L3f",null,{"backHref":"/gcse/chemistry/ocr/questions","basePath":"/gcse/chemistry/ocr/monitoring-chemical-reactions/questions","formulaUrl":null,"initialIndex":75,"initialQuestionWindow":[{"id":"a948dfd5-81f6-4ee7-9285-1a1f3caee0ee","revisionId":"82a42b3f-47dc-4d68-9811-b8916c952fb3","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"f1ed3c6d-e9b5-4dfc-ad20-04984491b8c8","label":null,"content":"Propan-1-ol, $\\text{C}_3\\text{H}_7\\text{OH}$, can be dehydrated to make propene, $\\text{C}_3\\text{H}_6$, according to the equation:\n\n$$\\text{C}_3\\text{H}_7\\text{OH} \\rightarrow \\text{C}_3\\text{H}_6 + \\text{H}_2\\text{O}$$\n\nCalculate the atom economy for the production of propene from propan-1-ol.\n\nRelative formula masses ($M_r$):\n- $\\text{C}_3\\text{H}_7\\text{OH} = 60.0$\n- $\\text{C}_3\\text{H}_6 = 42.0$","markscheme":"- Recalls or applies the formula: $\\text{Atom economy} = \\frac{\\text{Total } M_r \\text{ of desired product}}{\\text{Total } M_r \\text{ of all reactants}} \\times 100\\%_t$ M1\n- Substitutes the given values into the formula to write $\\frac{42.0}{60.0} \\times 100$ M1\n- Obtains the final answer $70.0\\%$ A1","marks":3,"order":0,"markingRubric":null,"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"b942ba66-6a2e-4dde-b706-f1a1acbd5b8a","revisionId":"2e64d6f2-80f5-4a2c-b77c-c0301a2b1eff","questionNumber":null,"questionType":"MC","specification":"Avogadro's constant has a value of $6.02 \\times 10^{23}$.\n\nHow many **hydrogen atoms** are in $0.75$ moles of hydrogen molecules ($\\text{H}_2$)?","difficulty":6,"difficultyBand":"medium","totalMarks":1,"sourceTitle":null,"calculatorUse":null,"parts":[],"options":[{"id":"14c02f04-337d-49b0-a926-4a52c73b8060","content":"$$4.515 \\times 10^{23}$","correct":false,"markscheme":"This value is obtained by multiplying $0.75$ moles directly by Avogadro's constant, forgetting that each $\\text{H}_2$ molecule contains two hydrogen atoms.","order":0},{"id":"e6121338-1a06-430f-ba80-00fb421fdcd7","content":"$$9.030 \\times 10^{23}$","correct":true,"markscheme":"Correct. Since each $\\text{H}_2$ molecule contains two hydrogen atoms, $0.75$ moles of molecules contain $1.50$ moles of atoms. Multiplying by Avogadro's constant gives $9.030 \\times 10^{23}$ atoms.","order":1},{"id":"22a6d259-535e-4a87-8d1c-2577fb5674c7","content":"$$1.806 \\times 10^{24}$","correct":false,"markscheme":"This value is obtained if the number of moles is incorrectly multiplied by $4$ instead of $2$ before multiplying by Avogadro's constant.","order":2},{"id":"d2fdfcae-5b02-493b-966d-2dfa401e9e51","content":"$$6.020 \\times 10^{23}$","correct":false,"markscheme":"This is Avogadro's constant itself, which represents the number of particles in exactly $1$ mole of substance.","order":3}],"explanation":null,"firstAttemptFullMarks":null},{"id":"c271ee41-ed16-4a01-9e81-51ca028f9fff","revisionId":"5e6b8c07-2765-4017-9c39-bd0f90a55811","questionNumber":null,"questionType":"LA","specification":"Using your knowledge of the reactivity series and chemical reactions:","difficulty":3,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"75524522-396f-4025-867d-ba3c7f447745","label":"a","content":"Lead can be extracted from lead(II) oxide, \$\\text{PbO}\$, by heating it with carbon. Explain why carbon can be used to extract lead from its oxide, referring to their relative reactivity.","markscheme":"- Carbon is more reactive than lead (or carbon is higher in the reactivity series than lead) M1\n- Carbon reduces lead oxide (or carbon displaces lead / removes oxygen from lead oxide) A1","marks":2,"order":0,"markingRubric":null,"explanation":null,"options":[]},{"id":"9a2d6008-93a8-4a2c-a2f9-8346357a2754","label":"b","content":"Magnesium reacts with dilute hydrochloric acid, \$\\text{HCl}\$, to produce magnesium ch

Monitoring chemical reactions

Question 48

Monitoring chemical reactions Questions

Monitoring chemical reactions

Question 48

Monitoring chemical reactions Questions