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MgO."},{"text":"Ionic formulae use the actual number of ions present.","correct":false,"explanation":"Writing the actual number of ions present would depend on the sample size and would not define the compound. Mg²⁺ and O²⁻ combine in the simplest 1:1 ratio."},{"text":"Ionic formulae use an equal number of each ion.","correct":false,"explanation":"Equal numbers are used here because the ion charges have equal magnitudes, not because ionic formulae always contain equal numbers. Formulae use the simplest ratio that gives zero charge."},{"text":"Ionic formulae use the relative masses of the ions.","correct":false,"explanation":"The ratio is determined by balancing ionic charges, not by comparing relative atomic masses. Mg²⁺ and O²⁻ require a simplest whole-number ratio of 1:1."}]},{"id":"d7b8b757-0f4a-4d07-a76d-06c5a3a6b738","cardType":"cloze","points":null,"front":"When balancing an equation, change [ ], not [ ].","back":"When balancing an equation, change coefficients, not subscripts.","mcqOptions":[{"text":"When balancing an equation, change coefficients, not subscripts.","correct":true,"explanation":"Coefficients alter the number of particles without changing their identities, allowing each element’s atom count to balance. Subscripts belong to the chemical formula and must remain fixed."},{"text":"When balancing an equation, change subscripts, not coefficients.","correct":false,"explanation":"This reverses the correct rule. Changing subscripts creates different substances, whereas changing coefficients adjusts particle numbers while preserving every reactant’s and product’s identity."},{"text":"When balancing an equation, change charges, not subscripts.","correct":false,"explanation":"Charges cannot generally be changed because they identify particular ions. Equations are balanced by changing coefficients while leaving subscripts, charges and chemical identities unchanged."},{"text":"When balancing an equation, change state symbols, not subscripts.","correct":false,"explanation":"State symbols describe physical states and do not control atom numbers. Balancing requires changes to coefficients, while subscripts in each substance’s formula stay unchanged."}]},{"id":"f41eea7b-f683-4fe7-97a3-5bfdabd26509","cardType":"basic","points":2,"front":"Why must subscripts not be altered when balancing an equation?","back":"Subscripts are part of the substance's identity.","mcqOptions":[{"text":"Subscripts are part of the substance's identity.","correct":true,"explanation":"A subscript fixes the number and ratio of atoms in a chemical species. Altering it changes the formula and therefore produces a different substance rather than balancing the original equation."},{"text":"Coefficients are part of the substance's identity.","correct":false,"explanation":"Coefficients specify relative amounts and may be changed during balancing without changing substance identities. Subscripts must remain fixed because they define each substance’s composition."},{"text":"State symbols are part of the substance's identity.","correct":false,"explanation":"State symbols identify physical state, but they are not the reason subscripts remain fixed. Subscripts determine a substance’s elemental composition and therefore its chemical identity."},{"text":"Reaction conditions are part of the substance's identity.","correct":false,"explanation":"Reaction conditions may be adjusted without redefining a chemical formula. Subscripts cannot be altered because they specify the atom ratio and hence the substance’s identity."}]},{"id":"0a32941e-d2e4-4d5c-9bae-da46d4adfe00","cardType":"cloze","points":null,"front":"Group 1 ions usually have charge [ ]; Group 2 ions usually have charge [ ].","back":"Group 1 ions usually have charge +1; Group 2 ions usually have charge +2.","mcqOptions":[{"text":"Group 1 ions usually have charge +1; Group 2 ions usually have charge +2.","correct":true,"explanation":"Group 1 atoms typically lose one outer electron to form +1 ions, while Group 2 atoms typically lose two outer electrons to form +2 ions."},{"text":"Group 1 ions usually have charge +2; Group 2 ions usually have charge +1.","correct":false,"explanation":"This reverses the usual charges. Group 1 atoms lose one electron to form +1 ions, whereas Group 2 atoms lose two electrons to form +2 ions."},{"text":"Group 1 ions usually have charge 0; Group 2 ions usually have charge +2.","correct":false,"explanation":"The group number does not become the ionic charge for Group 2 alone while Group 1 remains neutral. Their usual charges are +1 and +2 respectively."},{"text":"Group 1 ions usually have charge -1; Group 2 ions usually have charge -2.","correct":false,"explanation":"These signs incorrectly suggest both groups gain electrons. Group 1 and Group 2 metals lose one and two electrons respectively, producing +1 and +2 ions."}]},{"id":"5f3c8bea-9fa7-4939-82b0-973ebb76bb55","cardType":"cloze","points":null,"front":"Aluminium forms [ ] ions; sulfate is [ ].","back":"Aluminium forms Al³⁺ ions; sulfate is SO₄^2-.","mcqOptions":[{"text":"Aluminium forms Al³⁺ ions; sulfate is SO₄^2-.","correct":true,"explanation":"Aluminium loses three electrons to form Al³⁺. Sulfate is the polyatomic ion SO₄²⁻, containing one sulfur atom and four oxygen atoms with an overall -2 charge."},{"text":"Aluminium forms Al²⁺ ions; sulfate is SO₄^2-.","correct":false,"explanation":"Al²⁺ is a common charge-balancing misconception, but aluminium normally loses three electrons and forms Al³⁺. Sulfate is correctly represented by SO₄²⁻."},{"text":"Aluminium forms Al³⁺ ions; sulfate is SO₃^2-.","correct":false,"explanation":"SO₃²⁻ is sulfite, not sulfate; the names are easily confused. Sulfate contains four oxygen atoms and has the formula SO₄²⁻, while aluminium forms Al³⁺."},{"text":"Aluminium forms Al⁺ ions; sulfate is SO₄^-.","correct":false,"explanation":"This gives both ions incorrect charges. Aluminium normally forms Al³⁺ by losing three electrons, and sulfate is SO₄²⁻ with an overall charge of -2."}]},{"id":"73717c14-f641-4cad-8232-fcccd4d4d72b","cardType":"basic","points":1,"front":"What must the total charge be in an ionic compound?","back":"The charges must cancel to zero.","mcqOptions":[{"text":"The charges must cancel to zero.","correct":true,"explanation":"An ionic compound is electrically neutral overall, so the total positive charge from its cations must exactly balance the total negative charge from its anions."},{"text":"The charges must be removed completely.","correct":false,"explanation":"Ions do not lose their individual charges when forming an ionic compound. Instead, the positive and negative charges must balance so that their total is zero."},{"text":"The charges must occur in equal numbers.","correct":false,"explanation":"The numbers of positive and negative ions need not be equal when their charge magnitudes differ. Their charges, rather than their quantities, must cancel to zero."},{"text":"The charges must add to a positive value.","correct":false,"explanation":"Adding ionic charges would leave a charged species rather than a neutral compound. Cation and anion charges must have opposite signs and cancel to zero overall."}]},{"id":"8c53b8b6-7628-45af-bc27-f889d0aa3e56","cardType":"basic","points":1,"front":"What is the formula of aluminium sulfate?","back":"Al₂(SO₄)₃","mcqOptions":[{"text":"Al₂(SO₄)₃","correct":true,"explanation":"Two Al³⁺ ions give +6 and three SO₄²⁻ ions give -6, so Al₂(SO₄)₃ is electrically neutral and uses the simplest whole-number ratio."},{"text":"AlSO₄","correct":false,"explanation":"A 1:1 ratio leaves a net charge of +1 because Al³⁺ and SO₄²⁻ do not cancel. Two aluminium ions and three sulfate ions are required."},{"text":"Al₃(SO₄)₂","correct":false,"explanation":"The ion ratio has been reversed: three Al³⁺ ions give +9 while two sulfates give -4. Charge balance requires Al₂(SO₄)₃ instead."},{"text":"Al(SO₄)₃","correct":false,"explanation":"This incorrectly uses one aluminium ion with three sulfate ions, producing a large negative net charge. Neutrality requires two Al³⁺ ions for every three SO₄²⁻ ions."}]},{"id":"cb6c6f88-6012-434d-9fbc-292f38228fc8","cardType":"basic","points":2,"front":"Why are brackets used in Ca(OH)₂?","back":"They show two hydroxide ions; OH^- repeats as a group.","mcqOptions":[{"text":"They show two hydroxide ions; OH^- repeats as a group.","correct":true,"explanation":"The subscript 2 applies to the entire OH group, so the formula contains two O atoms and two H atoms as two hydroxide ions balancing Ca²⁺."},{"text":"They show two calcium ions; Ca²⁺ repeats as a group.","correct":false,"explanation":"The brackets do not indicate two calcium ions; calcium has no following subscript and appears once. They show that the complete OH⁻ ion occurs twice."},{"text":"They show two hydrogen atoms; H repeats within one hydroxide ion.","correct":false,"explanation":"The subscript outside brackets applies to both O and H, not only to hydrogen. Ca(OH)₂ therefore contains two complete hydroxide ions, each with charge -1."},{"text":"They show one hydroxide ion; OH^2- has a doubled charge.","correct":false,"explanation":"Brackets do not mean the hydroxide ion has a doubled charge. Each OH⁻ remains -1, and two separate hydroxide ions provide the total -2 charge needed."}]},{"id":"f0414a91-b437-49d6-9b7d-68e14786f341","cardType":"basic","points":1,"front":"How is relative formula mass, M_r, calculated from a formula?","back":"Add the A_r values for every atom in the formula.","mcqOptions":[{"text":"Add the A_r values for every atom in the formula.","correct":true,"explanation":"Relative formula mass is the sum of the relative atomic masses, A_r, of every atom represented in the chemical formula, including atoms indicated by subscripts."},{"text":"Multiply the A_r values for every atom in the formula.","correct":false,"explanation":"Multiplying all A_r values together does not account correctly for a compound’s composition. M_r is found by adding the A_r values for every atom in the formula."},{"text":"Average the A_r values for every element in the formula.","correct":false,"explanation":"Averaging treats the atoms as equally weighted and ignores how their masses combine. M_r is calculated by adding the A_r values for every atom shown."},{"text":"Add the A_r values for each element once in the formula.","correct":false,"explanation":"Adding each element’s A_r only once ignores subscripts, such as the two oxygen atoms in CO₂. The A_r value of every atom must be included."}]},{"id":"04cacb95-f986-450f-b68c-8bb1a7bdd2b3","cardType":"basic","points":2,"front":"Calculate M_r for Ca(NO₃)₂: Ca = 40.1, N = 14.0, O = 16.0.","back":"164.1","mcqOptions":[{"text":"164.1","correct":true,"explanation":"Ca(NO₃)₂ contains one Ca, two N and six O atoms, so M_r = 40.1 + 2(14.0) + 6(16.0) = 164.1."},{"text":"102.1","correct":false,"explanation":"This counts only one nitrate group, ignoring the outer subscript 2. Including two N and six O atoms gives 40.1 + 28.0 + 96.0 = 164.1."},{"text":"148.1","correct":false,"explanation":"This total omits one oxygen atom from the formula. There are six oxygen atoms altogether, so the correct relative formula mass is 164.1."},{"text":"204.2","correct":false,"explanation":"This effectively counts two calcium atoms, although Ca has no subscript and appears only once. Counting one Ca, two N and six O gives 164.1."}]},{"id":"eba56fa8-4987-452f-946f-5ba46d1ae7c7","cardType":"basic","points":3,"front":"How do empirical and molecular formulae differ?","back":"

Empirical: simplest whole-number atom ratio.
Molecular: actual atoms in one molecule.

","mcqOptions":[{"text":"

Empirical: simplest whole-number atom ratio.
Molecular: actual atoms in one molecule.

","correct":true,"explanation":"An empirical formula gives the lowest whole-number ratio of atoms, whereas a molecular formula states the actual number of each type of atom in one molecule."},{"text":"

Empirical: actual atoms in one molecule.
Molecular: simplest whole-number atom ratio.

","correct":false,"explanation":"These definitions are reversed. The empirical formula gives the simplest whole-number ratio, while the molecular formula gives the actual numbers of atoms in one molecule."},{"text":"

Empirical: simplest whole-number mass ratio.
Molecular: actual masses in one molecule.

","correct":false,"explanation":"Neither formula describes relative masses. The empirical formula shows the simplest whole-number atom ratio, and the molecular formula shows the actual atoms in a molecule."},{"text":"

Empirical: actual atom ratio in a compound.
Molecular: simplest atoms in one molecule.

","correct":false,"explanation":"A molecular formula does not necessarily show the simplest ratio. It gives the actual atom numbers, while the empirical formula reduces those numbers to their simplest whole-number ratio."}]},{"id":"fe20319d-231f-40c6-bb32-3c480fb7cd9f","cardType":"basic","points":1,"front":"What is the empirical formula of glucose, C₆H₁₂O₆?","back":"CH₂O","mcqOptions":[{"text":"CH₂O","correct":true,"explanation":"Dividing all subscripts in C₆H₁₂O₆ by their highest common factor, 6, gives the simplest whole-number ratio 1:2:1 and therefore CH₂O."},{"text":"C₆H₁₂O₆","correct":false,"explanation":"This is glucose’s molecular formula, not its empirical formula. Dividing the subscripts 6, 12 and 6 by their highest common factor gives CH₂O."},{"text":"C₂H₄O₂","correct":false,"explanation":"Dividing each glucose subscript by 3 gives this ratio, but it can still be simplified further. Dividing by the highest common factor, 6, gives CH₂O."},{"text":"CH₃O","correct":false,"explanation":"The hydrogen subscript has been reduced incorrectly. The ratio 6:12:6 simplifies by division by 6 to 1:2:1, so the empirical formula is CH₂O."}]},{"id":"7f9ee0e2-e480-49f7-9538-266ba2a00346","cardType":"basic","points":1,"front":"When using percentage composition to find an empirical formula, what mass is assumed?","back":"Assume a 100 g sample.","mcqOptions":[{"text":"Assume a 100 g sample.","correct":true,"explanation":"Assuming 100 g makes each percentage numerically equal to a mass in grams, allowing each element’s mass to be converted conveniently into moles."},{"text":"Assume a 1 g sample.","correct":false,"explanation":"A 1 g sample makes each percentage correspond to hundredths of a gram, adding unnecessary conversion. The standard assumption is 100 g, so percentages directly become gram masses."},{"text":"Assume a 10 g sample.","correct":false,"explanation":"With a 10 g sample, each percentage must be divided by 10 to obtain the element’s mass. Assuming 100 g makes percentage values equal directly to masses in grams."},{"text":"Assume a 1000 g sample.","correct":false,"explanation":"A 1000 g sample makes each percentage ten times the mass in grams, so an extra conversion is needed. A 100 g sample gives direct numerical equivalence."}]},{"id":"f4c7a11b-52c9-45c8-b4cc-2f6670d95810","cardType":"basic","points":1,"front":"After converting element masses to moles, how is the simplest atom ratio found?","back":"Divide each mole amount by the smallest mole amount.","mcqOptions":[{"text":"Divide each mole amount by the smallest mole amount.","correct":true,"explanation":"Dividing every mole amount by the smallest normalises that value to 1 and gives the simplest relative mole ratio, which corresponds to the atom ratio."},{"text":"Divide each mole amount by the largest mole amount.","correct":false,"explanation":"Dividing by the largest makes the largest value 1 but usually leaves awkward fractions below 1. Dividing every amount by the smallest gives the simplest atom ratio."},{"text":"Multiply each mole amount by the smallest mole amount.","correct":false,"explanation":"Multiplying by the smallest mole amount does not normalise the ratio and produces values dependent on sample size. Each amount should be divided by the smallest mole amount."},{"text":"Divide each mole amount by the mean mole amount.","correct":false,"explanation":"The mean has no role in simplifying mole ratios and can produce unsuitable fractional values. Divide every mole amount by the smallest mole amount to establish the relative ratio."}]},{"id":"fe71dc14-f3b9-49aa-acb8-b46313ea9ad5","cardType":"basic","points":2,"front":"How is a molecular formula found from an empirical formula and relative molecular mass?","back":"

Divide relative molecular mass by empirical formula mass.
Multiply every empirical subscript by that value.

","mcqOptions":[{"text":"

Divide relative molecular mass by empirical formula mass.
Multiply every empirical subscript by that value.

","correct":true,"explanation":"The molecular mass divided by the empirical formula mass gives a whole-number multiplier. Applying it to every empirical subscript produces the actual molecular formula."},{"text":"

Divide empirical formula mass by relative molecular mass.
Multiply every empirical subscript by that value.

","correct":false,"explanation":"The division has been reversed, giving the reciprocal of the required whole-number multiplier. Divide relative molecular mass by empirical formula mass, then multiply every subscript."},{"text":"

Divide relative molecular mass by empirical formula mass.
Divide every empirical subscript by that value.

","correct":false,"explanation":"The mass ratio is a multiplier, not a divisor. Divide relative molecular mass by empirical formula mass, then multiply every empirical subscript by the resulting whole number."},{"text":"

Multiply relative molecular mass by empirical formula mass.
Multiply every empirical subscript by that value.

","correct":false,"explanation":"Multiplying the two masses cannot give the number of empirical units in a molecule. Divide relative molecular mass by empirical formula mass, then multiply every subscript by that value."}]},{"id":"0cca98a8-e53c-491a-98b4-5b0baca29200","cardType":"basic","points":1,"front":"What must match on both sides of a balanced equation?","back":"The number of atoms of each element.","mcqOptions":[{"text":"The number of atoms of each element.","correct":true,"explanation":"A balanced equation obeys conservation of atoms: each element must have the same number of atoms before and after the reaction, although their chemical arrangements can change."},{"text":"The number of molecules of each substance.","correct":false,"explanation":"The number of molecules can differ because coefficients describe different reacting proportions. Balance requires the number of atoms of each element to match on both sides."},{"text":"The number of substances in each state.","correct":false,"explanation":"Reactants are converted into different products, so the number of substances need not match. It is the number of atoms of each element that must be conserved."},{"text":"The number of coefficients for each substance.","correct":false,"explanation":"Coefficients are deliberately adjusted during balancing and do not have to be identical on both sides. They must produce equal numbers of atoms of each element."}]},{"id":"34d1bd64-9871-48b4-b2b8-4d37909a1f23","cardType":"cloze","points":null,"front":"A stoichiometric coefficient multiplies the [ ]; 3CO₂ contains [ ].","back":"A stoichiometric coefficient multiplies the whole formula; 3CO₂ contains 6 oxygen atoms.","mcqOptions":[{"text":"A stoichiometric coefficient multiplies the whole formula; 3CO₂ contains 6 oxygen atoms.","correct":true,"explanation":"The coefficient 3 multiplies every atom in CO₂. Each formula contains two oxygen atoms, so three formula units contain six oxygen atoms."},{"text":"A stoichiometric coefficient multiplies the first element only; 3CO₂ contains 6 oxygen atoms.","correct":false,"explanation":"A coefficient does not apply only to the first element; it multiplies the whole formula. Therefore, 3CO₂ represents three carbon atoms and six oxygen atoms."},{"text":"A stoichiometric coefficient multiplies the whole formula; 3CO₂ contains 3 oxygen atoms.","correct":false,"explanation":"This overlooks the oxygen subscript of 2. The coefficient multiplies the whole formula, giving $3 \\times 2 = 6$ oxygen atoms in 3CO₂."},{"text":"A stoichiometric coefficient multiplies the following subscript; 3CO₂ contains 9 oxygen atoms.","correct":false,"explanation":"The coefficient multiplies the numbers of atoms represented by the formula; it is not added to or combined with the subscript. Thus, 3CO₂ contains six oxygen atoms."}]},{"id":"c84f5a30-8ac5-4ec2-b427-2a0afbcbfd55","cardType":"basic","points":3,"front":"Balance: Fe₂O₃ + CO → Fe + CO₂","back":"Fe₂O₃ + 3CO → 2Fe + 3CO₂","mcqOptions":[{"text":"Fe₂O₃ + 3CO → 2Fe + 3CO₂","correct":true,"explanation":"These coefficients give two iron atoms, three carbon atoms and six oxygen atoms on each side, so the equation conserves every element."},{"text":"Fe₂O₃ + 2CO → 2Fe + 2CO₂","correct":false,"explanation":"Balancing carbon this way leaves five oxygen atoms on the left but only four on the right. Three CO and three CO₂ are required."},{"text":"Fe₂O₃ + 3CO → Fe + 3CO₂","correct":false,"explanation":"Carbon and oxygen are balanced, but the two iron atoms in Fe₂O₃ require 2Fe on the right. Chemical equations must conserve all elements."},{"text":"Fe₂O₃ + CO → 2Fe + CO₂","correct":false,"explanation":"This balances iron and carbon but not oxygen: there are four oxygen atoms on the left and two on the right. The CO and CO₂ coefficients must both be 3."}]},{"id":"5582eee6-3706-43f6-8452-c4dce1882d1d","cardType":"basic","points":2,"front":"Why is water written H₂O(l), not H₂O(aq), in neutralisation equations?","back":"It is a liquid product, not a solute dissolved in water.","mcqOptions":[{"text":"It is a liquid product, not a solute dissolved in water.","correct":true,"explanation":"The state symbol (l) identifies water formed as a liquid product. The symbol (aq) is reserved for a substance dissolved in water, not water itself."},{"text":"It is an aqueous product, because the reaction occurs in water.","correct":false,"explanation":"The reaction mixture is aqueous, but that does not make the water product aqueous. Water is the liquid solvent itself, so its state symbol is (l)."},{"text":"It is a gaseous product, not a liquid formed during the reaction.","correct":false,"explanation":"Neutralisation under ordinary laboratory conditions produces liquid water, not water vapour. The product is therefore written H₂O(l), while (g) would indicate steam."},{"text":"It is a solid product, not an ion dissolved in water.","correct":false,"explanation":"Water produced in neutralisation is liquid under standard laboratory conditions and does not precipitate as a solid. Its correct formula and state are H₂O(l)."}]},{"id":"23b14813-a358-433c-b5e0-4a335ec39d00","cardType":"cloze","points":null,"front":"Ionic equations omit [ ], which are present but [ ].","back":"Ionic equations omit spectator ions, which are present but unchanged.","mcqOptions":[{"text":"Ionic equations omit spectator ions, which are present but unchanged.","correct":true,"explanation":"Spectator ions occur in the reaction mixture but undergo no chemical change. Cancelling them from both sides leaves only the species directly involved in the reaction."},{"text":"Ionic equations omit reacting ions, which are present but unchanged.","correct":false,"explanation":"Reacting ions must remain because they show the chemical change. Ionic equations omit spectator ions, which appear in the mixture but remain unchanged throughout the reaction."},{"text":"Ionic equations omit spectator ions, which are present but neutralised.","correct":false,"explanation":"Spectator ions are not neutralised or otherwise reacted; they remain unchanged. Their identical appearance on both sides allows them to be cancelled from the ionic equation."},{"text":"Ionic equations omit product ions, which are present but consumed.","correct":false,"explanation":"Products cannot be omitted merely because they are ions, and products are formed rather than consumed. Only spectator ions are removed because they remain unchanged."}]},{"id":"a3bd98a1-aa06-4eb8-b5c6-a66f56b90a73","cardType":"cloze","points":null,"front":"A correct ionic equation balances both [ ] and overall [ ].","back":"A correct ionic equation balances both atoms and overall charge.","mcqOptions":[{"text":"A correct ionic equation balances both atoms and overall charge.","correct":true,"explanation":"An ionic equation must conserve matter and electric charge. Each element must have equal atom totals on both sides, and the net charges must also match."},{"text":"A correct ionic equation balances both molecules and overall charge.","correct":false,"explanation":"Equations conserve atoms, not necessarily the number of molecules or particles, because particles can combine or split. Overall electrical charge must also be equal on both sides."},{"text":"A correct ionic equation balances both atoms and overall mass number.","correct":false,"explanation":"Balancing atoms already conserves mass in an ordinary chemical reaction. The separate second requirement for an ionic equation is conservation of overall electrical charge."},{"text":"A correct ionic equation balances both electrons and overall state symbols.","correct":false,"explanation":"State symbols describe physical states and are not quantities to balance. A correct ionic equation must have equal numbers of each atom and equal overall charge on both sides."}]},{"id":"59aa3ead-9333-45f1-9a2b-c71d21b733ce","cardType":"basic","points":2,"front":"Write the ionic equation for BaCl₂(aq) reacting with Na₂SO₄(aq).","back":"Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s)","mcqOptions":[{"text":"Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s)","correct":true,"explanation":"Ba²⁺ and SO₄^2- combine in a 1:1 ratio to form insoluble barium sulfate. Na⁺ and Cl^- are omitted as spectator ions."},{"text":"Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(aq)","correct":false,"explanation":"The ionic species and ratio are correct, but barium sulfate is insoluble and forms a precipitate. Its state symbol must therefore be (s), not (aq)."},{"text":"Ba⁺(aq) + SO₄^-(aq) → BaSO₄(s)","correct":false,"explanation":"This incorrectly assigns single charges to both ions. Barium forms Ba²⁺ and sulfate is SO₄^2-, which combine in a 1:1 ratio."},{"text":"2Ba²⁺(aq) + SO₄^2-(aq) → Ba₂SO₄(s)","correct":false,"explanation":"This uses the wrong ion ratio and produces a formula with a net positive charge. Ba²⁺ and SO₄^2- combine 1:1 as BaSO₄(s)."}]},{"id":"2c510a51-9fa9-453a-8ea6-ca997c83fbdf","cardType":"basic","points":1,"front":"Which arrow is used for reversible reactions and equilibria?","back":"Use ⇌.","mcqOptions":[{"text":"Use ⇌.","correct":true,"explanation":"The pair of opposing half-arrows shows that forward and reverse reactions occur and can establish dynamic equilibrium. It is the standard equilibrium symbol."},{"text":"Use →.","correct":false,"explanation":"A single forward arrow represents a reaction treated as proceeding in one direction. Reversible reactions and equilibria require the opposing half-arrows symbol ⇌."},{"text":"Use ←.","correct":false,"explanation":"A single left-pointing arrow only reverses the written reaction direction; it does not show that both directions occur. Equilibrium is represented by ⇌."},{"text":"Use ↔.","correct":false,"explanation":"A double-headed arrow is commonly associated with resonance or a general two-way relationship, not the required equilibrium notation. Reversible chemical reactions use ⇌."}]}]}],"examCountdown":{"examDate":"2027-05-11","paperLabel":"Paper 1"},"heading":"Formulae and equations","title":"Formulae and equations Flashcards","topbar":{"backHref":"/as-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"ad58eb9f-7f2f-5679-a747-6bb8d6cde0db","walled":"$L38"}],"$L39","$L3a"]