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btn-default btn-block" style="margin-bottom: 10px"><i class="fa fa-table"></i> Periodic table</button> <div class="hidden-xs hidden-sm"> <button class="btn btn-default btn-block text-xs-center" data-toggle="modal" data-target="#modal-feedback" style="margin-bottom: 10px"><i class="fa fa-send"></i> Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> Redox core (SL and HL) paper 1 questions <a href="#" class="mark-page-favorite pull-right" data-pid="2724" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../2075/paper-1-exam-questions.html">Paper 1 Exam Questions</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Redox core (SL and HL) paper 1 questions</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 30 minutes"><i class="fa fa-clock-o"></i> 30'</span> </ol> <article id="main-article"> <p><strong>Topic 9.1 and 9.2</strong></p> <p><strong>Paper 1 style questions </strong>are multiple choice. You are <strong>not permitted to use a calculator or the data book</strong> for these questions, but you should use a periodic table.</p> <p>A <strong>periodic table pop-up</strong> is available on the left hand menu.</p> <div class="greenBg"> <p> </p> <div class="tib-quiz" data-stats="7-816-2724"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>In which species does sulfur have the same oxidation state as in SOCl<sub>2</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> S<sub>2</sub>O<sub>3</sub><sup>2−</sup></label></p><p><label class="radio"> <input type="radio"> SO<sub>4</sub><sup>2−</sup></label></p><p><label class="radio"> <input class="c" type="radio"> SO<sub>3</sub><sup>2−</sup></label></p><p><label class="radio"> <input type="radio"> SO<sub>3</sub></label></p></div><div class="q-explanation"><p>In SOCl<sub>2</sub>, the oxygen atom has an oxidation state of −2, the chlorine atoms both have −1, therefore the sulfur atom has an oxidation state of +4.</p><p>The only other species in which the sulfur has a +4 oxidation state is SO<sub>3</sub><sup>2−</sup>. Three oxygen atoms gives (3 × −2) = −6, the overall charge is 2−, so the sulfur atom must be +4.</p><p><strong>SO<sub>3</sub><sup>2−</sup> </strong>is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>In the electrolysis of molten sodium chloride, what happens to the sodium ion?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> It <span>goes to the positive electrode where it is oxidised</span></label> </p><p><label class="radio"> <input type="radio"> It <span>goes to the negative electrode where it is oxidised</span></label> </p><p><label class="radio"> <input type="radio"> It <span>goes to the positive electrode where it is reduced</span></label> </p><p><label class="radio"> <input class="c" type="radio"> It <span>goes to the negative electrode where it is reduced </span></label> </p></div><div class="q-explanation"><p>In an electrolytic cell the external potential difference pushes electrons out to one electrode making it negative and pulls electrons in from another electrode, making it positive. The ions in the electrolyte are then attracted to the oppositely charged electrode where they gain electrons (if positive ions) or lose electrons (if negative ions) to form atoms.</p><p>The sodium ion, being Na<sup>+</sup> will go to the negative electrode and gain an electron to become Na, which is reduction; RIG.</p><p><strong>It goes to the negative electrode where it is reduced</strong><strong> </strong>is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>Where does reduction occur in a voltaic cell?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>At the anode, which is the negative electrode</span></label> </p><p><label class="radio"> <input type="radio"> <span>At the anode, which is the positive electrode</span></label> </p><p><label class="radio"> <input class="c" type="radio"> At the cathode, which is the positive electrode</label></p><p><label class="radio"> <input type="radio"> <span>At the cathode, which is the negative electrode</span></label> </p></div><div class="q-explanation"><p>Red Cat; An Ox. <strong>Red</strong>uction at the <strong>Cat</strong>hode; <strong>Ox</strong>idation at the <strong>An</strong>ode. This is true in both types of cell.</p><p><img alt="" src="../../images/redox-processes/redcatanox.jpg" style="width: 240px; height: 117px;"></p><p>In a voltaic cell the redox reaction generates the charges on the electrodes and causes the current to flow. The most active metal loses electrons (oxidation; atoms to ions) causing the electrode to be negative, and the least active metal gains electrons (reduction; ions to atoms) causing the electrode to be positive.</p><p>So in a voltaic cell metal ions will gain electrons (reduction) at the positive electrode. Cations go to the cathode and (in a voltaic cell) they make it positive.</p><p><strong>At the cathode, which is the positive electrode</strong> is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which chemical species is the oxidising agent in this reaction?</p><p>2KMnO<sub>4</sub> + 3Na<sub>2</sub>SO<sub>3</sub> + H<sub>2</sub>O → 2MnO<sub>2</sub> + 3Na<sub>2</sub>SO<sub>4</sub> + 2KOH</p><dl></dl></div><div class="q-answer"><p><label class="radio"> <input type="radio"> MnO<sub>2</sub><span></span></label></p><p><label class="radio"> <input type="radio"> Na<sub>2</sub>SO<sub>3</sub></label></p><p><label class="radio"> <input type="radio"> H<sub>2</sub>O</label></p><p><label class="radio"> <input class="c" type="radio"> KMnO<sub>4</sub></label></p></div><div class="q-explanation"><p>An oxidising agent oxidises another element and therefore is itself reduced (OILRIG: if it takes electrons away from another element, it has gained electrons).</p><p>Oxidation state changes are needed:</p><p>Manganese goes from +7 to +4 (KMnO<sub>4</sub> to MnO<sub>2</sub>).</p><p>Sulfur goes from +4 to +6 (Na<sub>2</sub>SO<sub>3</sub> to Na<sub>2</sub>SO<sub>4</sub>).</p><p>Thus sulfur is oxidised (OIL) and manganese is reduced (RIG). The KMnO<sub>4 </sub>is the oxidising agent.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>Consider the following cell:</p><p><img alt="" src="../../images/redox-processes/zncucell.png" style="width: 320px; height: 221px;"></p><p>The salt bridge is made from filter paper soaked in saturated potassium nitrate (KNO<sub>3</sub>).</p><p>What happens to the ions in the salt bridge when the current flows?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>K<sup>+</sup> ions and NO<sub>3</sub><sup>−</sup> ions flow to the zinc half cell</span></label></p><p><label class="radio"> <input class="c" type="radio"> K<sup>+</sup> ions flow to copper half cell and NO<sub>3</sub><sup>−</sup> ions flow to the zinc half cell</label></p><p><label class="radio"> <input type="radio"> <span>K<sup>+</sup> ions flow to zinc half cell and NO<sub>3</sub><sup>−</sup> ions flow to the copper half cell</span></label></p><p><label class="radio"> <input type="radio"> <span>K<sup>+</sup> ions and NO<sub>3</sub><sup>−</sup> ions flow to the copper half cell</span></label></p></div><div class="q-explanation"><p>In a voltaic cell the redox reaction generates the charges on the electrodes and causes the current to flow. The most active metal loses electrons (zinc) causing the electrode to be negative, and the least active metal (copper) gains electrons causing the electrode to be positive.</p><p>As electrons are flowing from zinc to copper, the charge will be equalised by negative ions flowing across the salt bridge to the zinc half-cell and positive ions flowing across to the copper half cell.</p><p>Thus <strong>K<sup>+</sup> ions flow to copper half cell and NO<sub>3</sub><sup>−</sup> ions flow to the zinc half </strong><strong>cell </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>The following reaction occurs in a voltaic cell:</p><p>Mg(s) + Zn<sup>2+</sup>(aq) → Mg<sup>2+</sup>(aq) + Zn(s)</p><p>Which reaction takes place at each electrode?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> N<span>egative electrode: Zn(s) → Zn<sup>2+</sup>(aq) + 2e<sup>−</sup> Positive electrode: Mg<sup>2+</sup>(aq) + 2e− → Mg(s)</span></label></p><p><label class="radio"> <input type="radio"> <span>Negative electrode: Mg<sup>2+</sup>(aq) + 2e− → Mg(s) Positive electrode: Zn(s) → Zn<sup>2+</sup>(aq) + 2e<sup>−</sup></span></label></p><p><label class="radio"> <input type="radio"> <span>Negative electrode: Zn<sup>2+</sup>(aq) + 2e<sup>−</sup> → Zn(s) Positive electrode: Mg(s) → Mg<sup>2+</sup>(aq) + 2e<sup>−</sup></span></label></p><p><label class="radio"> <input class="c" type="radio"> Negative electrode: Mg(s) → Mg<sup>2+</sup>(aq) + 2e<sup>−</sup> Positive electrode: Zn<sup>2+</sup>(aq) + 2e<sup>−</sup> → Zn(s)</label></p></div><div class="q-explanation"><p>In a voltaic cell the redox reaction generates the charges on the electrodes and causes the current to flow. The most active metal loses electrons (magnesium) causing the electrode to be negative, and the least active metal (zinc) gains electrons causing the electrode to be positive.</p><p><strong>Negative electrode: Mg(s) → Mg<sup>2+</sup>(aq) + 2e<sup>−</sup> Positive electrode: Zn<sup>2+</sup>(aq) + 2e<sup>−</sup> → Zn(s)</strong><strong> </strong>is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of these reactions is not a redox reaction?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>4Cl<sub>2</sub> + CH<sub>4</sub> → CCl<sub>4</sub> + 4HCl</span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>OH + [O] → CH<sub>3</sub>CHO + H<sub>2</sub>O</span></label> </p><p><label class="radio"> <input class="c" type="radio"> Pb(NO<sub>3</sub>)<sub>2</sub><span> + 2KI → PbI<sub>2</sub> + 2KNO<sub>3</sub></span></label></p><p><label class="radio"> <input type="radio"> <span>CO<sub>2</sub> + C → 2CO</span></label> </p></div><div class="q-explanation"><p>The reaction between lead nitrate and potassium iodide is not a redox reaction as it involves some ion exchange without a change in oxidation state.</p><p>Oxidation states change in the three other reactions:</p><p>4Cl<sub>2</sub> + CH<sub>4</sub> → CCl<sub>4</sub> + 4HCl (chlorine 0 to −1 and carbon −4 to +4)</p><p>CH<sub>3</sub>CH<sub>2</sub>OH + [O] → CH<sub>3</sub>CHO + H<sub>2</sub>O (carbon −2 to −1 and oxygen 0 to −2)</p><p>CO<sub>2</sub> + C → 2CO (carbon from both +4 and 0 to +2)</p><p>Therefore <strong>Pb(NO<sub>3</sub>)<sub>2</sub> + 2KI → PbI<sub>2</sub> + 2KNO<sub>3</sub></strong> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question">In a voltaic cell consisting of zinc and silver half-cells, which is the strongest reducing agent?</div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> Zn</label></p><p><label class="radio"> <input type="radio"> Ag</label></p><p><label class="radio"> <input type="radio"> Ag<sup>+</sup></label></p><p><label class="radio"> <input type="radio"> Zn<sup>2+</sup></label></p></div><div class="q-explanation"><p>This question refers to voltaic cells, but the only information needed is that zinc is more active than silver.</p><p>The strongest reducing agent (reduction is gain of electrons) is the species most likely to lose electrons and give them to something else, thereby reducing it. Zinc atoms have the greatest tendency to lose electrons as they are the most active (silver is less likely to lose electrons, and zinc and silver ions, having already lost electrons are unlikely to lose any more).</p><p>Thus the correct answer is <b>Zn</b></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>Balance the following redox equation to give the lowest whole number ratio:</p><p>MnO<sub>4</sub><sup>−</sup> + SO<sub>3</sub><sup>2</sup><sup>−</sup> + H<sub>2</sub>O → MnO<sub>2</sub> + SO<sub>4</sub><sup>2</sup><sup>−</sup> + OH<sup>−</sup></p><p>How many moles of SO<sub>4</sub><sup>2−</sup> ions are produced?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2</span></label> </p><p><label class="radio"> <input type="radio"> 1</label></p><p><label class="radio"> <input class="c" type="radio"> 3</label></p><p><label class="radio"> <input type="radio"> <span>4</span></label> </p></div><div class="q-explanation"><p>Always balance the electrons first by calculating changes in oxidation states:</p><p>Oxygen remains as −2 throughout.</p><p>Hydrogen remains as +1 throughout.</p><p>Sulfur (S) changes from +4 to +6 (loss of 2 electrons).</p><p>Manganese (Mn) changes from +7 to +4 (gain of 3 electrons).</p><p>To balance electrons lost and gained (6 electrons), <b>three </b>sulfur atoms and<b> two </b>manganese atoms are needed.</p><p><u><strong>2</strong></u>MnO<sub>4</sub><sup>−</sup> + <u><strong>3</strong></u>SO<sub>3</sub><sup>2</sup><sup>−</sup> + H<sub>2</sub>O → <u><strong>2</strong></u>MnO<sub>2</sub> + <u><strong>3</strong></u>SO<sub>4</sub><sup>2</sup><sup>−</sup> + OH<sup>−</sup></p><p>Now atoms of other elements can be balanced:</p><p>2MnO<sub>4</sub><sup>−</sup> + 3SO<sub>3</sub><sup>2</sup><sup>−</sup> + H<sub>2</sub>O → 2MnO<sub>2</sub> + 3SO<sub>4</sub><sup>2</sup><sup>−</sup> + <u><strong>2</strong></u>OH<sup>−</sup></p><p>So the correct answer is <strong>3</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the order of decreasing reactivity of the metals (most reactive first)?</p><p>Zn + CrSO<sub>4</sub> → ZnSO<sub>4</sub> + Cr</p><p>Ba + Zn(NO<sub>3</sub>)<sub>2</sub> → Ba(NO<sub>3</sub>)<sub>2</sub> + Zn</p><p>CrSO<sub>4</sub> + Sb → no reaction</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> Sb > Cr > Zn > Ba</label></p><p><label class="radio"><input class="c" type="radio"> Ba > Zn > Cr > Sb</label></p><p><label class="radio"><input type="radio"> Zn > Sb > Ba > Cr</label></p><p><label class="radio"><input type="radio"> Cr > Ba > Sb > Zn</label></p></div><div class="q-explanation"><p>These reactions are replacement/displacement reactions.</p><p>Zn displaces Cr so in reactivity Zn > Cr</p><p>Ba displaces Zn so in reactivity Ba > Zn</p><p>Sb cannot displace Cr so in reactivity Cr > Sb</p><p><strong>Ba > Zn > Cr > Sb</strong> is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">11</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which statements are correct for a<strong> voltaic</strong> cell?</p><p><strong>1: </strong>A spontaneous redox reaction produces electrical energy.</p><p><strong>2: </strong>Oxidation occurs at the anode (negative electrode).</p><p><strong>3: </strong>The electrons flow from the cathode (positive electrode) to the anode (negative electrode).</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1 and 2 only</span></label> </p></div><div class="q-explanation"><p>Voltaic cells use <strong>spontaneous redox reactions</strong> to generate electrical energy. (Electrolytic cells use electrical energy to drive non-spontaneous redox reactions.)</p><p>Thus <strong>statement 1</strong> <b>is correct.</b></p><p>In both types of electrolytic cell <strong>oxidation occurs at the anode </strong>and reduction occurs at the cathode. In a voltaic cell the anode is negative, and the cathode is positive. (In an electrolytic cell the anode is positive, and the cathode is negative.)</p><p>Thus <strong>statement 2 is correct.</strong></p><p>In a voltaic cell the redox reaction causes the electrodes to become charged and this generates electron flow from negative electrode (anode) to positive electrode (cathode).</p><p>Thus <strong>statement 3 is incorrect.</strong></p><p>Therefore <strong>1 and 2 only </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">12</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which statement is correct for the redox equation?</p><p>Sn<sup>2</sup><sup>+</sup>(aq) + Ni (s) → Sn (s) + Ni<sup>2+</sup> (aq)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Sn<sup>2+</sup> is oxidised and Ni is reduced</span></label> </p><p><label class="radio"> <input type="radio"> Sn<sup>2+</sup><span> is the reducing agent and Ni is oxidised</span></label></p><p><label class="radio"> <input class="c" type="radio"> Sn<sup>2+</sup><span> is reduced and Ni is the reducing agent</span></label> </p><p><label class="radio"> <input type="radio"> <span>Sn<sup>2+</sup> is the reducing agent and Ni is the oxidising agent</span></label> </p></div><div class="q-explanation"><p>In the reaction, tin (Sn) gains electrons. This means that it has been reduced (reduction is gain of electrons). Therefore, the nickel is the reducing agent, since it reduces the tin.</p><p>Each nickel atom loses two electrons to form a nickel ion. This means that it has been oxidised (oxidation is loss of electrons). Therefore, the tin (Sn) is the oxidising agent, since it oxidises the nickel.</p><p>Therefore <strong>Sn<sup>2+</sup> is reduced and Ni is the reducing agent </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </div> <div class="page-container panel-self-assessment" data-id="2724"> <div class="panel-heading">MY PROGRESS</div> <div class="panel-body understanding-rate"> <div class="msg"></div> <label class="label-lg">Self-assessment</label><p>How much of <strong>Redox core (SL and HL) paper 1 questions</strong> have you understood?</p><div class="slider-container text-center"><div id="self-assessment-slider" class="sib-slider self-assessment " data-value="1" data-percentage=""></div></div> <label class="label-lg">My notes</label> <textarea name="page-notes" class="form-control" rows="3" placeholder="Write your notes here..."></textarea> </div> <div class="panel-footer text-xs-center"> <span id="last-edited" class="mb-xs-3"> </span> <div class="actions mt-xs-3"> <button id="save-my-progress" type="button" class="btn btn-sm btn-primary text-center btn-xs-block"> <i class="fa fa-fw fa-floppy-o"></i> Save </button> </div> </div></div> <div id="modal-feedback" class="modal fade" tabindex="-1" role="dialog"> <div class="modal-dialog" role="document"> <div class="modal-content"> <div class="modal-header"> <h4 class="modal-title">Feedback</h4> <button type="button" class="close hidden-xs hidden-sm" data-dismiss="modal" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div> <div class="modal-body"> <div class="errors"></div> <p><strong>Which of the following best describes your feedback?</strong></p> <form method="post" style="overflow: hidden"> <div class="form-group"> <div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Recommendation"> Recommend</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Problem"> Report a problem</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Improvement"> Suggest an improvement</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Other"> Other</label></div> </div> <hr> <div class="row"> <div class="col-md-6"> <div class="form-group"> <label for="feedback-name">Name</label> <input type="text" class="form-control" name="feedback-name" placeholder="Name" value=" "> </div> </div> <div class="col-md-6"> <div class="form-group"> <label for="feedback-email">Email address</label> <input type="email" class="form-control" name="feedback-email" placeholder="Email" value="@airmail.cc"> </div> </div> </div> <div class="form-group"> <label for="feedback-comments">Comments</label> <textarea class="form-control" name="feedback-comments" style="resize: vertical;"></textarea> </div> <input type="hidden" name="feedback-ticket" value="082b9c9c4ae3624d"> <input type="hidden" name="feedback-url" value="https://studyib.net/chemistry/page/2724/redox-core-sl-and-hl-paper-1-questions"> <input type="hidden" name="feedback-subject" value="7"> <input type="hidden" name="feedback-subject-name" value="Chemistry"> <div class="pull-left"> </div> </form> </div> <div class="modal-footer"> <button type="button" class="btn btn-primary btn-xs-block feedback-submit mb-xs-3 pull-right"> <i class="fa fa-send"></i> Send </button> <button type="button" class="btn btn-default btn-xs-block m-xs-0 pull-left" data-dismiss="modal"> Close </button> </div> </div> </div></div> <style type="text/css" media="screen">/* Important part */
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