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before it and shows the relative number of atoms or ions of that element in the substance."},{"text":"The charge of atoms or ions of that element present in the substance.","correct":false,"explanation":"A subscript gives the relative number of particles, not their charge. Charge is shown separately by a superscript, such as $^{2+}$ or $^{3-}$."},{"text":"The mass of atoms or ions of that element present in the substance.","correct":false,"explanation":"A subscript represents the ratio of atoms or ions, not their mass. Atomic masses are found from relative atomic mass values, which are not written as subscripts."},{"text":"The state of atoms or ions of that element present in the substance.","correct":false,"explanation":"A subscript does not show physical state. It shows the ratio of atoms or ions, while state is written using symbols such as (s), (l), (g) or (aq)."}]},{"id":"2a6f0820-f1b7-408c-9391-87145af82207","cardType":"basic","points":1,"front":"What is the formula for the ammonium ion?","back":"$$\\text{NH}_4^+$","mcqOptions":[{"text":"$$\\text{NH}_4^+$","correct":true,"explanation":"Ammonium is the positively charged polyatomic ion formed from ammonia by gaining a proton, so its formula is $\\text{NH}_4^+$."},{"text":"$$\\text{NH}_4^-$","correct":false,"explanation":"The ammonium ion has a positive, not negative, charge. The correct formula is $\\text{NH}_4^+$; the negative sign reverses the ion’s charge."},{"text":"$$\\text{NO}_3^-$","correct":false,"explanation":"$$\\text{NO}_3^-$ is the nitrate ion, a common confusion because both ions contain nitrogen and oxygen or hydrogen. Ammonium is $\\text{NH}_4^+$."},{"text":"$$\\text{NH}_3^+$","correct":false,"explanation":"Ammonium contains four hydrogen atoms, not three. Removing or forgetting one hydrogen gives the incorrect formula $\\text{NH}_3^+$ instead of $\\text{NH}_4^+$."}]},{"id":"84b66928-2446-4813-8557-2d43c4755b30","cardType":"basic","points":1,"front":"What is the formula and charge of the phosphate ion?","back":"$$\\text{PO}_4^{3-}$","mcqOptions":[{"text":"$$\\text{PO}_4^{3-}$","correct":true,"explanation":"The phosphate ion contains one phosphorus atom and four oxygen atoms, with an overall charge of $3-$, giving $\\text{PO}_4^{3-}$."},{"text":"$$\\text{PO}_3^{3-}$","correct":false,"explanation":"The phosphite ion is commonly confused with phosphate because their names are similar. Phosphate contains four oxygen atoms, so its formula is $\\text{PO}_4^{3-}$."},{"text":"$$\\text{PO}_4^{2-}$","correct":false,"explanation":"The oxygen count is correct, but phosphate has a $3-$ charge rather than $2-$. The correct ion is $\\text{PO}_4^{3-}$."},{"text":"$$\\text{P}_2\\text{O}_7^{4-}$","correct":false,"explanation":"$$\\text{P}_2\\text{O}_7^{4-}$ is the pyrophosphate ion, not phosphate. Phosphate contains one phosphorus and four oxygen atoms: $\\text{PO}_4^{3-}$."}]},{"id":"36b29727-d902-4a4a-a69d-c503ca57e3d1","cardType":"basic","points":1,"front":"What is the formula for the dichromate(VI) ion?","back":"$$\\text{Cr}_2\\text{O}_7^{2-}$","mcqOptions":[{"text":"$$\\text{Cr}_2\\text{O}_7^{2-}$","correct":true,"explanation":"The dichromate(VI) ion contains two chromium atoms and seven oxygen atoms, with an overall charge of $2-$, so its formula is $\\text{Cr}_2\\text{O}_7^{2-}$."},{"text":"$$\\text{CrO}_4^{2-}$","correct":false,"explanation":"$$\\text{CrO}_4^{2-}$ is the chromate(VI) ion. Dichromate has two chromium atoms and seven oxygen atoms, giving $\\text{Cr}_2\\text{O}_7^{2-}$."},{"text":"$$\\text{Cr}_2\\text{O}_7^{3-}$","correct":false,"explanation":"The atom numbers match dichromate, but its charge is $2-$, not $3-$. The correct formula is $\\text{Cr}_2\\text{O}_7^{2-}$."},{"text":"$$\\text{Cr}_2\\text{O}_4^{2-}$","correct":false,"explanation":"Dichromate contains seven oxygen atoms, not four. Confusing the oxygen count with a simpler oxide ratio gives the wrong formula; the correct ion is $\\text{Cr}_2\\text{O}_7^{2-}$."}]},{"id":"4d4197d2-8f3c-41fe-80ca-3f75f8cc6ce6","cardType":"cloze","points":null,"front":"To find the formula of an ionic compound, choose the smallest [ ] of ions that results in a zero overall charge.","back":"To find the formula of an ionic compound, choose the smallest whole-number ratio of ions that results in a zero overall charge.","mcqOptions":[{"text":"whole-number ratio","correct":true,"explanation":"Ionic compounds must have no overall charge, so their ions are combined in the smallest whole-number ratio that balances the positive and negative charges."},{"text":"whole-number total","correct":false,"explanation":"The ions are not selected by adding their numbers to make a total. The correct description is the smallest whole-number ratio that gives a zero overall charge."},{"text":"lowest-charge ratio","correct":false,"explanation":"The charges determine the required ratio, but the ratio itself is not called a lowest-charge ratio. The correct term is whole-number ratio."},{"text":"balanced-charge number","correct":false,"explanation":"The aim is to balance charge, but the formula uses a ratio of ions rather than a single number. The correct fill-in is whole-number ratio."}]},{"id":"3f914296-760e-4b97-8a13-82864d1270c5","cardType":"cloze","points":null,"front":"To find the formula of an ionic compound, choose the smallest whole-number ratio of ions that results in a [ ].","back":"To find the formula of an ionic compound, choose the smallest whole-number ratio of ions that results in a zero overall charge.","mcqOptions":[{"text":"zero overall charge","correct":true,"explanation":"An ionic compound is electrically neutral, so its positive and negative charges must cancel. The smallest ratio of ions therefore produces zero overall charge."},{"text":"positive overall charge","correct":false,"explanation":"A positive overall charge would mean there are more positive charges than negative charges, so the compound would not be neutral. The correct result is zero overall charge."},{"text":"negative overall charge","correct":false,"explanation":"A negative overall charge would mean the negative ions are not balanced by enough positive charge. Ionic compounds require zero overall charge."},{"text":"equal numbers of ions","correct":false,"explanation":"The numbers of positive and negative ions need not be equal; their charges must balance. For example, $\\text{Ca}^{2+}$ and $\\text{Cl}^-$ require a 1:2 ratio."}]},{"id":"61735ea0-3bea-4303-a9b3-b8717ad496e7","cardType":"basic","points":2,"front":"Why must brackets be used in the formula $\\text{Al}_2(\\text{SO}_4)_3$?","back":"To show there are three units of the polyatomic sulfate ion.","mcqOptions":[{"text":"To show there are three units of the polyatomic sulfate ion.","correct":true,"explanation":"The brackets group the entire sulfate ion, $\\text{SO}_4^{2-}$. The subscript 3 outside them shows that three complete sulfate ions are present."},{"text":"To show there are three atoms of the polyatomic sulfate ion.","correct":false,"explanation":"The subscript 3 counts complete sulfate ions, not individual atoms. Each sulfate unit contains one sulfur and four oxygen atoms, so the correct description is three units."},{"text":"To show there are two units of the polyatomic sulfate ion.","correct":false,"explanation":"The subscript outside the brackets is 3, so it indicates three sulfate units. Two units would be written as $(\\text{SO}_4)_2$, not $(\\text{SO}_4)_3$."},{"text":"To show there are three units of the monatomic sulfate ion.","correct":false,"explanation":"Sulfate is polyatomic because it contains sulfur and oxygen atoms bonded together. The brackets show three units of this polyatomic sulfate ion."}]},{"id":"3f7d1723-352f-438e-ae15-b636b816b4b2","cardType":"basic","points":1,"front":"What is the oxidation number of any uncombined element, such as $\\text{Mg}$ or $\\text{Cl}_2$?","back":"Zero (0).","mcqOptions":[{"text":"Zero (0).","correct":true,"explanation":"Any uncombined element has no ionic charge and is electrically neutral, so each element in its elemental form is assigned an oxidation number of zero."},{"text":"One (+1).","correct":false,"explanation":"An uncombined element is neutral, so its oxidation number is zero, not $+1$. Positive oxidation numbers usually occur when an element is combined with another element."},{"text":"Two (+2).","correct":false,"explanation":"The $+2$ value is a common oxidation number for metals such as magnesium in compounds, but uncombined magnesium has oxidation number zero."},{"text":"Minus one (-1).","correct":false,"explanation":"The $-1$ value is often associated with halide ions such as $\\text{Cl}^-$. Uncombined chlorine, $\\text{Cl}_2$, is elemental and therefore has oxidation number zero."}]},{"id":"096a341a-360b-42ed-a5ab-61c22fd089d4","cardType":"basic","points":1,"front":"What is the rule for the sum of oxidation numbers in a neutral compound?","back":"The sum must be zero.","mcqOptions":[{"text":"The sum must be zero.","correct":true,"explanation":"A neutral compound has no overall electrical charge, so the positive and negative oxidation numbers of all its atoms must add up to zero."},{"text":"The sum must be +1.","correct":false,"explanation":"A sum of +1 would describe an ion with a single positive charge. Because the compound is neutral, all oxidation numbers must instead sum to zero."},{"text":"The sum must be -1.","correct":false,"explanation":"A sum of -1 would correspond to an ion with a single negative charge. In a neutral compound, the oxidation numbers must balance to give zero."},{"text":"The sum must equal the number of elements.","correct":false,"explanation":"The number of elements does not determine the oxidation-number total. Regardless of its composition, a neutral compound has oxidation numbers that sum to zero."}]},{"id":"c48a2493-b8fa-404f-9913-a02298432e7d","cardType":"cloze","points":null,"front":"The sum of oxidation numbers in a polyatomic ion is equal to the [ ].","back":"The sum of oxidation numbers in a polyatomic ion is equal to the charge on the ion.","mcqOptions":[{"text":"charge on the ion","correct":true,"explanation":"A polyatomic ion has an overall electrical charge, and the oxidation numbers assigned to all its constituent atoms must sum to that charge."},{"text":"number of atoms in the ion","correct":false,"explanation":"Counting atoms does not give the oxidation-number sum. The oxidation numbers of all atoms in a polyatomic ion add up to the ion’s overall charge."},{"text":"charge on the central atom","correct":false,"explanation":"The central atom has its own oxidation number, which is not necessarily the ion’s total. The sum for every atom equals the charge on the ion."},{"text":"number of bonds in the ion","correct":false,"explanation":"The number of bonds is unrelated to the required total. Oxidation numbers in a polyatomic ion must sum to the ion’s overall charge."}]},{"id":"223ba1dc-e14a-422d-9576-6465db6da3fd","cardType":"cloze","points":null,"front":"Oxygen usually has an oxidation number of -2, except in [ ] where it is -1.","back":"Oxygen usually has an oxidation number of -2, except in peroxides where it is -1.","mcqOptions":[{"text":"peroxides","correct":true,"explanation":"Peroxides contain an oxygen-oxygen single bond and the peroxide ion, O₂²⁻. Sharing the -2 charge between two equivalent oxygen atoms gives each oxygen an oxidation number of -1."},{"text":"oxides","correct":false,"explanation":"In ordinary oxides, oxygen follows its usual oxidation number of -2 rather than the -1 exception. Oxygen has an oxidation number of -1 specifically in peroxides."},{"text":"superoxides","correct":false,"explanation":"Superoxides contain O₂⁻, so the average oxidation number of oxygen is -½, not -1. The -1 oxidation number occurs in peroxides."},{"text":"ozonides","correct":false,"explanation":"Ozonides contain O₃⁻, giving oxygen an average oxidation number of -⅓ rather than -1. Oxygen has oxidation number -1 in peroxides."}]},{"id":"20a387ca-32c6-455a-9b66-b71a070c16b5","cardType":"cloze","points":null,"front":"Oxygen usually has an oxidation number of -2, except in peroxides where it is [ ].","back":"Oxygen usually has an oxidation number of -2, except in peroxides where it is -1.","mcqOptions":[{"text":"-1","correct":true,"explanation":"A peroxide contains the O₂²⁻ group. Its total oxidation number of -2 is shared equally between two oxygen atoms, making each oxygen -1."},{"text":"-2","correct":false,"explanation":"Using -2 applies oxygen’s usual oxidation number and overlooks the oxygen-oxygen bond in a peroxide. Each oxygen atom in a peroxide has oxidation number -1."},{"text":"+1","correct":false,"explanation":"The sign has been reversed: oxygen remains negative in ordinary peroxides. Each oxygen in the O₂²⁻ group has an oxidation number of -1."},{"text":"+2","correct":false,"explanation":"This incorrectly applies a positive oxidation number to peroxide oxygen. The O₂²⁻ group has a total of -2, so each oxygen is -1."}]},{"id":"27f576e2-0c9a-44a7-af44-ee08cd4cb8f9","cardType":"cloze","points":null,"front":"Hydrogen usually has an oxidation number of +1, except in [ ] where it is -1.","back":"Hydrogen usually has an oxidation number of +1, except in metal hydrides where it is -1.","mcqOptions":[{"text":"metal hydrides","correct":true,"explanation":"In metal hydrides, hydrogen is bonded to an electropositive metal and exists as the hydride ion, H⁻. It therefore has an oxidation number of -1."},{"text":"acids","correct":false,"explanation":"Hydrogen in acids generally has its usual oxidation number of +1, not -1. The -1 exception occurs when hydrogen forms a metal hydride."},{"text":"covalent hydrides","correct":false,"explanation":"Hydrogen in covalent hydrides with non-metals is generally assigned +1. It has oxidation number -1 specifically in hydrides of electropositive metals."},{"text":"metal hydroxides","correct":false,"explanation":"Hydrogen in a hydroxide group is assigned its usual oxidation number of +1. Hydrogen has oxidation number -1 in metal hydrides, not metal hydroxides."}]},{"id":"b45fd5d4-9d2e-42b6-82a2-2fbd544eabdd","cardType":"cloze","points":null,"front":"Hydrogen usually has an oxidation number of +1, except in metal hydrides where it is [ ].","back":"Hydrogen usually has an oxidation number of +1, except in metal hydrides where it is -1.","mcqOptions":[{"text":"-1","correct":true,"explanation":"Metals are generally less electronegative than hydrogen, so hydrogen in a metal hydride is treated as H⁻ and assigned an oxidation number of -1."},{"text":"+1","correct":false,"explanation":"This applies hydrogen’s usual value and misses the hydride exception. In metal hydrides, hydrogen gains formal electron ownership and has oxidation number -1."},{"text":"-2","correct":false,"explanation":"An oxidation number of -2 is not assigned to hydrogen in binary metal hydrides. Hydrogen is present as H⁻, giving it an oxidation number of -1."},{"text":"0","correct":false,"explanation":"Zero is used for an element in its uncombined state, such as H₂. In a metal hydride, bonded hydrogen has oxidation number -1."}]},{"id":"70d3e777-5924-488d-a553-b3b70a95c7b3","cardType":"basic","points":2,"front":"Calculate the oxidation number of manganese in $\\text{MnO}_4^-$.","back":"+7","mcqOptions":[{"text":"+7","correct":true,"explanation":"Each oxygen is -2, giving -8 in total. Since the ion’s oxidation numbers sum to -1, manganese must be +7 because $+7-8=-1$."},{"text":"+6","correct":false,"explanation":"This results from making the oxidation numbers sum to -2 rather than the ion’s -1 charge. Solving $x-8=-1$ gives manganese an oxidation number of +7."},{"text":"-7","correct":false,"explanation":"The sign has been reversed. Four oxygen atoms contribute -8, so manganese must contribute +7 for the oxidation numbers to total the ion’s -1 charge."},{"text":"+8","correct":false,"explanation":"This treats permanganate as neutral by balancing oxygen’s -8 contribution exactly. Because the ion has charge -1, manganese is +7 rather than +8."}]},{"id":"95a79a3d-45c8-4188-84a0-fe20b9372cf6","cardType":"basic","points":1,"front":"What does a Roman numeral in a compound name (e.g., iron(II) sulfate) indicate?","back":"The oxidation number of that specific element.","mcqOptions":[{"text":"The oxidation number of that specific element.","correct":true,"explanation":"The Roman numeral states the oxidation number of the named element in that compound. Thus iron(II) indicates that iron has an oxidation number of +2."},{"text":"The number of atoms of that specific element.","correct":false,"explanation":"A Roman numeral does not count atoms; subscripts in the formula perform that role. It specifies the oxidation number of the named element."},{"text":"The overall charge of the complete compound.","correct":false,"explanation":"The numeral is not the compound’s overall charge, which may be zero. It gives the oxidation number of the particular element immediately preceding it."},{"text":"The periodic table group of that specific element.","correct":false,"explanation":"The numeral does not identify the element’s periodic-table group. It distinguishes oxidation states by stating the oxidation number of the named element."}]},{"id":"d9fe5dfb-7ce3-4646-8301-4a8e552c07fc","cardType":"basic","points":2,"front":"In a chemical equation, what is the difference between a coefficient and a subscript regarding balancing?","back":"

Coefficients can be changed to balance atoms.
Subscripts must never be changed as they define the substance.

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

Coefficients can be changed to balance atoms.
Subscripts must never be changed as they define the substance.

","correct":true,"explanation":"Coefficients alter the number of particles and can balance each element, whereas subscripts belong to the chemical formula. Changing a subscript would create a different substance."},{"text":"

Subscripts can be changed to balance atoms.
Coefficients must never be changed as they define the substance.

","correct":false,"explanation":"This reverses the roles of the two numbers. Subscripts define a compound’s composition, while coefficients change particle amounts and are adjusted to balance atoms; the correct distinction is the reverse."},{"text":"

Coefficients can be changed to alter the substance.
Subscripts must never be changed when balancing.

","correct":false,"explanation":"Changing a coefficient changes how many particles react, not their identities. A subscript changes the compound itself, so subscripts must remain fixed while coefficients balance the equation."},{"text":"

Coefficients and subscripts can both be changed to balance atoms.
Neither number affects the substance.

","correct":false,"explanation":"Only coefficients may be adjusted during balancing. Altering subscripts changes the chemical formula and therefore the substance, so treating both numbers as interchangeable is incorrect."}]},{"id":"fdc67b41-3a6e-4450-b778-ca794f87f6de","cardType":"basic","points":1,"front":"What is the specific meaning of the state symbol $(aq)$?","back":"Aqueous: the substance is dissolved in water.","mcqOptions":[{"text":"Aqueous: the substance is dissolved in water.","correct":true,"explanation":"The state symbol $(aq)$ means aqueous. It indicates that the substance is present as a solute dissolved in water, often as separated ions if it is soluble."},{"text":"Aqueous: the substance is dissolved in an organic solvent.","correct":false,"explanation":"Aqueous specifically means dissolved in water, not in an unspecified or organic solvent. Organic solutions use different descriptions and do not receive the $(aq)$ state symbol."},{"text":"Liquid: the substance is in its pure liquid state.","correct":false,"explanation":"A pure liquid is represented by $(l)$. The symbol $(aq)$ means the substance has dissolved in water, so it describes a solution rather than the pure liquid state."},{"text":"Solid: the substance has precipitated from water.","correct":false,"explanation":"A precipitated solid is represented by $(s)$, not $(aq)$. Aqueous indicates material dissolved in water, whereas precipitation produces a separate solid phase."}]},{"id":"1ed4ce3a-4e44-4ace-b154-469cf166d2b4","cardType":"basic","points":1,"front":"Define \"spectator ions\" in the context of ionic equations.","back":"Ions that appear unchanged on both sides of the equation and do not take part in the reaction.","mcqOptions":[{"text":"Ions that appear unchanged on both sides of the equation and do not take part in the reaction.","correct":true,"explanation":"Spectator ions remain chemically unchanged during the reaction. They appear on both sides of the full ionic equation and cancel when it is simplified to the net ionic equation."},{"text":"Ions that are changed into products and appear only on the product side of the equation.","correct":false,"explanation":"Ions that change or form products are participating ions, not spectators. Spectator ions remain unchanged on both sides and cancel from the ionic equation."},{"text":"Neutral atoms that appear unchanged on both sides of the equation and do not take part in the reaction.","correct":false,"explanation":"Spectator ions must be charged species present in solution, not neutral atoms. Their defining feature is remaining unchanged on both sides and making no chemical contribution."},{"text":"Ions that appear unchanged on only one side of the equation and initiate the reaction.","correct":false,"explanation":"A spectator ion appears on both sides and does not initiate or drive the reaction. Species appearing on only one side are involved in the net chemical change."}]},{"id":"b040612b-1a9c-4e5b-a56e-3749bf85da5b","cardType":"basic","points":2,"front":"When writing an ionic equation, which four types of substances are NOT split into ions?","back":"

Solids $(s)$
Liquids $(l)$
Gases $(g)$
Covalent molecules

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

Solids $(s)$
Liquids $(l)$
Gases $(g)$
Covalent molecules

","correct":true,"explanation":"Only aqueous ionic substances are normally split into their constituent ions. Solids, liquids, gases and covalent molecules are kept intact when constructing an ionic equation."},{"text":"

Aqueous solutions $(aq)$
Liquids $(l)$
Gases $(g)$
Covalent molecules

","correct":false,"explanation":"Aqueous soluble ionic substances are the species that are split into ions. Solids, liquids, gases and covalent molecules remain intact, so replacing solids with aqueous solutions is the key error."},{"text":"

Solids $(s)$
Aqueous solutions $(aq)$
Gases $(g)$
Ionic compounds

","correct":false,"explanation":"Aqueous ionic compounds are split into ions, unlike insoluble solids and gases. Covalent molecules also remain together, so both the aqueous and ionic categories are misplaced here."},{"text":"

Solids $(s)$
Liquids $(l)$
Aqueous solutions $(aq)$
Covalent molecules

","correct":false,"explanation":"Aqueous ionic substances are dissociated when writing the full ionic equation. Gases remain unsplit, while solids, liquids and covalent molecules are kept as written."}]},{"id":"208b5a40-2da8-4f62-860c-5deffce4ac6e","cardType":"basic","points":1,"front":"State the net ionic equation for the precipitation of silver chloride.","back":"$$\\text{Ag}^+(aq) + \\text{Cl}^-(aq) \\to \\text{AgCl}(s)$","mcqOptions":[{"text":"$$\\text{Ag}^+(aq) + \\text{Cl}^-(aq) \\to \\text{AgCl}(s)$","correct":true,"explanation":"Silver ions and chloride ions combine in solution to form insoluble silver chloride. Atoms and charge balance, with the neutral precipitate shown as $(s)$."},{"text":"$$\\text{Ag}^+(aq) + \\text{Cl}^-(aq) \\to \\text{AgCl}(aq)$","correct":false,"explanation":"Silver chloride is insoluble, so it precipitates as a solid rather than remaining aqueous. The correct equation ends with $\\text{AgCl}(s)$, while the ions combine in a 1:1 ratio."},{"text":"$$\\text{Ag}^+(aq) + \\text{Cl}^-(aq) \\to \\text{AgCl}_2(s)$","correct":false,"explanation":"The formula of silver chloride is $\\text{AgCl}$ because $\\text{Ag}^+$ and $\\text{Cl}^-$ combine in a 1:1 ratio. Adding a second chloride gives the wrong composition and charge."},{"text":"$$\\text{Ag}^{2+}(aq) + 2\\text{Cl}^-(aq) \\to \\text{AgCl}_2(s)$","correct":false,"explanation":"Silver forms the $+1$ ion in this precipitation, not $+2$. One silver ion therefore combines with one chloride ion to produce neutral solid $\\text{AgCl}$."}]},{"id":"27d090e6-0c08-4265-a0c7-6214a329ebc4","cardType":"cloze","points":null,"front":"An ionic equation is only correct if both the [ ] and the total charge are balanced.","back":"An ionic equation is only correct if both the number of atoms and the total charge are balanced.","mcqOptions":[{"text":"number of atoms","correct":true,"explanation":"An ionic equation must conserve each element, so the number of atoms of every element must match on both sides. It must also conserve total electric charge."},{"text":"number of molecules","correct":false,"explanation":"Ionic equations may contain individual ions rather than molecules, so molecule count is not the required conservation rule. The number of atoms and total charge must both balance."},{"text":"number of electrons","correct":false,"explanation":"Electron transfer may be tracked in redox equations, but the standard ionic-equation check is conservation of atoms and total charge. Electron count is not the missing general condition here."},{"text":"number of compounds","correct":false,"explanation":"Ionic equations can involve ions and do not require the same number of compounds on each side. The essential checks are matching each element’s atoms and balancing charge."}]},{"id":"daa1b432-3f12-435e-b85f-ef2d21ce9f59","cardType":"cloze","points":null,"front":"An ionic equation is only correct if both the number of atoms and the [ ] are balanced.","back":"An ionic equation is only correct if both the number of atoms and the total charge are balanced.","mcqOptions":[{"text":"total charge","correct":true,"explanation":"Charge must be conserved in an ionic equation, so the algebraic total charge on the reactant side must equal that on the product side, alongside conservation of atoms."},{"text":"number of ions","correct":false,"explanation":"The number of ions can change when ions combine into a precipitate or dissociate. What must remain equal is total charge, together with the number of atoms of each element."},{"text":"mass of ions","correct":false,"explanation":"Mass conservation follows from conserving atoms, but ionic equations are specifically checked for total charge as well. Counting ionic mass is not the required second balancing condition."},{"text":"number of electrons transferred","correct":false,"explanation":"Electron transfer is relevant to balancing redox half-equations, but general ionic equations require equal total charge on both sides. The missing condition is total charge, not electron count."}]},{"id":"a9564313-0eb5-4f76-8fff-2940effb95c7","cardType":"basic","points":2,"front":"What are the chemical formulae for Sulfuric acid and Nitric acid?","back":"

Sulfuric acid: $\\text{H}_2\\text{SO}_4$
Nitric acid: $\\text{HNO}_3$

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

Sulfuric acid: $\\text{H}_2\\text{SO}_4$
Nitric acid: $\\text{HNO}_3$

","correct":true,"explanation":"Sulfuric acid contains two acidic hydrogens and the sulfate group, giving $\\text{H}_2\\text{SO}_4$. Nitric acid contains one hydrogen and nitrate, giving $\\text{HNO}_3$."},{"text":"

Sulfuric acid: $\\text{H}_2\\text{SO}_3$
Nitric acid: $\\text{HNO}_2$

","correct":false,"explanation":"These formulae are commonly confused with sulfurous acid and nitrous acid. Sulfuric acid is $\\text{H}_2\\text{SO}_4$, while nitric acid is $\\text{HNO}_3$."},{"text":"

Sulfuric acid: $\\text{HSO}_4$
Nitric acid: $\\text{H}_2\\text{NO}_3$

","correct":false,"explanation":"The hydrogen counts must match the charges of the acid anions. Sulfate requires two hydrogens and nitrate requires one, giving $\\text{H}_2\\text{SO}_4$ and $\\text{HNO}_3$."},{"text":"

Sulfuric acid: $\\text{H}_2\\text{S}_2\\text{O}_4$
Nitric acid: $\\text{HNO}_3$

","correct":false,"explanation":"The sulfuric acid formula contains one sulfur atom, not two; the subscript must not be duplicated. Its correct formula is $\\text{H}_2\\text{SO}_4$, while nitric acid is $\\text{HNO}_3$."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Component 1"},"heading":"Formulae and equations","title":"Formulae and equations Flashcards","topbar":{"backHref":"/a-level/chemistry/eduqas/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"c7578f81-a1b9-5d9a-89de-f07091574ffe","walled":"$L37"}],"$L38","$L39"]