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only) paper 1 questions</a></label></li></ul></div> <button id="show-periodic-table" class="btn 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"> Equilibrium core (SL and HL) paper 1 questions <a href="#" class="mark-page-favorite pull-right" data-pid="2155" 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">Equilibrium 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 7.1</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"> <div class="tib-quiz" data-stats="7-587-2155"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Given the equilibrium: CO<sub>(g)</sub> + H<sub>2</sub>O<sub>(g)</sub> ⇌ CO<sub>2(g)</sub> + H<sub>2</sub><sub>(g)</sub> ΔH<sup><s>o</s></sup> = −41.2 kJ mol<sup>−1</sup></p><p>What will happen to the position of equilibrium and to the value of K<sub>c</sub> when temperature increases?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>The position of equilibrium will shift left, and the value of K<sub>c</sub> will remain the same.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The position of equilibrium will shift right, and the value of K<sub>c</sub> will increase.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The position of equilibrium will shift right, and the value of K<sub>c</sub> will remain the same.</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>The position of equilibrium will shift left and the value of K<sub>c</sub> will decrease.</span></label> </p></div><div class="q-explanation"><p>Remember that K<sub>c</sub> is constant at constant temperature, but temperature is changing here so the value of K<sub>c</sub> will change.</p><p><span class="math-tex">\(K_c = {[CO_2][H_2]\over{[CO][H_2O]}}\)</span></p><p>Increasing temperature will shift the equilibrium in the endothermic direction (to oppose the change in temperature in accordance with Le Chatelier's principle). The forward reaction is exothermic (negative ΔH) so the equilibrium will therefore shift left.</p><p>In the K<sub>c</sub> expression, this will increase the quantity of reactants, increasing the denominator and therefore reducing the value of Kc.</p><p><strong>The position of equilibrium will shift left and the value of K<sub>c</sub> will decrease</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>Given the equilibrium: N<sub>2(g) </sub>+3H<sub>2(g) </sub>⇌ 2NH<sub>3(g)</sub> ΔH = −92 kJ</p><p>What will happen to the value of K<sub>c</sub> if both the pressure and the temperature are decreased?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> It will remain the same<span></span></label> </p><p><label class="radio"> <input type="radio"> <span>Cannot be predicted as the effects are opposite</span></label> </p><p><label class="radio"> <input type="radio"> <span>Decreases</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Increases</span></label> </p></div><div class="q-explanation"><p>Remember that K<sub>c</sub> is constant at constant temperature, so the change in pressure will not affect K<sub>c</sub> (although it will shift the position of equilibrium), however the change in temperature will affect K<sub>c</sub>.</p><p><span class="math-tex">\(K_c = {{[NH_3]^2} \over [N_2][H_2]^3}\)</span></p><p>Decreasing temperature will favour the <strong>exothermic </strong>direction, which is the forward direction in this case (a negative ΔH indicates an exothermic forward reaction). The equilibrium will shift to the right to give a greater proportion of products (NH<sub>3</sub>). A greater proportion of products will make the top of the fraction greater and will therefore increase the value of K<sub>c</sub>.</p><p><strong>Increases </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>What is the equilibrium constant expression for the equilibrium given below?</p><p>2NO + Cl<sub>2</sub> ⇌ 2NOCl</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span> <span class="math-tex">\(K_c = {{[NO]^2[Cl_2]} \over [NOCl]^2}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span> <span class="math-tex">\(K_c = {{2[NOCl]} \over 2[NO][Cl_2]}\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(K_c = {{[NOCl]^2} \over [NO]^2[Cl_2]}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(K_c = {{[NOCl]} \over [NO]^2[Cl_2]}\)</span></span></label> </p></div><div class="q-explanation"><p>The equilibrium constant is defined as being the product of the concentrations of the products (each to the power of the number of moles) over the product of the concentrations of the reactants (each to the power of the number of moles), e.g. for a reaction:</p><p>aA + bB ⇌ cC + dD</p><p><img alt="" height="45" src="../../images/equilibrium/generic-kc.png" width="576"></p><p>Therefore the correct answer is:</p><p><span class="math-tex">\(K_c = {{[NOCl]^2} \over [NO]^2[Cl_2]}\)</span></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>Consider the equilibrium: 2NO<sub>2</sub><sub>(g)</sub> ⇌ N<sub>2</sub>O<sub>4(g) </sub></p><p>The values of K<sub>c</sub> at different temperatures are:</p><table border="0" cellpadding="0" cellspacing="0" style="width:50%;"><tbody><tr><td style="text-align: center;">Temperature (K)</td><td style="text-align: center;">K<sub>c</sub></td></tr><tr><td style="text-align: center;">273</td><td style="text-align: center;">72.9</td></tr><tr><td style="text-align: center;">290</td><td style="text-align: center;">16.6</td></tr><tr><td style="text-align: center;">305</td><td style="text-align: center;">5.2</td></tr><tr><td style="text-align: center;">325</td><td style="text-align: center;">1.3</td></tr></tbody></table><p>Which statement is correct as temperature increases?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>The forward reaction is favoured.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The rate of reaction in both directions increases equally.</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>The reverse reaction is favoured.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The rate of forward reaction is greater that the rate of the reverse reaction.</span></label> </p></div><div class="q-explanation"><p>As temperature is increasing here so the value of K<sub>c</sub> decreases.</p><p><span class="math-tex">\(K_c = {[N_2O_4]\over{[NO_2]^2}}\)</span></p><p>That means that the bottom (denominator) of the fraction must be increasing, so the reaction must be shifting to the left; the reverse reaction is favoured as temperature increases.</p><p><strong>The reverse reaction is favoured </strong>is therefore the correct answer.</p><p>Note that increasing the temperature will increase the rate of reaction in both directions, but <strong>not </strong>equally. As the reverse reation is favoured here, an increase in temperature will increase the rate of the reverse reaction to a greater extent than the rate of the forward reaction (that is how the equilibrium shifts).</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>Given that the equilibrium constant for 2SO<sub>2(g)</sub> + O<sub>2(g)</sub> ⇌ 2SO<sub>3(g)</sub> is K.</p><p>What is the equilibrium constant for this equation?</p><p>4SO<sub>2(g)</sub> + 2O<sub>2(g)</sub> ⇌ 4SO<sub>3(g)</sub></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>K<sup>2</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>K</span></label> </p><p><label class="radio"> <input type="radio"> <span>2K<sup>2</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>2K</span></label> </p></div><div class="q-explanation"><p>K<sub>c</sub> is for 2SO<sub>2(g)</sub> + O<sub>2(g)</sub> ⇌ 2SO<sub>3(g) </sub>is:</p><p><span class="math-tex">\(K_c = {{[SO_3]^2} \over [SO_2]^2[O_2]}\)</span></p><p>K<sub>c</sub> is for 4SO<sub>2(g)</sub> + 2O<sub>2(g)</sub> ⇌ 4SO<sub>3(g) </sub>is:</p><p><span class="math-tex">\(K_c = {{[SO_3]^4} \over [SO_2]^4[O_2]^2}\)</span></p><p>It is important to know (law of indices) that when we multiply values with indices (powers) then we add the indices. Looking at the two K<sub>c </sub>expressions we can see that in the second expression, all of the terms are a square of the terms in the first expression e.g. [SO<sub>3</sub>]<sup>2</sup> squared is [SO<sub>3</sub>]<sup>2</sup> × [SO<sub>3</sub>]<sup>2</sup> = [SO<sub>3</sub>]<sup>4</sup>.</p><p>As all the terms are squared, the equilibrium constant for the second equation must be the square of the equilibrium constant for the first equation.</p><p><strong>K<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">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which factor does not affect the position of equilibrium in this reaction?</p><p>2SO<sub>2(g)</sub> + O<sub>2(g)</sub> ⇌ 2SO<sub>3(g)</sub> ΔH=−196kJ</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Change in temperature</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Addition of a catalyst</span></label> </p><p><label class="radio"> <input type="radio"> <span>Change in pressure</span></label> </p><p><label class="radio"> <input type="radio"> <span>Change in volume of the container</span></label> </p></div><div class="q-explanation"><p>Addition of a catalyst will not affect the position of an equilbrium. A catalyst will decrease the time it takes for a reaction to reach equilibrium, because it will speed up the rate of the foward and backward reactions equally, but will not alter the position of the equilibrium when it gets there.</p><p>All the other factors will affect the position of equilibrium in this case; temperature always changes the position of equilibrium; change in pressure and change in volume will both affect the position of equilibrium as there are different numbers of moles of gas on each side of the equation.</p><p><strong>Addition of a catalyst </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>What will happen if the pressure is decreased in the following reaction mixture at equilibrium?</p><p>CO<sub>2(g)</sub> + H<sub>2</sub>O<sub>(l) </sub>⇌ H<sup>+</sup><sub>(aq)</sub> + HCO<sub>3</sub><sup>−</sup><sub>(aq)</sub></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> The reaction will shift to the right and the pH will increase</label></p><p><label class="radio"><input type="radio"> The reaction will shift to the right and the pH will decrease</label></p><p><label class="radio"><input type="radio"> The reaction will shift to the left and the pH will decrease</label></p><p><label class="radio"><input class="c" type="radio"> The reaction will shift to the left and the pH will increase</label></p></div><div class="q-explanation"><p>If the pressure is decreased the equation will favour the side with greatest moles of gas (according to Le Chatelier's principle), which means the equilibrium will shift to the left (one mole of gas on the left and none on the right).</p><p>Shifting the equilibrium to the left will reduce the quantity of H<sup>+</sup> ions in the aqueous solution, which will increase the pH (make it less acidic) as pH=−log<sub>10</sub>[H<sup>+</sup>].</p><p><strong>The reaction will shift to the left and the pH will increase</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">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>Consider the equilibrium:</p><p>2NO<sub>(g)</sub> + Cl<sub>2(g)</sub> ⇌ 2NOCl<sub>(g)</sub> ΔH=−76kJ</p><p>Which changes shift the position of equilibrium to the right?</p><p><strong>1: </strong>Increasing the temperature</p><p><strong>2: </strong>Increasing the pressure</p><p><strong>3: </strong>Adding a catalyst</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>2 only</span></label> </p></div><div class="q-explanation"><p>Adding a catalyst will never change the position of equilibrium (only increase the rate of attainment of the equilibrium position).</p><p>Increasing the temperature will favour the endothermic direction (according to Le Chatelier's principle) which is the left hand side; the forward reaction is exothermic as indicated by the negative ΔH.</p><p>Increasing the pressure will favour the side with fewest moles of gas (according to Le Chatelier's principle) which is the right hand side, with 2 moles compared to 3 moles on the left.</p><p><strong>2 only</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">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>Consider the equilibrium: H<sub>2</sub><sub>(g) </sub>+ I<sub>2(g)</sub> ⇌ 2HI<sub>(g)</sub></p><p>The reaction is exothermic in the forward direction.</p><p>What happens when the temperature is increased?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>The rate of reaction in both directions decreases and the position of equilibrium shifts left..</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>The rate of reaction in both directions increases and the position of equilibrium shifts left.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The rate of reaction in both directions increases and the position of equilibrium shifts right.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The rate of reaction in both directions increases equally and the position of equilibrium does not change.</span></label> </p></div><div class="q-explanation">Increasing the temperature will increase the rate of reaction in both directions, but <strong>not </strong>equally. An increase in temperature will favour the endothermic direction according to Le Chatelier's principle; as the reverse reation is favoured here, an increase in temperature will increase the rate of the reverse reaction to a greater extent than the rate of the forward reaction (that is how the equilibrium shifts). The correct answer is therefore <strong>The rate of reaction in both directions increases and the position of equilibrium shifts left.</strong></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>Which change will favour the reverse reaction in this equilibrium?</p><p>2SO<sub>4</sub><sup>2−</sup><sub>(aq)</sub> + 4H<sup>+</sup><sub>(aq)</sub> + 2e<sup>−</sup> ⇌ H<sub>2</sub>SO<sub>3(aq)</sub> + H<sub>2</sub>O<sub>(l)</sub></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Addition of SO<sub>4</sub><sup>2−</sup><sub>(aq) </sub>ions</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Addition of OH<sup>−</sup><sub>(aq)</sub> ions</span></label> </p><p><label class="radio"> <input type="radio"> <span>Addition of H<sup>+</sup><sub>(aq)</sub> ions</span></label> </p><p><label class="radio"> <input type="radio"> <span>Adding a catalyst</span></label> </p></div><div class="q-explanation"><p>Addition of a catalyst will not affect the position of an equilbrium. A catalyst will decrease the time it takes for a reaction to reach equilibrium, because it will speed up the rate of the foward and backward reactions equally, but will not alter the position of the equilibrium.</p><p>Addition of H<sup>+</sup> (hydrogen) ions or SO<sub>4</sub><sup>2−</sup> (sulfate) ions will add ions to the left hand side and according to Le Chatelier's principle will therefore shift the equilibrium to the right; favouring the forward reaction.</p><p>Addition of OH<sup>−</sup><sub> </sub>ions will cause the H<sup>+</sup> ions to react: H<sup>+</sup> + OH<sup>−</sup> → H<sub>2</sub>O, so will remove H<sup>+</sup> ions from the left hand side and will therefore shift the equilibrium to the left; favouring the reverse reaction.</p><p><strong>Addition of OH<sup>−</sup><sub>(aq)</sub> ions</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">11</div><div class="exercise shadow-bottom"><div class="q-question"><p>Consider the equilibrium: 2NO<sub>2</sub><sub>(g)</sub> ⇌ N<sub>2</sub>O<sub>4(g)</sub> ΔH = −58 kJ</p><p>Which combination of temperature and pressure will give the greatest yield of dinitrogen tetroxide?</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> Low temperature and low pressure</label></p><p><label class="radio"><input class="c" type="radio"> Low temperature and high pressure</label></p><p><label class="radio"><input type="radio"> High temperature and low pressure</label></p><p><label class="radio"><input type="radio"> High temperature and high pressure</label></p></div><div class="q-explanation"><p>The highest yield will be obtained by shifting the equilibrium to the right hand side (products).</p><p>The reaction is exothermic (negative ΔH), so a low temperature will favour the exothermic/forward direction and increase the yield of product.</p><p>There are fewer moles of gas on the right hand side of the equation, so a high pressure will favour the forward direction and increase the yield of product.</p><p><strong>Low temperature and high pressure</strong> is therefore the correct answer.</p><p>(Note that these conditions will generate the highest yield, but may not be used in industry as a low temperature is likely to result in a slow rate of reaction.)</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>Given the equilibrium for the Haber process:</p><p>N<sub>2(g) </sub>+3H<sub>2(g) </sub>⇌ 2NH<sub>3(g)</sub></p><p>What will happen to the position of equilibrium and to the value of K<sub>c</sub> if the total volume of the reaction mixture is decreased at constant temperature?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>The position of equilibrium will shift right, and the value of K<sub>c</sub> will increase.</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>The position of equilibrium will shift right, and the value of K<sub>c</sub> will remain the same.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The position of equilibrium will shift left, value of K<sub>c</sub> will increase.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The position of equilibrium and the value of K<sub>c</sub> will remain the same.</span></label> </p></div><div class="q-explanation"><p>Remember that K<sub>c</sub> is <strong>constant at constant temperature</strong>.</p><p>The value of K<sub>c</sub> will therefore remain the same.</p><p>The decrease in volume causes an increase in pressure, but the increase in pressure will favour the side of the equilibrium with fewest moles of gas (2 moles of the right) so the equilibrium will shift right.</p><p>Therefore <strong>The position of equilibrium will shift right, and the value of K<sub>c</sub> will remain the same </strong>is the correct answer.</p><p>In explanation: decreasing the volume (and thus increasing total pressure) causes the concentration values to each increase. There are more concentration terms on the bottom than the top of the fraction, so the bottom becomes relatively bigger; thus decreasing the volume throws the reaction out of equilibrium and Q becomes smaller than K<sub>c</sub>. Thus the equilibrium shifts to increase the top of the fraction (more ammonia, NH<sub>3</sub>); that is a shift to the right, until Q once more equals K<sub>c</sub> (K<sub>c</sub> remains the same).</p><p><span class="math-tex">\(Q = {{[NH_3]^2} \over [N_2][H_2]^3} = K_c\)</span></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="2155"> <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>Equilibrium 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/2155/equilibrium-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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