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href="../19287/proteins-enzymes--1.html">Proteins & enzymes </a></li><li><a href="../19291/lipids-1.html">Lipids</a></li><li><a href="../19299/carbohydrates--1.html">Carbohydrates </a></li><li><a href="../19305/vitamins--1.html">Vitamins </a></li><li><a href="../19313/biochemistry-the-environment--1.html">Biochemistry & the environment </a></li><li><a href="../19336/proteins-enzymes-ahl--1.html">Proteins & enzymes (AHL) </a></li><li><a href="../19347/nucleic-acids--1.html">Nucleic acids </a></li><li><a href="../19361/biological-pigments--1.html">Biological pigments </a></li><li><a href="../19366/stereochemistry-in-biomolecules-1.html">Stereochemistry in biomolecules</a></li></ul></li><li><a href="../19390/option-c--1.html" class="father">Option C </a><ul class="level-2"><li><a href="../19391/energy-sources-1.html">Energy sources</a></li><li><a href="../19407/fossil-fuels-1.html">Fossil fuels</a></li><li><a href="../19411/nuclear-fusion-nuclear-fission-reactions-1.html">Nuclear fusion & 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href="../19194/antiviral-medications--1.html">Antiviral medications </a></li><li><a href="../19211/environmental-impact-of-some-medications-1.html">Environmental impact of some medications</a></li><li><a href="../19254/taxol-a-chiral-auxiliary-case-study-1.html">Taxol - a chiral auxiliary case study</a></li><li><a href="../19258/nuclear-medicine--1.html">Nuclear medicine </a></li><li><a href="../19273/drug-detection-analysis--1.html">Drug detection & analysis </a></li></ul></li></ul></li><li><a href="../16592/practical-scheme-of-work-ia--1.html">Practical scheme of work & IA </a><ul class="level-1"><li><a href="../16682/internal-assessment-1.html" class="father">Internal Assessment</a><ul class="level-2"><li><a href="../18892/scaffolding-the-investigation-1.html">'Scaffolding' the investigation</a></li><li><a href="../18896/timing-organisation--1.html">Timing & organisation </a></li><li><a href="../18905/choosing-the-research-question-1.html">Choosing the research question</a></li><li><a href="../18946/personal-engagement-1.html">Personal engagement</a></li><li><a href="../18950/exploration-1.html">Exploration</a></li><li><a href="../18957/analysis-1.html">Analysis</a></li><li><a href="../18965/evaluation-1.html">Evaluation</a></li><li><a href="../18966/communication-1.html">Communication</a></li><li><a href="../19882/internal-standardization-of-the-ia-1.html">Internal standardization of the IA</a></li><li><a href="../19901/submitting-the-samples-for-moderation-2.html">Submitting the samples for moderation</a></li><li><a href="../33754/ten-suggestions-for-ias-using-secondary-data-1.html">Ten suggestions for IAs using secondary data</a></li><li><a href="../37544/gaining-full-marks-for-a-databased-ia--1.html">Gaining full marks for a databased IA </a></li><li><a href="../19506/ia-example-marking-exercise--1.html">IA example & marking exercise </a></li><li><a href="../23929/genuine-examples-of-moderated-ia-reports-1.html">Genuine examples of moderated IA reports</a></li><li><a href="../25234/examples-of-teacher-marked-ia-reports--1.html">Examples of teacher-marked IA reports </a></li><li><a href="../37542/history-of-internal-assessment-1.html">History of Internal Assessment</a></li></ul></li><li><a href="../16598/mandatory-laboratory-components-1.html" class="father">Mandatory laboratory components</a><ul class="level-2"><li><a href="../16613/formula-of-magnesium-oxide--1.html">Formula of magnesium oxide </a></li><li><a href="../16612/determining-the-mr-of-an-unknown-gas-1.html">Determining the <i>M</i><sub>r</sub> of an unknown gas</a></li><li><a href="../16615/acid-base-titrations--1.html">Acid-base titrations </a></li><li><a href="../16616/a-green-acid-base-practical-1.html">A green acid-base practical</a></li><li><a href="../16617/analysis-of-aspirin-tablets-1.html">Analysis of aspirin tablets</a></li><li><a href="../16618/caco3-in-egg-shells-1.html">CaCO<sub>3</sub> in egg shells</a></li><li><a href="../16644/enthalpy-changes-1.html">Enthalpy changes</a></li><li><a href="../16631/reaction-rates-1.html">Reaction rates</a></li><li><a href="../16635/rate-dependent-factors-1.html">Rate-dependent factors</a></li><li><a href="../16636/determining-ea-for-a-reaction-1.html">Determining <i>E</i><sub>a</sub> for a reaction</a></li><li><a href="../16637/iodination-of-propanone-1.html">Iodination of propanone</a></li><li><a href="../18144/titrations-with-a-ph-meter-1.html">Titrations with a pH meter</a></li><li><a href="../18355/redox-titration-with-kmno4--1.html">Redox titration with KMnO<sub>4</sub> </a></li><li><a href="../16640/voltaic-cells-1.html">Voltaic cells</a></li><li><a href="../16659/3-d-molecular-modelling--1.html">3-D molecular modelling </a></li></ul></li><li><a href="../17168/other-good-practicals-1.html" class="father">Other good practicals</a><ul class="level-2"><li><a href="../17196/common-chemical-reactions-1.html">Common chemical reactions</a></li><li><a href="../227/elements-oxides-of-the-third-period--1.html">Elements & oxides of the third period </a></li><li><a href="../17239/the-halogens-1.html">The halogens</a></li><li><a href="../17162/boiling-points-of-mixtures-1.html">Boiling points of mixtures</a></li><li><a href="../17170/polarity-of-molecules-1.html">Polarity of molecules</a></li><li><a href="../18083/le-chateliers-principle--1.html">Le Chatelier's principle </a></li><li><a href="../19582/determining-kc-for-an-esterification-reaction-1.html">Determining <i>K</i><sub>c</sub> for an esterification reaction</a></li><li><a href="../18164/redox-reactions-of-vanadium--1.html">Redox reactions of vanadium </a></li><li><a href="../18351/chlorine-in-swimming-pools--1.html">Chlorine in swimming pools </a></li><li><a href="../18178/analysis-of-cuii-ions-in-solution--1.html">Analysis of Cu(II) ions in solution </a></li><li><a href="../18179/percentage-of-copper-in-brass-1.html">Percentage of copper in brass</a></li><li><a href="../18356/electrolytic-cells--1.html">Electrolytic cells </a></li><li><a href="../18598/reactions-of-organic-compounds-1.html">Reactions of organic compounds</a></li><li><a href="../18752/hydrolysis-of-halogenoalkanes-1.html">Hydrolysis of halogenoalkanes</a></li><li><a href="../18784/preparation-of-13-dinitrobenzene--1.html">Preparation of 1,3-dinitrobenzene </a></li><li><a href="../19022/determination-of-an-organic-structure-1.html">Determination of an organic structure</a></li><li><a href="../19887/preparation-of-nylon-66-1.html">Preparation of nylon 6,6</a></li><li><a href="../19343/determination-of-vitamin-c-content--1.html">Determination of vitamin C content </a></li><li><a href="../19342/hydrolysis-of-starch--1.html">Hydrolysis of starch </a></li><li><a href="../19159/preparation-purification-of-aspirin--1.html">Preparation & purification of aspirin </a></li></ul></li><li><a href="../18851/ict-in-practical-work-1.html" class="father">ICT in practical work</a><ul 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concepts</a></li><li><a href="../3507/why-tok-chemistry-1.html">Why TOK & Chemistry?</a></li><li><a href="../3505/a-modern-paradigm-1.html">A modern paradigm</a></li></ul></li></ul></li><li><a href="../17783/fast-track-to-tests-questions-1.html">Fast track to tests & questions</a><ul class="level-1"><li><a href="../17784/sl-multiple-choice-tests-on-individual-topics-1.html">SL Multiple choice tests on individual topics</a></li><li><a href="../24351/sl-multiple-choice-quizzes-on-sub-topics-1.html">SL Multiple choice 'quizzes' on sub-topics</a></li><li><a href="../31519/sl-practice-paper-1-exams-1.html">SL Practice Paper 1 exams</a></li><li><a href="../17786/sl-questions-on-each-sub-topic-1.html">SL Questions on each sub-topic</a></li><li><a href="../20431/sl-paper-3-section-a-questions-1.html">SL Paper 3 Section A questions</a></li><li><a href="../17792/sl-questions-on-the-options-1.html">SL Questions on the options</a></li><li><a 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</div><section id="main-content"><h1><img alt="" class="noborder" src="../../../ib/chemistry/images/Exams/Taking%20exams.jpg" style="width: 1px; height: 1px;"><img alt="" height="21" src="../../../img/tib-icons/standard-level-32-1.png" title="" width="21"> 'Mock' Paper 1 multiple choice exam (2)</h1><section class="tib-teacher-only" readonly="true" title="This box is not visible to students"><div class="header" readonly="true"><img class="icon" src="../../../img/icons/teacher-only.svg"> Teacher only box</div><div class="content" readonly="false"><p>To give your students access to this page you will need to change the access from "Filtered student access" to "Direct student access".</p></div></section><hr class="hidden"><div class="greenBg"><h2>Instructions</h2><ul><li>Time allowed: 45 minutes</li><li>Answer all the questions.</li><li>For each question choose the answer you consider to be the best.</li><li>Use the periodic table from Section 6 of the data booklet as your 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data-structure="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" data-score-answers="6d4a526a457a48584f763758344a7649472f76324c53614b78524c393439595570496d5a5971374b38686f3d"><div class="exercise"><div class="q-question"><p>Which is a mixture?</p></div><div class="q-answer"><p><label class="radio" data-answer="88336b5bb2a1cc21bac7cf33fd451270"><input type="radio"><span> sand</span></label></p><p><label class="radio" data-answer="cb925101f59ddb76ce2431b14a181a3f"><input type="radio"><span> sucrose</span></label></p><p><label class="radio" data-answer="3eb38fa8079f887324e8ee1d92b3da12"><input type="radio"><span> steel</span></label></p><p><label class="radio" data-answer="9460370bb0ca1c98a779b1bcc6861c2c"><input type="radio"><span>water</span></label></p></div><div class="q-explanation"><p>Water, H<sub>2</sub>O and sucrose, C<sub>12</sub>H<sub>22</sub>O<sub>12</sub> are both pure single substances. Steel is an alloy in which iron is bonded chemically to other elements such as carbon. Sand is a mixture as its different constituents are not bonded chemically to each other and retain their individual properties.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>How many atoms of oxygen are present in 12.6 g of hydrated oxalic acid, (COOH)<sub>2</sub>.2H<sub>2</sub>O?<br>Avogadro's constant, <em>L</em> or <em>N</em><sub>A</sub> = 6.02 x 10<sup>23</sup> mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio" data-answer="a2195f6787b24968c82efa2c0fe7b175"><input type="radio"><span> 3.6 x 10<sup>24</sup></span></label></p><p><label class="radio" data-answer="e95e1ca27d0e39aa03eb5a611ce4122f"><input type="radio"><span> 0.6</span></label></p><p><label class="radio" data-answer="1679091c5a880faf6fb5e6087eb1b2dc"><input type="radio"><span> 6</span></label></p><p><label class="radio" data-answer="587a38867c15b3a4a00c853112c8a829"><input type="radio"><span> 3.6 x 10<sup>23</sup></span></label></p></div><div class="q-explanation"><p><em>M</em> of (COOH)<sub>2</sub>.2H<sub>2</sub>O = (2 x 12) + (6 x 16) + (6 x 1) = 126 g mol<sup>−1</sup>, so the amount of (COOH)<sub>2</sub>.2H<sub>2</sub>O = 0.1 mol<br>Each (COOH)<sub>2</sub>.2H<sub>2</sub>O contains six oxygen atoms, so the number of O atoms = 6 x 0.1 x 6.02 x 10<sup>23</sup> = 3.6 x 10<sup>23</sup></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What volume of carbon dioxide measured at STP will be evolved when 4.0 g of calcium carbonate is added to 20 cm<sup>3</sup> of 2.0 mol dm<sup>−3</sup> hydrochloric acid solution?<br>Molar volume of a gas at STP = 22.7 dm<sup>3</sup> mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio" data-answer="769f5794bb7e44c10b70dcdcb7103f41"><input type="radio"><span> 114 cm<sup>3</sup></span></label></p><p><label class="radio" data-answer="3de3a5a18143e66894ee0f5bff3b7d26"><input type="radio"><span> 227 cm<sup>3</sup></span></label></p><p><label class="radio" data-answer="79ddfa8e18346903d5fa8d661ff6446b"><input type="radio"><span> 908 cm<sup>3</sup></span></label></p><p><label class="radio" data-answer="6c7fb1adc7762982279d3090d8129600"><input type="radio"><span> 454 cm<sup>3</sup></span></label></p></div><div class="q-explanation"><p>CaCO<sub>3</sub>(s) + 2HCl(aq) → CO<sub>2</sub>(g) + H<sub>2</sub>O(l) + CaCl<sub>2</sub>(aq)<br>Amount of CaCO<sub>3</sub> = 4/(40 + 12 + 48) = 0.04 mol; Amount of HCl = 20/1000 x 2.0 = 0.04 mol so HCl is the limiting reagent as 2 mol of HCl are required to react with 1 mol of CaCO<sub>3</sub> to produce 1 mol of CO<sub>2</sub>.<br>Amount of CO<sub>2</sub> produced = 0.02 mol, so volume of CO<sub>2 </sub>evolved = 0.02 x 22.7 = 0.454 dm<sup>3</sup> = 454 cm<sup>3</sup></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>A gas occupies 100 cm<sup>3</sup> at a temperature of 36 <sup>o</sup>C and a pressure of 9.63 x 10<sup>4 </sup>Pa. What volume (in cm<sup>3</sup>) will the same amount of gas occupy if the pressure is increased to 9.89 x 10<sup>4</sup> Pa and the temperature is increased to 63 <sup>o</sup>C?</p></div><div class="q-answer"><p><label class="radio" data-answer="b389e20d3835601f861fec0a7b7fd62b"><input type="radio"><span> 100 x (336 ÷ 309) x (9.89 ÷ 9.63) </span></label></p><p><label class="radio" data-answer="5e2e45566c64005549a087516ee94f36"><input type="radio"><span> 100 x (336 ÷ 309) x (9.63 ÷ 9.89)</span></label></p><p><label class="radio" data-answer="24fb591cdda7d043fe94237802aa4709"><input type="radio"><span> 100 x (309 ÷ 336) x (9.63 ÷ 9.89)</span></label></p><p><label class="radio" data-answer="24bfac5af22dfab26be1226e1266267d"><input type="radio"><span> 100 x (63 ÷ 36) x (9.89 ÷ 9.63)</span></label></p></div><div class="q-explanation"><p>Increasing the temperature increases the volume whereas increasing the pressure decreases the volume. The temperature must be measured in Kelvin so 36 <sup>o</sup>C and 63 <sup>o</sup>C is 309 K and 336 K respectively.</p><p>P<sub>1</sub>V<sub>1</sub>/T<sub>1</sub> = P<sub>2</sub>V<sub>2</sub>/T<sub>2</sub> so V<sub>2</sub> = V<sub>1 </sub>x T<sub>2</sub>/T<sub>1</sub> x P<sub>1</sub>/P<sub>2</sub> = 100 x (336 ÷ 309) x (9.63 x 10<sup>4</sup> ÷ 9.89 x 10<sup>4</sup>) = 100 x (336 ÷ 309) x (9.63 ÷ 9.89)</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which transition in the hydrogen emission spectrum will emit light in the visible region of the electromagnetic spectrum?</p></div><div class="q-answer"><p><label class="radio" data-answer="72c8ba9262a5142f1db31015e6886f3c"><input type="radio"><span> <em>n</em> = 4 to <em>n</em> = 2</span></label></p><p><label class="radio" data-answer="4c56d302e2284de82509467262cffa8d"><input type="radio"><span> <em>n</em> = 2 to <em>n</em> = 4</span></label></p><p><label class="radio" data-answer="c890eba337da134b7810da64c0e9e565"><input type="radio"><span> <em>n</em> = 1 to <em>n</em> = 3</span></label></p><p><label class="radio" data-answer="17bed0ba636c246f599334cd5b9e7747"><input type="radio"><span> <em>n</em> = 3 to <em>n</em> = 1</span></label></p></div><div class="q-explanation"><p>In an emission spectrum light is emitted when excited electrons fall to lower levels. When they fall to the <em>n </em>= 1 level the light emitted is in the ultraviolet region of the spectrum. When they fall from higher levels to the <em>n</em> = 2 level the light emitted is in the visible region of the spectrum.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What is the composition of the <sup>37</sup>Cl<sup>+</sup> ion?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/composition-of-cl-.png" style="width: 500px; height: 138px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p>Chlorine has an atomic number of 17, so contains 17 protons. This isotope has a mass number of 37 which is equal to the number of protons and neutrons, so it contains 20 neutrons. A neutral chlorine atom contains the same number of electrons as protons but the Cl<sup>+</sup> ion has lost one electron, so it contains 16 electrons.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which electron configuration is possessed by an element in group 13 and period 3 in the periodic table?</p></div><div class="q-answer"><p><label class="radio" data-answer="3ed3d5c8ea402de865d693270ec979fe"><input type="radio"><span> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>3d<sup>10</sup>4s<sup>2</sup>4p<sup>1</sup></span></label><span></span></p><p><label class="radio" data-answer="1036ffac144ba0aeed5eb3afadbde9dc"><input type="radio"><span> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>3d<sup>5</sup></span></label></p><p><label class="radio" data-answer="63be74f41ddacd0340299b2d1c025ecb"><input type="radio"><span> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>3</sup></span></label></p><p><label class="radio" data-answer="2f2ee86fea2fa8fc13e92a124743f3ca"><input type="radio"><span> 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>1</sup></span></label></p></div><div class="q-explanation"><p>If it is in the third period the outer level of electrons must be occupying the 3s or 3s and 3p sub-levels and as it is in group 13 there must be three electrons in the outer level, so the correct electron configuration is 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>1</sup> (which is the electron configuration of aluminium).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which are the correct trends as group 17 in the periodic table is descended (F → I)?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/group-17-trends.png" style="width: 500px; height: 139px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p>Atomic radius increases as the outer electrons become further from the nucleus. ionization energy decreases as it requires less energy to remove the outermost electron and melting point increases as the intermolecular forces get stronger due to an increase in mass and number of electrons.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Europium, Eu is a lanthanoid and shows the same oxidation state in its sulfate and nitrate salts. The formula of europium sulfate is Eu<sub>2</sub>(SO<sub>4</sub>)<sub>3</sub>. What will be the formula of europium nitrate?</p></div><div class="q-answer"><p><label class="radio" data-answer="6ccda41be9ac09498f20f71afc43a582"><input type="radio"><span> EuNO<sub>3</sub></span></label></p><p><label class="radio" data-answer="786450f23bffb543e7308e5dfbb50a56"><input type="radio"><span> Eu<sub>2</sub>(NO<sub>3</sub>)<sub>3</sub></span></label></p><p><label class="radio" data-answer="798b0234cbf5ff82e8b74659ebb83847"><input type="radio"><span> Eu(NO<sub>3</sub>)<sub>3</sub></span></label></p><p><label class="radio" data-answer="53c8c0c3c7cb461d26327d5597486c52"><input type="radio"><span> Eu<sub>3</sub>(NO<sub>3</sub>)<sub>2</sub></span></label></p></div><div class="q-explanation"><p>The oxidation state of Eu in Eu<sub>2</sub>(SO<sub>4</sub>)<sub>3</sub> is +3 so Eu forms the Eu<sup>3+</sup> ion. The formula for the nitrate ion is NO<sub>3</sub><sup>−</sup> so the formula of Eu(III) nitrate is Eu(NO<sub>3</sub>)<sub>3</sub>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which contains at least one coordinate covalent bond?</p></div><div class="q-answer"><p><label class="radio" data-answer="67b2079e2cae80b6d80c63e9a6d4959e"><input type="radio"><span> the carbonate ion, CO<sub>3</sub><sup>2−</sup></span></label></p><p><label class="radio" data-answer="02927405a2ffacb5edbe5e4465346b98"><input type="radio"><span> carbon dioxide, CO<sub>2</sub></span></label></p><p><label class="radio" data-answer="f84ddecf38831bd4ca904b3252687198"><input type="radio"><span> carbon monoxide, CO</span></label></p><p><label class="radio" data-answer="51e1b92f02db8ff8d2d8b2e38490b8ff"><input type="radio"><span> methanoic acid, HCOOH</span></label></p></div><div class="q-explanation"><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/lewis-structure-of-co.png" style="width: 300px; height: 27px;"><br>In carbon monoxide the triple bond between the C and O atom is formed by the carbon atom sharing two electrons with two electrons from the O and the O donating a further pair of electrons. In the other three species all the bonds are normal covalent single or double bonds, even in the carbonate ion which forms a resonance hybrid.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which compound has molecules with a bond angle less than 109<sup>o</sup>?</p></div><div class="q-answer"><p><label class="radio" data-answer="eb967d83215992e6c8ef32fb10412778"><input type="radio"><span> boron trichloride, BCl<sub>3</sub></span></label></p><p><label class="radio" data-answer="98df826b324907e620082409996e8f4e"><input type="radio"><span> beryllium chloride, BeCl<sub>2</sub></span></label></p><p><label class="radio" data-answer="97274c6cd502bd438497b5fb8e2dea3d"><input type="radio"><span> water, H<sub>2</sub>O</span></label></p><p><label class="radio" data-answer="02927405a2ffacb5edbe5e4465346b98"><input type="radio"><span> carbon dioxide, CO<sub>2</sub></span></label></p></div><div class="q-explanation"><p>Water contains two non-bonding pairs of electrons which repel the electrons in the two bonding pairs making the bond angle less than 109<sup>o</sup>. BeCl<sub>2</sub> and CO<sub>2</sub> both have two electron domains making the molecules linear with bond angles of 180<sup>o</sup> and BCl<sub>3</sub> has three electron domains so has a trigonal planar shape with bond angles of 120<sup>o</sup>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which describes the strongest type of intermolecular attractive forces between molecules in the compound?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/hydrogen-bonding-etc-(1).png" style="width: 500px; height: 154px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p>Both propan-2-ol and propanoic acid have a hydrogen atom bonded directly to an electronegative oxygen atom so form hydrogen bonds between molecules whereas in propanal the hydrogen atom is bonded to a carbon atom and so only dipole-dipole attractions occur.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>The boiling point of propanone is 56 <sup>o</sup>C. Which reaction has an enthalpy change equal to the enthalpy of formation of propanone?</p></div><div class="q-answer"><p><label class="radio" data-answer="db6c8f5fa6ae335c76c91fa54cc6c6fe"><input type="radio"><span> 6C(s) + 6H<sub>2</sub>(g) + O<sub>2</sub>(g) → 2CH<sub>3</sub>COCH<sub>3</sub>(l)</span></label></p><p><label class="radio" data-answer="143879536b7254dfbf60fda17e0947a6"><input type="radio"><span> 3C(s) + 3H<sub>2</sub>(g) + ½O<sub>2</sub>(g) → CH<sub>3</sub>COCH<sub>3</sub>(g)</span></label></p><p><label class="radio" data-answer="2da5afc89ce687c1d1a374ac1089c0de"><input type="radio"><span> 3C(s) + 3H<sub>2</sub>(g) + ½O<sub>2</sub>(g) → CH<sub>3</sub>COCH<sub>3</sub>(l)</span></label></p><p><label class="radio" data-answer="4bae4de179a05bec610624aa35261c01"><input type="radio"><span> 6C(s) + 6H<sub>2</sub>(g) + O<sub>2</sub>(g)→ 2CH<sub>3</sub>COCH<sub>3</sub>(g)</span></label></p></div><div class="q-explanation"><p>Enthalpy of formation refers to the enthalpy change when one mole of the compound is formed from its elements with all the substances being in their standard state at 298 K (25 <sup>o</sup>C).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>The superscript symbol, <sup>⦵</sup> placed after Δ<em>H</em> means that the enthalpy change for the reaction is the standard enthalpy change. Which statement must be true for the standard enthalpy change of any chemical reaction?</p></div><div class="q-answer"><p><label class="radio" data-answer="994f4ae927c23cfd9003f82a66fb95ab"><input type="radio"><span> The value should always be given in kJ mol<sup>−1</sup></span></label></p><p><label class="radio" data-answer="7232d0155fd17965dcfbc8fadd4d658a"><input type="radio"><span> The enthalpy change must be measured at 298 K</span></label></p><p><label class="radio" data-answer="3afccf4ba1e8e0e3b006597067997606"><input type="radio"><span> Δ<em>H</em><sup>⦵ </sup>reaction = Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> reactants) − Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> products)</span></label></p><p><label class="radio" data-answer="c2e16a093da585a9d261fb862bb1f215"><input type="radio"><span> Reactants and products must be in their normal most pure state measured at 100 kPa.</span></label></p></div><div class="q-explanation"><p>Temperature is not a part of the definition of standard state (although 298 K is commonly given as the temperature of interest).<br>Δ<em>H</em><sup>⦵ </sup>reaction = Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> products) − Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> reactants) not Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> reactants) − Σ(Δ<em>H</em><sup>⦵</sup><sub>f</sub> products).<br>Although Δ<em>H</em><sup>⦵</sup><sub>f </sub>and<sup> </sup>Δ<em>H</em><sup>⦵</sup><sub>c</sub> are normally given in kJ mol<sup>−1</sup> as they refer to one particular substance, Δ<em>H</em><sup>⦵</sup>reaction is normally given in kJ, not kJ mol<sup>−1</sup> as, for example in the reaction N<sub>2</sub>(g) + 3<sub>2</sub>(g) ⇌ 2NH<sub>3</sub>(g), it is not clear if the mol<sup>−1</sup> is referring to N<sub>2</sub>(g), H<sub>2</sub>(g) or NH<sub>3</sub>(g).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What is the enthalpy change in kJ mol<sup>−1</sup> for the combustion of propan-2-ol, CH<sub>3</sub>CH(OH)CH<sub>3</sub>?</p><p>2CH<sub>3</sub>CH(OH)CH<sub>3</sub>(l) + 9O<sub>2</sub>(g) → 6CO<sub>2</sub>(g) + 8H<sub>2</sub>O(l)</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/propan-2-ol-data.png" style="width: 350px; height: 91px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="edfbbc6dc576e9dda40f1dad01aeafbc"><input type="radio"><span> [(2 x − 318) − (6 x − 394) − (8 x − 286)]</span></label></p><p><label class="radio" data-answer="aa1ebca58c42aae200e6ace5da24d4fc"><input type="radio"><span> ½ [(2 x − 318) − (6 x − 394) − (8 x − 286)]</span></label></p><p><label class="radio" data-answer="d3c8eedf6160ad2fa8c047b2ccbb09bd"><input type="radio"><span> [(6 x − 394) + (8 x − 286) − (2 x − 318)]</span></label></p><p><label class="radio" data-answer="b1e3a64370857b3c8188c4a540cf4b25"><input type="radio"><span> ½ [(6 x − 394) + (8 x − 286) − (2 x − 318)]</span></label></p></div><div class="q-explanation"><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/propan-2-ol-combustion-cycle.png" style="width: 350px; height: 142px;"></p><p><em>x</em> = [(6 x −394) + (8 x − 286) − (2 x − 318)] kJ</p><p>But <em>x</em> is the enthalpy change for the combustion of 2 mol of propan-2-ol</p><p>so Δ<em>H</em> = ½ [(6 x −394) + (8 x − 286) − (2 x − 318)] kJ mol<sup>−1</sup></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>The graph shows the fraction of reacting particles with a particular energy for an uncatalysed reaction at two temperatures, T<sub>1</sub> and T<sub>2</sub>. The activation energy for the reaction is shown as <em>E</em><sub>a</sub>.</p><p> </p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/maxwell-boltzman-graph.png" style="width: 400px; height: 281px;"></p><p> </p><p>Which statement is correct?</p></div><div class="q-answer"><p><label class="radio" data-answer="c22b3709afd5ba0fac592dfd1442183a"><input type="radio"><span> Adding a catalyst moves the peak of T<sub>1</sub> and the peak of T<sub>2</sub> further to the right.</span></label></p><p><label class="radio" data-answer="ff950bd7a1320a09a53ee64d231771af"><input type="radio"><span> T<sub>1</sub> > T<sub>2</sub></span></label></p><p><label class="radio" data-answer="f98979065c8df0fad23fd48910d216ea"><input type="radio"><span> The area under T<sub>2</sub> > the area under T<sub>1</sub>.</span></label></p><p><label class="radio" data-answer="3dadd4316eb6315e5c4257a23f3dcc34"><input type="radio"><span> The activation energy for the catalysed reaction will be less than <em>E</em><sub>a</sub>.</span></label></p></div><div class="q-explanation"><p>Increasing the temperature increases the average energy of the reacting particles so T<sub>2</sub> > T<sub>1</sub> but does not change the number of reacting particles so the area under the graph remains constant. Adding a catalyst does not change the energy of the reacting particles but does provide a different pathway for the reaction with a lower activation energy.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>100 cm<sup>3</sup> of 2.0 mol dm<sup>−3</sup> hydrochloric acid was added to 1.0 g of a strip of magnesium metal. A plot of volume of hydrogen evolved against time gave graph A.</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/mg-hcl-graph(1).png" style="width: 300px; height: 224px;"></p><p>Assuming all the other conditions were kept the same which changes would have given graphs B and C?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/data-for-mg-hcl(2).png" style="width: 600px; height: 211px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p>100 cm<sup>3</sup> of 2.0 mol dm<sup>−3</sup> HCl(aq) = 0.20 mol, 1.0 g of Mg = 1.0/24.3 = < 0.05 mol so the acid is in excess and Mg is the limiting reagent. If only 0.5 g of magnesium is used half the volume of hydrogen will be evolved so C and D cannot be correct. Using powdered Mg instead of a single piece of Mg will increase the rate whereas using 1.0 mol dm<sup>−3</sup> HCl(aq) instead of 2.0 mol dm<sup>−3</sup> HCl(aq) will decrease the rate but the same volume of H<sub>2</sub> will eventually be evolved.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which change will increase the value of the equilibrium constant for the reaction of sulfur dioxide with oxygen to produce sulfur trioxide?</p><p style="margin-left: 120px;">2SO<sub>2</sub>(g) + O<sub>2</sub>(g) ⇌ 2SO<sub>3</sub>(g) Δ<em>H</em><sup>⦵</sup> = − 196 kJ</p></div><div class="q-answer"><p><label class="radio" data-answer="9c451630bad73003fc529c19cafc3a67"><input type="radio"><span> lowering the temperature</span></label></p><p><label class="radio" data-answer="71c9b84347d182391ffbfd31bd594b96"><input type="radio"><span> removing some of the sulfur trioxide as it is formed</span></label></p><p><label class="radio" data-answer="bc8a272a6a8a67fa152d4b9a0f5a006b"><input type="radio"><span> decreasing the pressure</span></label></p><p><label class="radio" data-answer="46771f43ea41d7ca1eab3cc0c4531b0d"><input type="radio"><span> increasing the pressure</span></label></p></div><div class="q-explanation"><p>All of the changes will alter the position of equilibrium but the equilibrium constant is a constant at a particular temperature so only a change in temperature will change the value of <em>K</em><sub>c</sub>. As Δ<em>H</em><sup>⦵</sup> has a negative value the reaction is exothermic so lowering the temperature will increase the value of <em>K</em><sub>c</sub>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What will be the pH of a solution made by adding 90 cm<sup>3</sup> of water to 10 cm<sup>3</sup> of an aqueous solution of a strong base with a pH of 12?</p></div><div class="q-answer"><p><label class="radio" data-answer="aab3238922bcc25a6f606eb525ffdc56"><input type="radio"><span> 14</span></label></p><p><label class="radio" data-answer="d3d9446802a44259755d38e6d163e820"><input type="radio"><span> 10</span></label></p><p><label class="radio" data-answer="c20ad4d76fe97759aa27a0c99bff6710"><input type="radio"><span> 12</span></label></p><p><label class="radio" data-answer="6512bd43d9caa6e02c990b0a82652dca"><input type="radio"><span> 11</span></label></p></div><div class="q-explanation"><p>In the original solution with a pH of 12 [H<sup>+</sup>(aq)] = 1 x 10<sup>−12</sup> mol dm<sup>−3</sup> and [OH<sup>−</sup>(aq)] = 1 x 10<sup>−2</sup> mol dm<sup>−3</sup>. Adding 90 cm<sup>3</sup> of water dilutes the solution ten times so [OH<sup>−</sup>(aq)] = 1 x 10<sup>−3</sup> mol dm<sup>−3 </sup>and [H<sup>+</sup>(aq)] = 1 x 10<sup>−11</sup> mol dm<sup>−3 </sup>so pH = 11.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Hydrogen chloride gas dissolves in water to form hydrochloric acid.</p><p>HCl(g) + H<sub>2</sub>O(l) → H<sub>3</sub>O<sup>+</sup>(aq) + Cl<sup>−</sup>(aq)</p><p>What is the role of the water?</p></div><div class="q-answer"><p><label class="radio" data-answer="ac445bdc12d86258fd1891a371d41803"><input type="radio"><span> It is only acting as a base</span></label></p><p><label class="radio" data-answer="c52c04ce0ee93a381f1a34a671454d05"><input type="radio"><span> It is only acting as a solvent</span></label></p><p><label class="radio" data-answer="107a27a9680c7fcd6890750ac6954efd"><input type="radio"><span> It is acting as an acid and a base and a solvent</span></label></p><p><label class="radio" data-answer="737f1c88c6641981da9aed5d5264eb9c"><input type="radio"><span> It is acting as a base and a solvent</span></label></p></div><div class="q-explanation"><p>Water is amphiprotic so depending upon the reaction can act as an acid or a base by losing or gaining a proton. In this reaction it is not acting as an acid as it is not losing a proton to form OH<sup>−</sup> but is acting as a Brønsted-Lowry base by gaining a proton to form H<sub>3</sub>O<sup>+</sup>. Since the gas is dissolving and the ions formed are hydrated it is also acting as a solvent.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What is the oxidation state of nitrogen in HNO<sub>3</sub>?</p></div><div class="q-answer"><p><label class="radio" data-answer="44904c488b9ca058f08c76986f3c414f"><input type="radio"><span> 5+</span></label></p><p><label class="radio" data-answer="e4da3b7fbbce2345d7772b0674a318d5"><input type="radio"><span> 5</span></label></p><p><label class="radio" data-answer="568576e1139107c85be264b010528e59"><input type="radio"><span> +5</span></label></p><p><label class="radio" data-answer="ac20b0a9f2323ab099161d58790c9850"><input type="radio"><span> (V)</span></label></p></div><div class="q-explanation"><p>Oxidation states should be represented with the sign given before the number, e.g. +5 not 5+. Roman numerals, e.g. (V) refer to oxidation numbers. When the plus sign is after the number it refers to the charge on a positive ion e.g. Fe<sup>2+</sup>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>The reactions of zinc metal with nickel ions and nickel metal with lead ions occur spontaneously. </p><p>Zn(s) + Ni<sup>2+</sup>(aq) → Zn<sup>2+</sup>(aq) + Ni(s)</p><p>Ni(s) + Pb<sup>2+</sup>(aq) → Ni<sup>2+</sup>(aq) + Pb(s)</p><p>Which is the best oxidizing agent?</p></div><div class="q-answer"><p><label class="radio" data-answer="68d502c5dd8bdd76ec6725557ed219b4"><input type="radio"><span> Ni(s)</span></label></p><p><label class="radio" data-answer="7bc89c6ae00019e7be38e3251e906434"><input type="radio"><span> Ni<sup>2+</sup>(aq)</span></label></p><p><label class="radio" data-answer="4408fda3ac4861a41062182963aac121"><input type="radio"><span> Zn(s)</span></label></p><p><label class="radio" data-answer="c0ef8306b7dd2c167aad83133d5fe0ad"><input type="radio"><span> Pb<sup>2+</sup>(aq)</span></label></p></div><div class="q-explanation"><p>Ni<sup>2+</sup>(aq) ions are a better oxidizing agent than Zn(s) as they remove electrons from Zn(s) to form Ni(s), but Pb<sup>2+</sup>(aq) is an even stronger oxidizing agent than Ni<sup>2+</sup>(aq) as it can remove electrons from Ni(s). Zn(s) is the best reducing agent.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which is a tertiary alcohol?</p></div><div class="q-answer"><p><label class="radio" data-answer="9d9cd2eda036d4ecd51a57559b51e9a4"><input type="radio"><span> CH<sub>2</sub>(OH)−CH(OH)−CH<sub>2</sub>(OH)</span></label></p><p><label class="radio" data-answer="8a2b2849e83bad4cad9c632896040eac"><input type="radio"><span> C(CH<sub>3</sub>)<sub>3</sub>OH</span></label></p><p><label class="radio" data-answer="36a012a29c9dc556ba6a623d9ab5ec1c"><input type="radio"><span> CH<sub>3</sub>−CH<sub>2</sub>−CH(OH)−CH<sub>3</sub></span></label></p><p><label class="radio" data-answer="facd93f8edb009b7db7b095859e9830f"><input type="radio"><span> CH<sub>3</sub>−CH<sub>2</sub>−CH<sub>2</sub>OH</span></label></p></div><div class="q-explanation"><p>Tertiary alcohols contains three R− groups bonded to the carbon atom to which the −OH is bonded. Compounds such as propan−1,2,3−triol (also known as glycerol), which contain three −OH groups, are known as trihydric alcohols.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which compounds belong to the same homologous series?</p></div><div class="q-answer"><p><label class="radio" data-answer="59831ffb8117174682745edc409b81c2"><input type="radio"><span> CHCH and CH<sub>2</sub>CH<sub>2</sub></span></label></p><p><label class="radio" data-answer="c524f62900b7e52477a540fcfa6d491b"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>OCH<sub>2</sub>CH<sub>3</sub> and CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CH<sub>2</sub>OH</span></label></p><p><label class="radio" data-answer="66050c0a12ca176106b191e1af185ed1"><input type="radio"><span> CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CHO and CH<sub>3</sub>CH<sub>2</sub>COCH<sub>3</sub></span></label></p><p><label class="radio" data-answer="905dfbdd8baa2394144eaa1cc80d527a"><input type="radio"><span> CH<sub>2</sub>CH<sub>2</sub> and CH<sub>2</sub>CHCH<sub>3</sub></span></label></p></div><div class="q-explanation"><p>A homologous series is a series of compounds of the same family, with the same general formula, which differ from each other by a common structural unit. Ethene, CH<sub>2</sub>CH<sub>2</sub> and propene, CH<sub>2</sub>CHCH<sub>3</sub> are both members of the alkene homologous series with the general formula C<sub>n</sub>H<sub>2n</sub>. Ethyne, CHCH is an alkyne whereas ethene, CH<sub>2</sub>CH<sub>2</sub> is an alkene. Ethoxyethane, CH<sub>3</sub>CH<sub>2</sub>OCH<sub>2</sub>CH<sub>3</sub> and butan-1-ol, CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CH<sub>2</sub>OH are an ether and an alcohol respectively. They are isomers as they both have the same molecular formula, C<sub>4</sub>H1<sub>0</sub>O but are members of different classes of compounds. Butanal, CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>CHO and butan-2-one, CH<sub>3</sub>CH<sub>2</sub>COCH<sub>3</sub> are also isomers, one is an aldehyde and the other is a ketone.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which is a propagation step in the mechanism for the free radical substitution reaction between methane and chlorine in ultraviolet light?</p></div><div class="q-answer"><p><label class="radio" data-answer="00dd19d3f438cd38bf7d382e44eb4af6"><input type="radio"><span> CH<sub>3</sub><strong>· </strong>+ Cl<strong>·<sup> </sup></strong>→ CH<sub>3</sub>Cl</span></label></p><p><label class="radio" data-answer="fc89534a4ac0a37ac36cf6aeaa0144c7"><input type="radio"><span> Cl<sub>2</sub> → Cl<strong>·</strong> + Cl<strong>·</strong></span></label></p><p><label class="radio" data-answer="4e85b3ef29373475e57228b784a1088c"><input type="radio"><span> CH<sub>3</sub><strong>·</strong> + Cl<sub>2</sub> → CH<sub>3</sub>Cl + Cl<strong>·</strong></span></label></p><p><label class="radio" data-answer="8b45968a3543013ace786a5605ff9418"><input type="radio"><span> CH<sub>4</sub> + Cl<strong>·</strong>→ CH<sub>3</sub>Cl + H<strong>·</strong></span></label></p></div><div class="q-explanation"><p>The initiation step is the homolytic fission of the Cl−Cl bond by ultraviolet light to give chlorine free radicals. These radicals then react to produce more radicals in propagation steps such as CH<sub>4</sub> + Cl<strong>·</strong> → CH<sub>3</sub><strong>·</strong> + HCl and CH<sub>3</sub><strong>· </strong>+ Cl<sub>2</sub> → CH<sub>3</sub>Cl + Cl<strong>·</strong>. At no time are hydrogen, H<strong>·</strong> radicals formed. Termination steps such as CH<sub>3</sub><strong>·</strong> + Cl<strong>·</strong> → CH<sub>3</sub>Cl result in a product that is not a free radical.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which compound will rapidly decolourise an aqueous solution of bromine in the dark?</p></div><div class="q-answer"><p><label class="radio" data-answer="134701a579e5349f720af10f3782b879"><input type="radio"><span> butanal</span></label></p><p><label class="radio" data-answer="26b5276f3061c111b5deb68dfd6a7f24"><input type="radio"><span> benzene</span></label></p><p><label class="radio" data-answer="df92cbf550462954aca00b78261a2153"><input type="radio"><span> butane</span></label></p><p><label class="radio" data-answer="7749379ec28e4bcc4ef8590350c2e8b7"><input type="radio"><span> but-2-ene</span></label></p></div><div class="q-explanation"><p>Bromine adds across the C=C double bond of alkenes such as but-2-ene and the organic product is colourless. Benzene does not readily undergo addition reactions. Butane is saturated so cannot undergo addition reactions and butanal is an aldehyde and has no C=C double bond.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>A piece of magnesium of known mass was combusted in air in a crucible and the product was weighed afterwards to determine the empirical formula of magnesium oxide. Which could explain why the result obtained was Mg<sub>4</sub>O<sub>5</sub> rather than the expected result of MgO?</p></div><div class="q-answer"><p><label class="radio" data-answer="ed71309f2351a35495885a3c9dc392ab"><input type="radio"><span> Not all of the magnesium combusted.</span></label></p><p><label class="radio" data-answer="c4c770ef902daf057bf7dad659580806"><input type="radio"><span> The crucible was not inert and also reacted with the magnesium.</span></label></p><p><label class="radio" data-answer="ce1059ea41f6813967f511ebb7b1106d"><input type="radio"><span> Some of the product escaped before it could be weighed.</span></label></p><p><label class="radio" data-answer="5b158521690cda6034fb0031d0becc20"><input type="radio"><span> The piece of magnesium used already had a coating of the oxide layer. </span></label></p></div><div class="q-explanation"><p>If some of the product was lost, not all the magnesium burned or if some of the magnesium burned was already magnesium oxide then the mass of the weighed product would be less than expected, so the amount of oxygen combining with the magnesium would be less and the empirical formula would be richer in Mg compared to O. If the crucible reacted with some of the magnesium then the product might have a mass more than if it had just combined with oxygen so the formula would be richer in O compared to Mg.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>The concentration of a reactant, A was measured as a reaction proceeded.</p><p>It was found that the expression</p><p>ln[A] = − <em>k</em>t + ln[A<sub>o</sub>] (where [A<sub>o</sub>] is the initial concentration of A, [A] is the concentration of A after time t and<em> k</em> is the rate constant) gave a straight line when a graph of the natural logarithm of the concentration of A, ln[A] was plotted against time, t.</p><p>Which is correct for the values of the intercept and the gradient?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/intercept---gradient(1).png" style="width: 440px; height: 143px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="7fc56270e7a70fa81a5935b72eacbe29"><input type="radio"><span> A</span></label></p><p><label class="radio" data-answer="9d5ed678fe57bcca610140957afab571"><input type="radio"><span> B</span></label></p><p><label class="radio" data-answer="0d61f8370cad1d412f80b84d143e1257"><input type="radio"><span> C</span></label></p><p><label class="radio" data-answer="f623e75af30e62bbd73d6df5b50bb7b5"><input type="radio"><span> D</span></label></p></div><div class="q-explanation"><p>Straight line graphs follow the general equation y = mx + c, where m is the gradient and c is the intercept.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>What is the index of hydrogen deficiency of the amino acid phenylalanine, C<sub>9</sub>H<sub>11</sub>NO<sub>2</sub>?</p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/quiz/mc-test/phenylalanine.png" style="width: 130px; height: 147px;"></p></div><div class="q-answer"><p><label class="radio" data-answer="e4da3b7fbbce2345d7772b0674a318d5"><input type="radio"><span> 5</span></label></p><p><label class="radio" data-answer="1679091c5a880faf6fb5e6087eb1b2dc"><input type="radio"><span> 6</span></label></p><p><label class="radio" data-answer="a87ff679a2f3e71d9181a67b7542122c"><input type="radio"><span> 4</span></label></p><p><label class="radio" data-answer="eccbc87e4b5ce2fe28308fd9f2a7baf3"><input type="radio"><span> 3</span></label></p></div><div class="q-explanation"><p>When determining IHD O atoms count as zero and for a N atom add one to the number of C atoms and add one to the number of H atoms. This gives the equivalent of C<sub>10</sub>H<sub>12</sub> which needs 10 more hydrogen atoms (5 units of H<sub>2</sub>) to become saturated as C<sub>10</sub>H<sub>22</sub>, hence the IHD is 5. This can also be seen from the phenyl group which has an IHD of 4 due to the 'three double bonds' and the ring plus the IHD of 1 for the C=O in the carboxyl functional group to give a total IHD of 5.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="exercise"><div class="q-question"><p>Which compound will give three signals in its <sup>1</sup>H NMR spectrum with integration traces in the ratio of 3:3:2?</p></div><div class="q-answer"><p><label class="radio" data-answer="bb07d76740952e3e24ddaf8ef6877db0"><input type="radio"><span> CH<sub>3</sub>−CH(OH)−CH<sub>3</sub></span></label></p><p><label class="radio" data-answer="f8b5ed59d57dd6a895ceeee0b3028028"><input type="radio"><span> C<sub>2</sub>H<sub>5</sub>−CH<sub>2</sub>−CH<sub>3</sub></span></label></p><p><label class="radio" data-answer="910f13f0a7172ffdeb91e0d00de67413"><input type="radio"><span> CH<sub>3</sub>−CO−CH<sub>2</sub>−CH<sub>3</sub></span></label></p><p><label class="radio" data-answer="9d331a147e38e1df456bb471c6e5783e"><input type="radio"><span> CH<sub>3</sub>−CH<sub>2</sub>−COOH</span></label></p></div><div class="q-explanation"><p>C<sub>2</sub>H<sub>5</sub>− is actually an ethyl group, CH<sub>3</sub>−CH<sub>2</sub>−, so pentan-2-one, C<sub>2</sub>H<sub>5</sub>−CH<sub>2</sub>−CO−CH<sub>3</sub><sub> </sub>will give 4 signals. The other three compounds will give three signals as the hydrogen atoms are in three different chemical environments. The ratio of the integration traces will be 3:2:1 for propanoic acid, CH<sub>3</sub>CH<sub>2</sub>COOH, 6:1:1 for propan-2-ol, CH<sub>3</sub>CH(OH)CH<sub>3</sub> and 3:3:2 for butan-2-one, CH<sub>3</sub>COCH<sub>2</sub>CH<sub>3</sub>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn check"><i class="fa fa-check-square-o"></i> Check</button></div></div><p> </p><div class="totals"><span class="score">Total Score: </span><button class="btn btn-success check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr></section></article><section id="media-extras"><div class="page-actions no-print navbar inline hidden-desktop"><div class="navbar-inner"><ul class="nav"><li><a class="presentation" href="#" onclick="return false;"><i class="fa fa-desktop"></i></a></li><li><a class="print-section-blog" href="#" onclick="return false;"><i class="fa fa-print"></i></a></li><li><a class="page-bookmarker" data-ticket="IB Docs (2) Team" data-pid="31677" href="#" onclick="return false;"><i class="fa fa-star"></i></a></li><li><a class="personal-notes" href="#" onclick="return false;"><i class="fa fa-file-text"></i></a></li><li><a class="" data-toggle="modal" href="#modal-feedback" onclick="return false;"><i class="fa fa-envelope-o"></i></a></li><li class="dropdown"><a class="dropdown-toggle" data-toggle="dropdown" data-target="#" href="#"><i class="fa fa-share-alt"></i></a><ul class="dropdown-menu" role="menu"><li><a class="" target="_blank" title="Share on Twitter" href="https://twitter.com/share?text=%22SL+Practice+Paper+1+%282%29+%22+-+via+%40InThinker+%23chemistry%0A&url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F31677%2Fsl-practice-paper-1-2-"><i class="fa fa-twitter-square"></i><span>Twitter</span></a></li><li><a class="" target="_blank" title="Share on Facebook" href="https://www.facebook.com/sharer.php?u=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F31677%2Fsl-practice-paper-1-2-&t=SL+Practice+Paper+1+(2)+"><i class="fa fa-facebook-square"></i><span>Facebook</span></a></li><li><a class="" href="http://www.linkedin.com/shareArticle?mini=true&url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F31677%2Fsl-practice-paper-1-2-"><i class="fa fa-linkedin-square"></i><span>LinkedIn</span></a></li><li><a class="" href="https://plus.google.com/share?url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F31677%2Fsl-practice-paper-1-2-"><i class="fa fa-google-plus-square"></i><span>Google +</span></a></li><li><a class="" href="mailto:?subject=SL Practice Paper 1 (2) &body=To give your students access to this page you will need to change the access from " filtered="" student="" access"="" to="" "direct="" student="" access".="" %0ahttps%3a%2f%2fwww.thinkib.net%2fchemistry%2fpage%2f31677%2fsl-practice-paper-1-2-"=""><i class="fa fa-envelope"></i><span>Email</span></a></li></ul></li><li><a class="" href="chemistry/teaching-materials"><i class="fa fa-puzzle-piece colored"></i></a></li></ul></div></div><div style="border-top: solid 1px #eee;border-bottom: solid 1px #eee;padding: 4px 0 4px 0;line-height: 1em;color: #666;font-size: .8em"><small><em>All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. 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