount of reactant, how much product will form?&#8221;<br \/>\n<strong>Steps in solving a mass-mass problem:<\/strong><\/p>\n<ol>\n<li>\n<div>Write a balanced equation for the reaction.\n<\/div>\n<\/li>\n<li>\n<div>Write the given mass on a factor-label form.\n<\/div>\n<\/li>\n<li>\n<div>Convert mass of reactant to moles of reactant.\n<\/div>\n<\/li>\n<li>\n<div>Convert moles of reactant to moles of product.\n<\/div>\n<\/li>\n<li>\n<div>Convert moles of product to grams of product.\n<\/div>\n<\/li>\n<li>\n<div>Pick up the calculator and do the math.\n<\/div>\n<\/li>\n<\/ol>\n<p>\u00a0 <strong>Mass-Mass Sample Problem:<\/strong><br \/>\n\t\tIf iron pyrite, FeS<sub>2<\/sub>, is not removed from coal, oxygen from the air will combine with both the iron and the sulfur as coal burns. If a furnace burns an amount of coal containing 125 g of FeS<sub>2<\/sub>,<br \/>\nhow much SO<sub>2<\/sub> (an air pollutant) is produced?<br \/>\n1. Write a balanced equation showing the formation of iron (III) oxide and sulfur dioxide.<br \/>\n4 FeS<sub>2<\/sub> + 11 O<sub>2<\/sub> \u2192 2 Fe<sub>2<\/sub>O<sub>3<\/sub> + 8 SO<sub>2<\/sub><br \/>\n\t\t2. Write the mass information given in the problem.<br \/>\n3. Convert grams of FeS<sub>2<\/sub> to moles of FeS<sub>2<\/sub>.<br \/>\n4. Changes moles of FeS<sub>2<\/sub> (reactant) to moles of SO<sub>2<\/sub> (product).<br \/>\nThis ratio comes from the coefficients in the balanced equation. Notice that the ratio was reduced from 8 : 4 to 2 : 1 when placed in the dimensional analysis form. While reducing is not <\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0absolutely necessary (the ratio will cancel properly even if not reduced), a good chemistry student notices such things and will do it.<br \/>\n5. Convert moles of SO<sub>2<\/sub> to grams of SO<sub>2<\/sub> .<br \/>\n6. All units have been canceled except for grams of SO<sub>2<\/sub> (product). The problem has been solved. Pick up the calculator and do the math. <\/p>\n<p>\u00a0Stoichiometry<br \/>\n\t\t\u00a0 The <strong>limiting reactant<\/strong> is the reactant that is completely consumed in the reaction. <\/p>\n<ul>\n<li>\n<div>The limiting reactant is not present in sufficient quantity to react with all other reactants. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>The reaction <strong>stops<\/strong> when the limiting reactant is completely consumed. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Any remaining reactants are considered <strong>&#8220;excess reactants&#8221;<\/strong>. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>The amount of product formed is determined by the &#8220;limiting reactant&#8221;.\n<\/div>\n<\/li>\n<\/ul>\n<p>\u00a0 <strong>Steps in solving a limiting reactant problem:<br \/>\n<\/strong><\/p>\n<ol>\n<li>\n<div>Write a balanced equation for the reaction. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Convert both reactant quantities to moles. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Determine the moles of product that could be formed by each reactant.\n\t\t\t\t<\/div>\n<\/li>\n<li>\n<div>The least amount in step #3 identifies the limiting reactant. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Use that number of moles of product to determine the mass produced.\n<\/div>\n<\/li>\n<\/ol>\n<p>\u00a0 <strong>A limiting reactant problem example:<\/strong> What mass of water can be produced by 4 grams of hydrogen gas reacting with 16 grams of oxygen gas? <strong><br \/>\n\t\t\t<\/strong><br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0The problem solution:<br \/>\n1. Write a balanced equation for the reaction.<br \/>\n2 H<sub>2<\/sub> + O<sub>2<\/sub> \u2192 2 H<sub>2<\/sub>O<br \/>\n2. Convert both reactant quantities to moles.<br \/>\nUsing the mole ratio from the equation, determine the moles of water that could be formed by each reactant. <\/p>\n<p>\u00a04. Oxygen produces the least amount of water. <\/p>\n<ul>\n<li>\n<div>16 grams of oxygen cannot produce as much water as 4 grams of hydrogen. In other words, 16 grams of oxygen will be used up in the reaction before 4 grams of hydrogen. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Oxygen is the &#8220;limiting&#8221; reactant. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Use oxygen for the calculation of product amount.\n<\/div>\n<\/li>\n<\/ul>\n<p>5. Complete the problem by converting moles of H<sub>2<\/sub>O to mass of H<sub>2<\/sub>O.<br \/>\nThe theoretical yield for this problem is 18 grams. If you performed this reaction in the lab, your actual yield might be less. Can you think of reasons why?<br \/>\nLimiting Reactant Problems<br \/>\n\t\t<strong>Percent Yield<\/strong><\/p>\n<ul>\n<li>\n<div>The quantity of product that is calculated to form when all the limiting reactant is used up is called the <strong>theoretical yield<\/strong>. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>The amount of product actually obtained in a reaction is called the <strong>actual yield<\/strong>. \u00a0\n<\/div>\n<\/li>\n<li>\n<div>The actual yield is almost always less than (and never greater than) the theoretical yield.\n<\/div>\n<\/li>\n<\/ul>\n<p><img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1203_Week3SS1Se2.gif\" alt=\"\"\/><\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>Sample problem:<\/strong><br \/>\n\t\tGiven the reaction:<br \/>\nFe<sub>2<\/sub>O<sub>3(<em>s<\/em>)<\/sub> + 3CO<sub>(<em>g<\/em>)<\/sub> \u2192 2Fe<sub>(<em>s<\/em>)<\/sub> + 3CO<sub>2(<em>aq<\/em>)<\/sub><br \/>\n\t\tA. If you start with 155 g of Fe<sub>2<\/sub>O<sub>3<\/sub> as the limiting reactant, what is the theoretical yield of Fe?<br \/>\n<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1203_Week3SS1Se3.gif\" alt=\"\"\/><br \/>\n\t\tB. If the actual yield of Fe was 87.9 g, what was the percent yield?<br \/>\n<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1203_Week3SS1Se4.gif\" alt=\"\"\/><br \/>\n\t\t\u00a0 EVALUATION<\/p>\n<ol>\n<li>\n<div>S<sub>8<\/sub> + O<sub>2<\/sub> \u2192 SO<sub>3<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>HgO \u2192 Hg + O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Zn + HCl \u2192 H<sub>2<\/sub> + ZnCl<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>Na + H<sub>2<\/sub>O \u2192 NaOH + H<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>C<sub>10<\/sub>H<sub>16<\/sub> + Cl \u2192 C + HCl \u00a0\n<\/div>\n<\/li>\n<li>\n<div> 5.Si<sub>2<\/sub>H<sub>3<\/sub> + O<sub>2<\/sub> \u2192 SiO<sub>2<\/sub> + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Fe + O \u2192 Fe<sub>2<\/sub>O<sub>3<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>FeS<sub>2<\/sub> + O<sub>2<\/sub> \u2192 Fe<sub>2<\/sub>O<sub>3<\/sub> + SO<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>Fe<sub>2<\/sub>O<sub>3<\/sub> + H<sub>2<\/sub> \u2192 Fe + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>K + Br \u2192 \u00a0\n<\/div>\n<\/li>\n<li>\n<div>C<sub>2<\/sub>H<sub>2<\/sub> + O<sub>2<\/sub> \u2192 \u00a0\n<\/div>\n<\/li>\n<li>\n<div>H<sub>2<\/sub>O<sub>2<\/sub> \u2192 H<sub>2<\/sub>O + O<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>C<sub>7<\/sub>H<sub>16<\/sub> + O<sub>2<\/sub> \u2192 CO<sub>2<\/sub> + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>SiO<sub>2<\/sub> + HF \u2192 \u00a0\n<\/div>\n<\/li>\n<li>\n<div>KClO<sub>3<\/sub> \u2192 KCl + O<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>KClO<sub>3<\/sub> \u2192 KClO<sub>4<\/sub> + KCl \u00a0\n<\/div>\n<\/li>\n<li>\n<div>P<sub>4<\/sub>O<sub>10<\/sub> + H<sub>2<\/sub>O \u2192 H<sub>3<\/sub>PO<sub>4<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>Sb + O \u2192 Sb<sub>4<\/sub>O<sub>6<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>Fe<sub>2<\/sub>O<sub>3<\/sub> + CO \u2192 Fe + CO<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>PCl<sub>5<\/sub> + H<sub>2<\/sub>O \u2192 HCl + H<sub>3<\/sub>PO<sub>4<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>H<sub>2<\/sub>S + Cl \u2192 S<sub>8<\/sub> + HCl \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Fe + H<sub>2<\/sub>O \u2192 Fe<sub>3<\/sub>O<sub>4<\/sub> + H<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>N + H \u2192 NH<sub>3<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>N<sub>2<\/sub> + O<sub>2<\/sub> \u2192 N<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>CO<sub>2<\/sub> + H<sub>2<\/sub>O \u2192 C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub> + O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>SiCl<sub>4<\/sub> + H<sub>2<\/sub>O \u2192 H<sub>4<\/sub>SiO<sub>4<\/sub> + HCl \u00a0\n<\/div>\n<\/li>\n<li>\n<div>H<sub>3<\/sub>PO<sub>4<\/sub> \u2192 H<sub>4<\/sub>P<sub>2<\/sub>O<sub>7<\/sub> + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Al(OH)<sub>3<\/sub> + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub> + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Fe<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub> + KOH \u2192 K<sub>2<\/sub>SO<sub>4<\/sub> + Fe(OH)<sub>3<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>H<sub>2<\/sub>SO<sub>4<\/sub> + HI \u2192 H<sub>2<\/sub>S + I + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Al + FeO \u2192 \u00a0\n<\/div>\n<\/li>\n<li>\n<div>P<sub>4<\/sub> + O<sub>2<\/sub> \u2192 P<sub>2<\/sub>O<sub>5<\/sub>\u00a0\n<\/div>\n<\/li>\n<\/ol>\n<p>\u00a0<br \/>\n\u00a0K<sub>2<\/sub>O + H<sub>2<\/sub>O \u2192 KOH \u00a0 <\/p>\n<ol>\n<li>\n<div>Na<sub>2<\/sub>O<sub>2<\/sub> + H<sub>2<\/sub>O \u2192 NaOH + O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>C + H<sub>2<\/sub>O \u2192 CO + H \u00a0\n<\/div>\n<\/li>\n<li>\n<div>H<sub>3<\/sub>AsO<sub>4<\/sub> \u2192 As<sub>2<\/sub>O<sub>5<\/sub> + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub> + Ca(OH)<sub>2<\/sub> \u2192 \u00a0\n<\/div>\n<\/li>\n<li>\n<div>FeCl<sub>3<\/sub> + NH<sub>4<\/sub>OH \u2192 \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> + SiO<sub>2<\/sub> \u2192 P<sub>4<\/sub>O<sub>10<\/sub> + CaSiO<sub>3<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>N<sub>2<\/sub>O<sub>5<\/sub> + H<sub>2<\/sub>O \u2192 HNO<sub>3<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>Al + HCl <img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1203_Week3SS1Se5.jpg\" alt=\"\"\/>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>H<sub>3<\/sub>BO<sub>3<\/sub> \u2192 H<sub>4<\/sub>B<sub>6<\/sub>O<sub>11<\/sub> + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Mg + N \u2192 \u00a0\n<\/div>\n<\/li>\n<li>\n<div>NaOH + Cl \u2192 NaCl + NaClO + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Li<sub>2<\/sub>O + H<sub>2<\/sub>O \u2192 LiOH \u00a0\n<\/div>\n<\/li>\n<li>\n<div>CaC<sub>2<\/sub> + H<sub>2<\/sub>O \u2192 C<sub>2<\/sub>H<sub>2<\/sub> + Ca(OH)<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>Fe(OH)<sub>3<\/sub> \u2192 Fe<sub>2<\/sub>O<sub>3<\/sub> + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Pb(NO<sub>3<\/sub>)<sub>2<\/sub> \u2192 PbO + NO<sub>2<\/sub> + O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Ca + AlCl<sub>3<\/sub> \u2192 CaCl<sub>2<\/sub> + Al \u00a0\n<\/div>\n<\/li>\n<li>\n<div>NH<sub>3<\/sub> + NO \u2192 N + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>H<sub>3<\/sub>PO<sub>3<\/sub> \u2192 H<sub>3<\/sub>PO<sub>4<\/sub> + PH<sub>3<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>Fe<sub>2<\/sub>O<sub>3<\/sub> + C \u2192 CO + Fe \u00a0\n<\/div>\n<p>\u00a0<\/li>\n<\/ol>\n<p>\u00a0<\/p>\n<ol>\n<li>\n<div>FeS + O<sub>2<\/sub> \u2192 Fe<sub>2<\/sub>O<sub>3<\/sub> + SO<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>NH<sub>3<\/sub> + O \u2192 NO + H<sub>2<\/sub>O \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Hg<sub>2<\/sub>CO<sub>3<\/sub> \u2192 Hg + HgO + CO<sub>2<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>SiC + Cl \u2192 SiCl<sub>4<\/sub> + C \u00a0\n<\/div>\n<\/li>\n<li>\n<div>Al<sub>4<\/sub>C<sub>3<\/sub> + H<sub>2<\/sub>O \u2192 CH<sub>4<\/sub> + Al(OH)<sub>3<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>Ag<sub>2<\/sub>S + KCN \u2192 KAg(CN)<sub>2<\/sub> + K<sub>2<\/sub>S\n<\/div>\n<\/li>\n<li>\n<div>Au<sub>2<\/sub>S<sub>3<\/sub> + H \u2192 Au + H<sub>2<\/sub>S \u00a0\n<\/div>\n<\/li>\n<li>\n<div>ClO<sub>2<\/sub> + H<sub>2<\/sub>O \u2192 HClO<sub>2<\/sub> + HClO<sub>3<\/sub>\u00a0\n<\/div>\n<\/li>\n<li>\n<div>MnO<sub>2<\/sub> + HCl \u2192 MnCl<sub>2<\/sub> + H<sub>2<\/sub>O + Cl \u00a0\n<\/div>\n<p>\u00a0<br \/>\n\u00a0<\/li>\n<\/ol>\n<p>1.From XNH<sub>3<\/sub>(g)+YO<sub>2<\/sub>-ZNO(g)+QH<sub>2<\/sub>O(g)<br \/>\nThe value of Z is<br \/>\n(A) .4   (B) .7   (C).6   (D).5<br \/>\n2.One molecule of oxygen atoms<br \/>\n(A).has a molar mass of 32g  (B).has 6.02x 10<sup>23<\/sup><br \/>\n\t\t(C).can be represented as O2 (D) .has a formula mass of 16<br \/>\n (E) Contains Avogadro&#8217;s number of atom<br \/>\n3.The numerical coefficients in a balanced equation give<br \/>\n(A).the number of moles of reactants and products<br \/>\n(B).the molar mass of the reactants and products<br \/>\n(C).the number of moles of reactants only<br \/>\n(D).the number of molecules and atoms of products<br \/>\n(E).the mass ratio of the reactants<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong><br \/>\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0 \u00a0WEEK 3 \u00a0Balancing Chemical Equations \u00a0\u00a0 Atoms are neither created nor destroyed during any&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,196],"tags":[],"class_list":["post-2298","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss1-chemistry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2298","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=2298"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2298\/revisions"}],"predecessor-version":[{"id":2299,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2298\/revisions\/2299"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2298"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2298"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2298"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}