answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
The law of conservation of mass states matter is neither created nor destroyed in reactions. | null | che_0451 | concept | en | What is the law of conservation of mass? | [
"Atoms rearrange only.",
"Total mass stays fixed.",
"Basis of balancing."
] | chemistry | Answer: Mass is conserved in reactions.
Steps:
1. Atoms rearrange only.
2. Total mass stays fixed.
3. Basis of balancing. | stoichiometry |
One mole of helium holds 6.022 x 10^23 atoms. | {
"A": "6.022 x 10^23",
"B": "1.204 x 10^24",
"C": "3.011 x 10^23",
"D": "1.00 x 10^23"
} | che_0452 | multiple_choice | en | How many atoms are in 1.00 mol of helium? | [
"One mole means Avogadro's count.",
"Value 6.022 x 10^23.",
"Select option A."
] | chemistry | Answer: 6.022 x 10^23.
Steps:
1. One mole means Avogadro's count.
2. Value 6.022 x 10^23.
3. Select option A.
Answer: (A) | stoichiometry |
The mass is 19.70 g, from 0.100 mol times 196.97 g/mol. | null | che_0453 | worked_example | en | What mass is 0.100 mol of gold (M = 196.97 g/mol)? | [
"Write: m = 0.100 x 196.97.",
"Multiply.",
"m = 19.70 g."
] | chemistry | Answer: 19.70 g.
Steps:
1. Write: m = 0.100 x 196.97.
2. Multiply.
3. m = 19.70 g. | stoichiometry |
Relativní atomová hmotnost je průměrná hmotnost atomu vůči 1/12 uhlíku 12. | null | che_0454 | concept | cs | Co je relativní atomová hmotnost? | [
"Průměr přes izotopy.",
"Jednotka 1/12 12C.",
"Bezrozměrné číslo."
] | chemistry | Answer: Průměrná hmotnost atomu.
Steps:
1. Průměr přes izotopy.
2. Jednotka 1/12 12C.
3. Bezrozměrné číslo. | stoichiometry |
There are 2.00 mol, from 8.00 g divided by 4.003 g/mol. | null | che_0455 | worked_example | en | How many moles are in 8.00 g of helium (M = 4.003 g/mol)? | [
"Write: n = 8.00/4.003.",
"Divide.",
"n = 2.00 mol."
] | chemistry | Answer: 2.00 mol.
Steps:
1. Write: n = 8.00/4.003.
2. Divide.
3. n = 2.00 mol. | stoichiometry |
Sulfuric acid totals 98.08 g/mol from 2 H, 1 S, and 4 O. | {
"A": "49.04 g/mol",
"B": "98.08 g/mol",
"C": "114.08 g/mol",
"D": "82.08 g/mol"
} | che_0456 | multiple_choice | en | What is the molar mass of H2SO4? | [
"2x1.008 + 32.06 + 4x16.00.",
"Total 98.08.",
"Select option B."
] | chemistry | Answer: 98.08 g/mol.
Steps:
1. 2x1.008 + 32.06 + 4x16.00.
2. Total 98.08.
3. Select option B.
Answer: (B) | stoichiometry |
Formula mass is per formula unit in amu while molar mass is per mole in g/mol. | null | che_0457 | concept | en | How do formula mass and molar mass differ? | [
"Formula mass: one unit, amu.",
"Molar mass: one mole, g/mol.",
"Numerically equal."
] | chemistry | Answer: Per unit versus per mole.
Steps:
1. Formula mass: one unit, amu.
2. Molar mass: one mole, g/mol.
3. Numerically equal. | stoichiometry |
The mass is 150.1 g, from 1.50 mol times 100.09 g/mol. | null | che_0458 | worked_example | en | What mass is 1.50 mol of CaCO3 (M = 100.09 g/mol)? | [
"Write: m = 1.50 x 100.09.",
"Multiply.",
"m = 150.1 g."
] | chemistry | Answer: 150.1 g.
Steps:
1. Write: m = 1.50 x 100.09.
2. Multiply.
3. m = 150.1 g. | stoichiometry |
Je to 0,50 molu, protože 20 děleno 40,08 se rovná 0,50. | null | che_0459 | worked_example | cs | Kolik molů je ve 20 g vápníku (M = 40,08 g/mol)? | [
"Napiš vzorec: n = 20/40,08.",
"Vyděl.",
"n = 0,50 mol."
] | chemistry | Answer: 0,50 mol.
Steps:
1. Napiš vzorec: n = 20/40,08.
2. Vyděl.
3. n = 0,50 mol. | stoichiometry |
The balanced equation is C3H8 + 5O2 -> 3CO2 + 4H2O. | {
"A": "1, 3, 3, 4",
"B": "2, 5, 3, 4",
"C": "1, 5, 3, 4",
"D": "1, 5, 4, 3"
} | che_0460 | multiple_choice | en | What coefficients balance C3H8 + O2 -> CO2 + H2O? | [
"C: 3 CO2. H: 4 H2O.",
"O: 5 O2.",
"Select option C."
] | chemistry | Answer: 1, 5, 3, 4.
Steps:
1. C: 3 CO2. H: 4 H2O.
2. O: 5 O2.
3. Select option C.
Answer: (C) | stoichiometry |
The stoichiometric map routes mass to moles, across the ratio, then back to mass. | null | che_0461 | concept | en | What is the stoichiometric map? | [
"Mass to moles via M.",
"Moles across via ratio.",
"Moles to mass via M."
] | chemistry | Answer: Mass-mole-mole-mass route.
Steps:
1. Mass to moles via M.
2. Moles across via ratio.
3. Mole to mass via M. | stoichiometry |
Three moles of H2 are needed, from 2 mol NH3 times the 3/2 ratio. | null | che_0462 | worked_example | en | How much H2 is needed to make 2.0 mol NH3 (N2 + 3H2 -> 2NH3)? | [
"Ratio H2:NH3 = 3:2.",
"Multiply: 2.0 x 3/2.",
"H2 = 3.0 mol."
] | chemistry | Answer: 3.0 mol.
Steps:
1. Ratio H2:NH3 = 3:2.
2. Multiply: 2.0 x 3/2.
3. H2 = 3.0 mol. | stoichiometry |
Hydrogen limits because 3 mol H2 pairs with only 1 mol N2, leaving N2 over. | {
"A": "H2 limits",
"B": "N2 limits",
"C": "Neither limits",
"D": "Both limit"
} | che_0463 | multiple_choice | en | With 3 mol N2 and 3 mol H2 for N2 + 3H2 -> 2NH3, what limits? | [
"3 H2 needs 1 N2.",
"N2 is in excess.",
"Select option A."
] | chemistry | Answer: H2 limits.
Steps:
1. 3 H2 needs 1 N2.
2. N2 is in excess.
3. Select option A.
Answer: (A) | stoichiometry |
Nadbytek reaktantu je množství zbývající po spotřebování limitujícího reaktantu. | null | che_0464 | concept | cs | Co je nadbytek reaktantu? | [
"Více, než reakce spotřebuje.",
"Zbude po reakci.",
"Opak limitujícího reaktantu."
] | chemistry | Answer: Nespotřebovaný zbytek.
Steps:
1. Více, než reakce spotřebuje.
2. Zbude po reakci.
3. Opak limitujícího reaktantu. | stoichiometry |
Excess N2 left is 0.333 mol, from 1.00 minus 0.667 mol consumed. | null | che_0465 | worked_example | en | 1.00 mol N2 reacts with 2.00 mol H2 (N2 + 3H2 -> 2NH3). How much N2 remains? | [
"H2 limits: needs 2.00/3 = 0.667 N2.",
"Subtract: 1.00 - 0.667.",
"N2 left = 0.333 mol."
] | chemistry | Answer: 0.333 mol.
Steps:
1. H2 limits: needs 2.00/3 = 0.667 N2.
2. Subtract: 1.00 - 0.667.
3. N2 left = 0.333 mol. | stoichiometry |
The molarity is 0.40 M, from 0.200 mol divided by 0.500 L. | {
"A": "0.10 M",
"B": "0.40 M",
"C": "1.00 M",
"D": "2.50 M"
} | che_0466 | multiple_choice | en | What is the molarity of 0.200 mol in 0.500 L? | [
"Write: M = 0.200/0.500.",
"Divide: 0.40.",
"Select option B."
] | chemistry | Answer: 0.40 M.
Steps:
1. Write: M = 0.200/0.500.
2. Divide: 0.40.
3. Select option B.
Answer: (B) | stoichiometry |
The volume is 0.200 L, from 0.100 mol divided by 0.500 M. | null | che_0467 | worked_example | en | What volume of 0.500 M solution holds 0.100 mol? | [
"Write: V = n/M.",
"Substitute: 0.100/0.500.",
"V = 0.200 L."
] | chemistry | Answer: 0.200 L.
Steps:
1. Write: V = n/M.
2. Substitute: 0.100/0.500.
3. V = 0.200 L. | stoichiometry |
Koncentrace je 0,2 M, protože 0,3 děleno 1,5 se rovná 0,2. | null | che_0468 | worked_example | cs | Jaká je koncentrace 0,3 molu v 1,5 l? | [
"Napiš vzorec: M = 0,3/1,5.",
"Vyděl.",
"M = 0,2 mol/l."
] | chemistry | Answer: 0,2 M.
Steps:
1. Napiš vzorec: M = 0,3/1,5.
2. Vyděl.
3. M = 0,2 mol/l. | stoichiometry |
The volume is 67.2 L, from 3.00 mol times 22.414 L/mol. | {
"A": "22.4 L",
"B": "44.8 L",
"C": "33.6 L",
"D": "67.2 L"
} | che_0469 | multiple_choice | en | What volume does 3.00 mol of gas occupy at STP? | [
"Multiply: 3.00 x 22.414.",
"Result 67.2 L.",
"Select option D."
] | chemistry | Answer: 67.2 L.
Steps:
1. Multiply: 3.00 x 22.414.
2. Result 67.2 L.
3. Select option D.
Answer: (D) | stoichiometry |
Arrhenius acids release H+ and bases release OH- in water. | null | che_0470 | concept | en | What is the Arrhenius definition? | [
"Acid gives H+ in water.",
"Base gives OH- in water.",
"Limited to aqueous solutions."
] | chemistry | Answer: H+ versus OH- in water.
Steps:
1. Acid gives H+ in water.
2. Base gives OH- in water.
3. Limited to aqueous solutions. | acids_bases |
An Arrhenius base yields hydroxide ions in water. | {
"A": "H+",
"B": "Electrons",
"C": "OH-",
"D": "Salts"
} | che_0471 | multiple_choice | en | What does an Arrhenius base produce in water? | [
"Recall OH- release.",
"Eliminate H+, electrons, salts.",
"Select option C."
] | chemistry | Answer: OH-.
Steps:
1. Recall OH- release.
2. Eliminate H+, electrons, salts.
3. Select option C.
Answer: (C) | acids_bases |
The pH is 3.70, from -log(2.0 x 10^-4). | null | che_0472 | worked_example | en | Compute the pH of 2.0 x 10^-4 M HCl. | [
"[H+] = 2.0 x 10^-4 M.",
"pH = -log(2.0 x 10^-4).",
"pH = 3.70."
] | chemistry | Answer: 3.70.
Steps:
1. [H+] = 2.0 x 10^-4 M.
2. pH = -log(2.0 x 10^-4).
3. pH = 3.70. | acids_bases |
Arrheniova kyselina uvolňuje ve vodě ionty H+. | null | che_0473 | concept | cs | Co je Arrheniova kyselina? | [
"Uvolňuje H+ ve vodě.",
"Příklad: HCl.",
"Platí jen ve vodném prostředí."
] | chemistry | Answer: Zdroj H+ ve vodě.
Steps:
1. Uvolňuje H+ ve vodě.
2. Příklad: HCl.
3. Platí jen ve vodném prostředí. | acids_bases |
Hydrofluoric acid ionizes only partly, so it is weak. | {
"A": "HCl",
"B": "HF",
"C": "HNO3",
"D": "HBr"
} | che_0474 | multiple_choice | en | Which of these is a weak acid? | [
"Strong: HCl, HNO3, HBr.",
"HF is famously weak.",
"Select option B."
] | chemistry | Answer: HF.
Steps:
1. Strong: HCl, HNO3, HBr.
2. HF is famously weak.
3. Select option B.
Answer: (B) | acids_bases |
Percent ionization is dissociated acid over initial concentration times 100. | null | che_0475 | concept | en | What is percent ionization? | [
"Divide [H+] by initial C.",
"Multiply by 100.",
"Higher means stronger acid."
] | chemistry | Answer: Share of acid dissociated.
Steps:
1. Divide [H+] by initial C.
2. Multiply by 100.
3. Higher means stronger acid. | acids_bases |
Percent ionization is 1.34%, from 1.34 x 10^-3 divided by 0.10. | null | che_0476 | worked_example | en | 0.10 M acetic acid gives [H+] = 1.34 x 10^-3 M. Compute percent ionization. | [
"Divide: 1.34 x 10^-3 / 0.10 = 0.0134.",
"Multiply by 100.",
"Ionization = 1.34%."
] | chemistry | Answer: 1.34%.
Steps:
1. Divide: 1.34 x 10^-3 / 0.10 = 0.0134.
2. Multiply by 100.
3. Ionization = 1.34%. | acids_bases |
The first dissociation dominates pH because later Ka values are tiny. | null | che_0477 | concept | en | Why does the first step dominate polyprotic pH? | [
"Ka1 far exceeds Ka2.",
"First H+ dwarfs the rest.",
"Approximate with Ka1 only."
] | chemistry | Answer: Ka1 dwarfs later steps.
Steps:
1. Ka1 far exceeds Ka2.
2. First H+ dwarfs the rest.
3. Approximate with Ka1 only. | acids_bases |
Ka2 of carbonic acid is far smaller than Ka1, so the second proton is very weak. | {
"A": "Large",
"B": "Equal to Ka1",
"C": "Strong",
"D": "Much smaller than Ka1"
} | che_0478 | multiple_choice | en | How does Ka2 of H2CO3 compare to Ka1? | [
"Second proton harder to lose.",
"Ka2 << Ka1.",
"Select option D."
] | chemistry | Answer: Much smaller than Ka1.
Steps:
1. Second proton harder to lose.
2. Ka2 << Ka1.
3. Select option D.
Answer: (D) | acids_bases |
pH je 1,70, protože -log(0,02) = 1,70. | null | che_0479 | worked_example | cs | Vypočítej pH roztoku 0,02 M HCl. | [
"HCl disociuje plně.",
"pH = -log(0,02).",
"pH = 1,70."
] | chemistry | Answer: 1,70.
Steps:
1. HCl disociuje plně.
2. pH = -log(0,02).
3. pH = 1,70. | acids_bases |
The leveling effect caps acid strength in water at H3O+. | null | che_0480 | concept | en | What is the leveling effect of water? | [
"Strong acids protonate water.",
"All become H3O+.",
"No stronger acid survives."
] | chemistry | Answer: Water caps strength at H3O+.
Steps:
1. Strong acids protonate water.
2. All become H3O+.
3. No stronger acid survives. | acids_bases |
Hydronium is the strongest acid that can exist in water. | {
"A": "H3O+",
"B": "HCl",
"C": "H2SO4",
"D": "CH3COOH"
} | che_0481 | multiple_choice | en | What is the strongest acid in water? | [
"Stronger acids level to H3O+.",
"H3O+ is the cap.",
"Select option A."
] | chemistry | Answer: H3O+.
Steps:
1. Stronger acids level to H3O+.
2. H3O+ is the cap.
3. Select option A.
Answer: (A) | acids_bases |
The base-to-acid ratio is 10:1, since log(ratio) = 5.76 - 4.76 = 1. | null | che_0482 | worked_example | en | An acetate buffer (pKa = 4.76) has pH 5.76. Find the base/acid ratio. | [
"Write: 5.76 = 4.76 + log(ratio).",
"log(ratio) = 1.",
"Ratio = 10:1."
] | chemistry | Answer: 10:1.
Steps:
1. Write: 5.76 = 4.76 + log(ratio).
2. log(ratio) = 1.
3. Ratio = 10:1. | acids_bases |
Buffer capacity is the amount of acid or base a buffer absorbs before pH shifts sharply. | null | che_0483 | concept | en | What is buffer capacity? | [
"Depends on concentrations.",
"Higher means tougher buffer.",
"Exhausted when a partner runs out."
] | chemistry | Answer: Absorbable acid/base amount.
Steps:
1. Depends on concentrations.
2. Higher means tougher buffer.
3. Exhausted when a partner runs out. | acids_bases |
Disociační konstanta Ka vyjadřuje sílu kyseliny v rovnováze. | null | che_0484 | concept | cs | Co je disociační konstanta? | [
"Rovnováha disociace.",
"Větší Ka = silnější kyselina.",
"pKa = -log Ka."
] | chemistry | Answer: Konstanta síly kyseliny.
Steps:
1. Rovnováha disociace.
2. Větší Ka = silnější kyselina.
3. pKa = -log Ka. | acids_bases |
Back-titration handles slow or insoluble analytes by reacting excess reagent first. | {
"A": "Strong-strong only",
"B": "Slow or insoluble analytes",
"C": "Gases only",
"D": "Never used"
} | che_0485 | multiple_choice | en | When is back-titration used? | [
"Direct endpoint unclear.",
"Excess reagent added first.",
"Select option B."
] | chemistry | Answer: Slow or insoluble analytes.
Steps:
1. Direct endpoint unclear.
2. Excess reagent added first.
3. Select option B.
Answer: (B) | acids_bases |
There are 7.50 mmol, from 15.0 mL times 0.500 M. | null | che_0486 | worked_example | en | How many millimoles are in 15.0 mL of 0.500 M base? | [
"Write: mmol = 15.0 x 0.500.",
"Multiply.",
"mmol = 7.50."
] | chemistry | Answer: 7.50 mmol.
Steps:
1. Write: mmol = 15.0 x 0.500.
2. Multiply.
3. mmol = 7.50. | acids_bases |
[H+] je 1 x 10^-4 M, protože 10 na -4 odpovídá pH 4. | null | che_0487 | worked_example | cs | Roztok má pH 4. Vypočítej [H+]. | [
"Napiš vzorec: [H+] = 10^(-4).",
"Dosaď.",
"[H+] = 1 x 10^-4 M."
] | chemistry | Answer: 1 x 10^-4 M.
Steps:
1. Napiš vzorec: [H+] = 10^(-4).
2. Dosaď.
3. [H+] = 1 x 10^-4 M. | acids_bases |
The solution is basic because Kb of NH3 exceeds Ka of HCN. | {
"A": "Acidic",
"B": "Neutral",
"C": "Basic",
"D": "Cannot tell"
} | che_0488 | multiple_choice | en | Is NH4CN solution acidic or basic (Kb NH3 = 1.8e-5, Ka HCN = 6.2e-10)? | [
"Compare Ka and Kb.",
"Kb larger means basic.",
"Select option C."
] | chemistry | Answer: Basic.
Steps:
1. Compare Ka and Kb.
2. Kb larger means basic.
3. Select option C.
Answer: (C) | acids_bases |
Alkali metals always take oxidation state +1 in compounds. | null | che_0489 | concept | en | What oxidation state do alkali metals take? | [
"Group 1 loses one electron.",
"State is +1 always.",
"Example: Na in NaCl."
] | chemistry | Answer: +1 always.
Steps:
1. Group 1 loses one electron.
2. State is +1 always.
3. Example: Na in NaCl. | redox |
Sodium is +1 in sodium peroxide, as in all its compounds. | {
"A": "+1",
"B": "-1",
"C": "0",
"D": "+2"
} | che_0490 | multiple_choice | en | What is the oxidation state of Na in Na2O2? | [
"Alkali metals are +1.",
"Applies here too.",
"Select option A."
] | chemistry | Answer: +1.
Steps:
1. Alkali metals are +1.
2. Applies here too.
3. Select option A.
Answer: (A) | redox |
Chlorine is +5, since Cl + 3(-2) = -1 gives Cl = +5. | null | che_0491 | worked_example | en | Find the oxidation number of Cl in ClO3-. | [
"Assign O -2.",
"Equation: Cl - 6 = -1.",
"Cl = +5."
] | chemistry | Answer: +5.
Steps:
1. Assign O -2.
2. Equation: Cl - 6 = -1.
3. Cl = +5. | redox |
Oxidace kovů vzduchem je jejich samovolná reakce s kyslíkem. | null | che_0492 | concept | cs | Co je oxidace kovů vzduchem? | [
"Kov odevzdá elektrony.",
"Kyslík se redukuje.",
"Vznikají oxidy, např. rez."
] | chemistry | Answer: Tvorba oxidů na kovu.
Steps:
1. Kov odevzdá elektrony.
2. Kyslík se redukuje.
3. Vznikají oxidy, např. rez. | redox |
Iron drops from +3 to 0, so it is reduced. | {
"A": "Carbon",
"B": "Iron",
"C": "Oxygen",
"D": "CO"
} | che_0493 | multiple_choice | en | In Fe2O3 + 3CO -> 2Fe + 3CO2, what is reduced? | [
"Fe goes +3 to 0.",
"That is reduction.",
"Select option B."
] | chemistry | Answer: Iron.
Steps:
1. Fe goes +3 to 0.
2. That is reduction.
3. Select option B.
Answer: (B) | redox |
The balanced ionic equation is Zn + 2Ag+ -> Zn2+ + 2Ag. | null | che_0494 | worked_example | en | Balance: Zn + Ag+ -> Zn2+ + Ag. | [
"Zn loses 2 e-.",
"Need 2 Ag+ to gain them.",
"Result: Zn + 2Ag+ -> Zn2+ + 2Ag."
] | chemistry | Answer: Zn + 2Ag+ -> Zn2+ + 2Ag.
Steps:
1. Zn loses 2 e-.
2. Need 2 Ag+ to gain them.
3. Result: Zn + 2Ag+ -> Zn2+ + 2Ag. | redox |
The oxidation number method tracks electron loss and gain to find coefficients. | null | che_0495 | concept | en | What is the oxidation number method? | [
"Assign all oxidation states.",
"Find changes up and down.",
"Scale to equalize electrons."
] | chemistry | Answer: Balance via electron changes.
Steps:
1. Assign all oxidation states.
2. Find changes up and down.
3. Scale to equalize electrons. | redox |
Six electrons are lost as sulfur rises from -2 to +4. | null | che_0496 | worked_example | en | How many electrons are lost in H2S -> SO2 (S: -2 to +4)? | [
"Change: -2 to +4.",
"Difference = 6 up.",
"Electrons lost = 6."
] | chemistry | Answer: 6 electrons.
Steps:
1. Change: -2 to +4.
2. Difference = 6 up.
3. Electrons lost = 6. | redox |
Sodium metal donates electrons eagerly, making it a strong reducing agent. | {
"A": "Na",
"B": "Cl2",
"C": "O2",
"D": "F2"
} | che_0497 | multiple_choice | en | Which is the best reducing agent? | [
"Reducers donate electrons.",
"Alkali metal Na tops list.",
"Select option A."
] | chemistry | Answer: Na.
Steps:
1. Reducers donate electrons.
2. Alkali metal Na tops list.
3. Select option A.
Answer: (A) | redox |
Železo má oxidační číslo +2, protože Fe + (-2) = 0. | null | che_0498 | worked_example | cs | Urči oxidační číslo železa v FeO. | [
"Přiřaď O -2.",
"Rovnice: Fe - 2 = 0.",
"Fe = +2."
] | chemistry | Answer: +2.
Steps:
1. Přiřaď O -2.
2. Rovnice: Fe - 2 = 0.
3. Fe = +2. | redox |
Corrosion cells pair anodic metal dissolving with cathodic oxygen reduction. | null | che_0499 | concept | en | What is a corrosion cell? | [
"Anode: metal oxidizes.",
"Cathode: O2 reduces.",
"Moisture conducts ions."
] | chemistry | Answer: Anode-cathode rust pair.
Steps:
1. Anode: metal oxidizes.
2. Cathode: O2 reduces.
3. Moisture conducts ions. | redox |
Paint blocks air and water from reaching the iron surface. | {
"A": "Oxidizes Fe faster",
"B": "Adds salt",
"C": "Blocks air and water",
"D": "Heats the metal"
} | che_0500 | multiple_choice | en | How does painting prevent rust? | [
"Rust needs O2 and H2O.",
"Paint seals them out.",
"Select option C."
] | chemistry | Answer: Blocks air and water.
Steps:
1. Rust needs O2 and H2O.
2. Paint seals them out.
3. Select option C.
Answer: (C) | redox |
The balanced equation is 2Al + 6HCl -> 2AlCl3 + 3H2. | null | che_0501 | worked_example | en | Balance: Al + HCl -> AlCl3 + H2. | [
"Balance Cl and Al with 2 AlCl3, 6 HCl, 2 Al.",
"Check H: 6 each side with 3 H2.",
"Result: 2Al + 6HCl -> 2AlCl3 + 3H2."
] | chemistry | Answer: 2Al + 6HCl -> 2AlCl3 + 3H2.
Steps:
1. Balance Cl and Al with 2 AlCl3, 6 HCl, 2 Al.
2. Check H: 6 each side with 3 H2.
3. Result: 2Al + 6HCl -> 2AlCl3 + 3H2. | redox |
Redoxní pár tvoří oxidovaná a redukovaná forma téže látky. | null | che_0502 | concept | cs | Co je redoxní pár? | [
"Dvě formy jedné látky.",
"Liší se o elektrony.",
"Příklad: Fe3+/Fe2+."
] | chemistry | Answer: Oxidovaná a redukovaná forma.
Steps:
1. Dvě formy jedné látky.
2. Liší se o elektrony.
3. Příklad: Fe3+/Fe2+. | redox |
Carbon is +4, since C + 2(-2) = 0 gives C = +4. | {
"A": "-4",
"B": "0",
"C": "+2",
"D": "+4"
} | che_0503 | multiple_choice | en | What is the oxidation state of carbon in CO2? | [
"Assign O -2.",
"Solve: C - 4 = 0.",
"C = +4; select D."
] | chemistry | Answer: +4.
Steps:
1. Assign O -2.
2. Solve: C - 4 = 0.
3. C = +4; select D.
Answer: (D) | redox |
Hydrogen peroxide disproportionates into water and oxygen: 2H2O2 -> 2H2O + O2. | null | che_0504 | concept | en | How does hydrogen peroxide disproportionate? | [
"O is -1 in H2O2.",
"Part goes to -2 in water.",
"Part goes to 0 in O2."
] | chemistry | Answer: 2H2O2 -> 2H2O + O2.
Steps:
1. O is -1 in H2O2.
2. Part goes to -2 in water.
3. Part goes to 0 in O2. | redox |
Vyčíslená rovnice je 6Li + N2 -> 2Li3N. | null | che_0505 | worked_example | cs | Vyčísli: Li + N2 -> Li3N. | [
"Dusík: 2 Li3N a 1 N2.",
"Lithium: 6 Li.",
"Výsledek: 6Li + N2 -> 2Li3N."
] | chemistry | Answer: 6Li + N2 -> 2Li3N.
Steps:
1. Dusík: 2 Li3N a 1 N2.
2. Lithium: 6 Li.
3. Výsledek: 6Li + N2 -> 2Li3N. | redox |
Oxygen with fluorine is +2, since O + 2(-1) = 0. | {
"A": "-2",
"B": "+2",
"C": "0",
"D": "-1"
} | che_0506 | multiple_choice | en | What is oxygen's oxidation state in OF2? | [
"F is -1 each.",
"Solve: O - 2 = 0.",
"O = +2; select B."
] | chemistry | Answer: +2.
Steps:
1. F is -1 each.
2. Solve: O - 2 = 0.
3. O = +2; select B.
Answer: (B) | redox |
One electron is lost as Fe2+ becomes Fe3+. | null | che_0507 | worked_example | en | How many electrons are lost in Fe2+ -> Fe3+? | [
"Change: +2 to +3.",
"Difference = 1 up.",
"Electrons lost = 1."
] | chemistry | Answer: 1 electron.
Steps:
1. Change: +2 to +3.
2. Difference = 1 up.
3. Electrons lost = 1. | redox |
Open systems exchange matter and energy; closed systems exchange energy only. | null | che_0508 | concept | en | How do open and closed systems differ? | [
"Open: matter crosses.",
"Closed: only energy crosses.",
"Isolated: neither crosses."
] | chemistry | Answer: Matter exchange versus energy only.
Steps:
1. Open: matter crosses.
2. Closed: only energy crosses.
3. Isolated: neither crosses. | thermochemistry |
A closed system exchanges energy but not matter. | {
"A": "Matter only",
"B": "Energy only",
"C": "Neither",
"D": "Both freely"
} | che_0509 | multiple_choice | en | What does a closed system exchange? | [
"Matter sealed inside.",
"Heat and work pass.",
"Select option B."
] | chemistry | Answer: Energy only.
Steps:
1. Matter sealed inside.
2. Heat and work pass.
3. Select option B.
Answer: (B) | thermochemistry |
The internal energy change is -150 J, from -300 plus +150 J. | null | che_0510 | worked_example | en | A system releases 300 J of heat while 150 J of work is done on it. Find ΔU. | [
"Write: ΔU = -300 + 150.",
"Add.",
"ΔU = -150 J."
] | chemistry | Answer: -150 J.
Steps:
1. Write: ΔU = -300 + 150.
2. Add.
3. ΔU = -150 J. | thermochemistry |
Vnitřní energie je součet kinetické a potenciální energie částic soustavy. | null | che_0511 | concept | cs | Co je vnitřní energie? | [
"Energie částic soustavy.",
"Mění se teplem a prací.",
"ΔU = q + w."
] | chemistry | Answer: Energie částic soustavy.
Steps:
1. Energie částic soustavy.
2. Mění se teplem a prací.
3. ΔU = q + w. | thermochemistry |
Heat leaving the system is negative by convention. | {
"A": "Negative",
"B": "Positive",
"C": "Zero",
"D": "Infinite"
} | che_0512 | multiple_choice | en | What sign does released heat (q) have? | [
"System loses energy.",
"Loss is negative.",
"Select option A."
] | chemistry | Answer: Negative.
Steps:
1. System loses energy.
2. Loss is negative.
3. Select option A.
Answer: (A) | thermochemistry |
The water releases 62.8 kJ, from 250 x 4.184 x (-60) J. | null | che_0513 | worked_example | en | 250 g of water cools from 90 C to 30 C. Find q. | [
"ΔT = 30 - 90 = -60 C.",
"q = 250 x 4.184 x (-60).",
"q = -62,760 J = -62.8 kJ."
] | chemistry | Answer: -62.8 kJ.
Steps:
1. ΔT = 30 - 90 = -60 C.
2. q = 250 x 4.184 x (-60).
3. q = -62,760 J = -62.8 kJ. | thermochemistry |
Thermal equilibrium means no net heat flow because temperatures are equal. | null | che_0514 | concept | en | What is thermal equilibrium? | [
"Bodies in contact.",
"Temperatures equalize.",
"Net heat flow stops."
] | chemistry | Answer: Equal temperatures, no flow.
Steps:
1. Bodies in contact.
2. Temperatures equalize.
3. Net heat flow stops. | thermochemistry |
The final temperature is 50 C, the average of 80 and 20. | null | che_0515 | worked_example | en | 100 g of water at 80 C mixes with 100 g at 20 C. Find the final T. | [
"Equal masses and material.",
"Average: (80 + 20)/2.",
"T = 50 C."
] | chemistry | Answer: 50 C.
Steps:
1. Equal masses and material.
2. Average: (80 + 20)/2.
3. T = 50 C. | thermochemistry |
Heat scales with mass, so doubling mass doubles q for the same ΔT. | {
"A": "Halves q",
"B": "No change",
"C": "Doubles q",
"D": "Quadruples q"
} | che_0516 | multiple_choice | en | Doubling mass with the same ΔT does what to q? | [
"q = m c ΔT.",
"m doubles, rest fixed.",
"Select option C."
] | chemistry | Answer: Doubles q.
Steps:
1. q = m c ΔT.
2. m doubles, rest fixed.
3. Select option C.
Answer: (C) | thermochemistry |
Potřebné teplo je 12 552 J, protože 300 x 4,184 x 10 se rovná 12 552. | null | che_0517 | worked_example | cs | Kolik tepla ohřeje 300 g vody o 10 °C? | [
"Napiš vzorec: q = 300 x 4,184 x 10.",
"Vynásob.",
"q = 12 552 J."
] | chemistry | Answer: 12 552 J.
Steps:
1. Napiš vzorec: q = 300 x 4,184 x 10.
2. Vynásob.
3. q = 12 552 J. | thermochemistry |
Molar heat capacity is the heat to warm one mole by one degree. | null | che_0518 | concept | en | What is molar heat capacity? | [
"Per mole basis.",
"Times ΔT gives q.",
"Water is 75.4 J/molC."
] | chemistry | Answer: Heat per mole per degree.
Steps:
1. Per mole basis.
2. Times ΔT gives q.
3. Water is 75.4 J/molC. | thermochemistry |
Molar heat capacity uses joules per mole per degree. | {
"A": "J/mol C",
"B": "J/g only",
"C": "kJ/L",
"D": "mol/J"
} | che_0519 | multiple_choice | en | What are molar heat capacity units? | [
"Energy per mole per degree.",
"That is J/mol C.",
"Select option A."
] | chemistry | Answer: J/mol C.
Steps:
1. Energy per mole per degree.
2. That is J/mol C.
3. Select option A.
Answer: (A) | thermochemistry |
The ΔH is -110.5 kJ, from -393.5 minus -283.0 kJ. | null | che_0520 | worked_example | en | Given C+O2->CO2 (-393.5) and CO+1/2O2->CO2 (-283.0 kJ), find C+1/2O2->CO. | [
"Subtract second from first.",
"-393.5 - (-283.0).",
"ΔH = -110.5 kJ."
] | chemistry | Answer: -110.5 kJ.
Steps:
1. Subtract second from first.
2. -393.5 - (-283.0).
3. ΔH = -110.5 kJ. | thermochemistry |
Slučovací teplo je teplo vzniku 1 molu sloučeniny z prvků. | null | che_0521 | concept | cs | Co je slučovací teplo? | [
"Vznik 1 molu z prvků.",
"Standardní stavy.",
"Příklad: voda -285,8 kJ/mol."
] | chemistry | Answer: Teplo vzniku z prvků.
Steps:
1. Vznik 1 molu z prvků.
2. Standardní stavy.
3. Příklad: voda -285,8 kJ/mol. | thermochemistry |
Aqueous H+ is the reference with ΔHf of zero. | {
"A": "100 kJ/mol",
"B": "0 kJ/mol",
"C": "-50 kJ/mol",
"D": "285.8 kJ/mol"
} | che_0522 | multiple_choice | en | What is ΔHf of H+(aq)? | [
"Reference ion defined zero.",
"All ions relative to it.",
"Select option B."
] | chemistry | Answer: 0 kJ/mol.
Steps:
1. Reference ion defined zero.
2. All ions relative to it.
3. Select option B.
Answer: (B) | thermochemistry |
ΔS universe is +20 J/K, so the process is spontaneous. | null | che_0523 | worked_example | en | ΔS sys = +50 J/K and ΔS surr = -30 J/K. Compute ΔS universe. | [
"Write: ΔS = 50 + (-30).",
"Add.",
"ΔS = +20 J/K, spontaneous."
] | chemistry | Answer: +20 J/K.
Steps:
1. Write: ΔS = 50 + (-30).
2. Add.
3. ΔS = +20 J/K, spontaneous. | thermochemistry |
Druhý zákon říká, že entropie izolované soustavy samovolně roste. | null | che_0524 | concept | cs | Co je druhý termodynamický zákon? | [
"Entropie vesmíru roste.",
"Samovolné děje ji zvyšují.",
"Teplo neteče samo do teplejšího."
] | chemistry | Answer: Entropie roste.
Steps:
1. Entropie vesmíru roste.
2. Samovolné děje ji zvyšují.
3. Teplo neteče samo do teplejšího. | thermochemistry |
Absolute zero is 0 K, the unreachable minimum temperature. | {
"A": "0 C",
"B": "-100 C",
"C": "273 K",
"D": "0 K"
} | che_0525 | multiple_choice | en | What is absolute zero? | [
"Kelvin scale starts there.",
"Equals -273.15 C.",
"Select option D."
] | chemistry | Answer: 0 K.
Steps:
1. Kelvin scale starts there.
2. Equals -273.15 C.
3. Select option D.
Answer: (D) | thermochemistry |
Boiling needs 38.6 kJ, from 1.00 mol times 38.6 kJ/mol. | null | che_0526 | worked_example | en | How much heat boils 46.1 g of ethanol (ΔHvap = 38.6 kJ/mol, M = 46.07 g/mol)? | [
"Moles = 46.1/46.07 = 1.00.",
"Multiply: 1.00 x 38.6.",
"Heat = 38.6 kJ."
] | chemistry | Answer: 38.6 kJ.
Steps:
1. Moles = 46.1/46.07 = 1.00.
2. Multiply: 1.00 x 38.6.
3. Heat = 38.6 kJ. | thermochemistry |
Haloalkane class counts carbons on the halogen-bearing carbon: one is primary, three is tertiary. | null | che_0527 | concept | en | How are alkyl halides classified? | [
"Count carbons on C-X carbon.",
"1 primary, 2 secondary, 3 tertiary.",
"Reactivity rises with class in SN1."
] | chemistry | Answer: Primary, secondary, tertiary.
Steps:
1. Count carbons on C-X carbon.
2. 1 primary, 2 secondary, 3 tertiary.
3. Reactivity rises with class in SN1. | organic_naming |
(CH3)3CBr has three carbons on the C-Br carbon, so it is tertiary. | {
"A": "Primary",
"B": "Secondary",
"C": "Tertiary",
"D": "Aryl"
} | che_0528 | multiple_choice | en | How is (CH3)3CBr classified? | [
"Count carbons on C-Br: three.",
"Three means tertiary.",
"Select option C."
] | chemistry | Answer: Tertiary.
Steps:
1. Count carbons on C-Br: three.
2. Three means tertiary.
3. Select option C.
Answer: (C) | organic_naming |
The name is 2-bromopropane: bromine on carbon 2 of three. | null | che_0529 | worked_example | en | Name CH3CHBrCH3. | [
"Count 3 carbons.",
"Br on carbon 2.",
"Name: 2-bromopropane."
] | chemistry | Answer: 2-bromopropane.
Steps:
1. Count 3 carbons.
2. Br on carbon 2.
3. Name: 2-bromopropane. | organic_naming |
Primární alkohol má -OH na uhlíku s jedním dalším uhlíkem. | null | che_0530 | concept | cs | Co je primární alkohol? | [
"-OH na koncovém uhlíku.",
"Jeden sousední uhlík.",
"Příklad: ethanol."
] | chemistry | Answer: -OH na primárním uhlíku.
Steps:
1. -OH na koncovém uhlíku.
2. Jeden sousední uhlík.
3. Příklad: ethanol. | organic_naming |
The -OH sits on a carbon with one neighbor, so it is primary. | {
"A": "Primary",
"B": "Secondary",
"C": "Tertiary",
"D": "Phenol"
} | che_0531 | multiple_choice | en | How is CH3CH2CH2OH classified? | [
"C-OH has one carbon neighbor.",
"One neighbor means primary.",
"Select option A."
] | chemistry | Answer: Primary.
Steps:
1. C-OH has one carbon neighbor.
2. One neighbor means primary.
3. Select option A.
Answer: (A) | organic_naming |
Alkenes number the chain to give the double bond the lowest locant. | null | che_0532 | concept | en | How are alkenes numbered? | [
"Find the double bond.",
"Number from nearer end.",
"Locant goes before -ene."
] | chemistry | Answer: Lowest locant for C=C.
Steps:
1. Find the double bond.
2. Number from nearer end.
3. Locant goes before -ene. | organic_naming |
The name is but-2-ene: double bond starting at carbon 2. | null | che_0533 | worked_example | en | Name CH3CH=CHCH3. | [
"Count 4 carbons.",
"Double bond at carbon 2.",
"Name: but-2-ene."
] | chemistry | Answer: But-2-ene.
Steps:
1. Count 4 carbons.
2. Double bond at carbon 2.
3. Name: but-2-ene. | organic_naming |
Cis-trans isomers differ in substituent sides across a rigid double bond. | null | che_0534 | concept | en | What is cis-trans isomerism? | [
"Double bond blocks rotation.",
"Same side is cis.",
"Opposite side is trans."
] | chemistry | Answer: Sides across C=C.
Steps:
1. Double bond blocks rotation.
2. Same side is cis.
3. Opposite side is trans. | organic_naming |
Cis means both methyl groups sit on the same side of the bond. | {
"A": "Opposite sides",
"B": "Same side",
"C": "No stereochemistry",
"D": "Ring form"
} | che_0535 | multiple_choice | en | What does cis-2-butene look like? | [
"Cis means same side.",
"Both CH3 together.",
"Select option B."
] | chemistry | Answer: Same side.
Steps:
1. Cis means same side.
2. Both CH3 together.
3. Select option B.
Answer: (B) | organic_naming |
CH2=CH2 je ethen. | null | che_0536 | worked_example | cs | Pojmenuj sloučeninu CH2=CH2. | [
"Dva uhlíky.",
"Dvojná vazba = -en.",
"Název: ethen."
] | chemistry | Answer: Ethen.
Steps:
1. Dva uhlíky.
2. Dvojná vazba = -en.
3. Název: ethen. | organic_naming |
Alkynes use the -yne suffix with the lowest locant for the triple bond. | null | che_0537 | concept | en | How are alkynes named? | [
"Find the triple bond.",
"Lowest locant applies.",
"Suffix is -yne."
] | chemistry | Answer: Lowest locant plus -yne.
Steps:
1. Find the triple bond.
2. Lowest locant applies.
3. Suffix is -yne. | organic_naming |
HC triple CH with two carbons is ethyne. | {
"A": "Ethene",
"B": "Ethane",
"C": "Benzene",
"D": "Ethyne"
} | che_0538 | multiple_choice | en | What is the name of HC≡CH? | [
"Two carbons present.",
"Triple bond gives -yne.",
"Select option D."
] | chemistry | Answer: Ethyne.
Steps:
1. Two carbons present.
2. Triple bond gives -yne.
3. Select option D.
Answer: (D) | organic_naming |
The name is propyne: three carbons with a triple bond. | null | che_0539 | worked_example | en | Name CH3C≡CH. | [
"Count 3 carbons.",
"Triple bond gives -yne.",
"Name: propyne."
] | chemistry | Answer: Propyne.
Steps:
1. Count 3 carbons.
2. Triple bond gives -yne.
3. Name: propyne. | organic_naming |
Phenyl is the C6H5- group used as a substituent name. | null | che_0540 | concept | en | What does phenyl mean in naming? | [
"Benzene minus one H.",
"Attaches as substituent.",
"Example: phenylamine."
] | chemistry | Answer: C6H5- substituent.
Steps:
1. Benzene minus one H.
2. Attaches as substituent.
3. Example: phenylamine. | organic_naming |
Fenyl je skupina C6H5- odvozená od benzenu. | null | che_0541 | concept | cs | Co je fenyl? | [
"Benzen bez jednoho H.",
"Substituent v názvech.",
"Příklad: fenylamin."
] | chemistry | Answer: Skupina C6H5-.
Steps:
1. Benzen bez jednoho H.
2. Substituent v názvech.
3. Příklad: fenylamin. | organic_naming |
The C6H5- fragment is called phenyl. | {
"A": "Phenyl",
"B": "Benzyl",
"C": "Methyl",
"D": "Ethyl"
} | che_0542 | multiple_choice | en | What is the C6H5- group called? | [
"Benzene ring fragment.",
"Named phenyl.",
"Select option A."
] | chemistry | Answer: Phenyl.
Steps:
1. Benzene ring fragment.
2. Named phenyl.
3. Select option A.
Answer: (A) | organic_naming |
The name is butanoic acid: four carbons ending in -COOH. | null | che_0543 | worked_example | en | Name CH3CH2CH2COOH. | [
"Count 4 carbons.",
"-COOH gives -oic acid.",
"Name: butanoic acid."
] | chemistry | Answer: Butanoic acid.
Steps:
1. Count 4 carbons.
2. -COOH gives -oic acid.
3. Name: butanoic acid. | organic_naming |
C6H6 je benzen. | null | che_0544 | worked_example | cs | Pojmenuj sloučeninu C6H6. | [
"Šestičlenný kruh.",
"Aromatické elektrony.",
"Název: benzen."
] | chemistry | Answer: Benzen.
Steps:
1. Šestičlenný kruh.
2. Aromatické elektrony.
3. Název: benzen. | organic_naming |
Ethers name the smaller group as alkoxy on the larger chain. | null | che_0545 | concept | en | How are ethers named systematically? | [
"Smaller part is alkoxy.",
"Larger part is parent.",
"Example: methoxyethane."
] | chemistry | Answer: Alkoxy plus parent.
Steps:
1. Smaller part is alkoxy.
2. Larger part is parent.
3. Example: methoxyethane. | organic_naming |
Nucleophilic substitution swaps a leaving group for a nucleophile. | null | che_0546 | concept | en | What is nucleophilic substitution? | [
"Nucleophile attacks carbon.",
"Leaving group departs.",
"SN1 or SN2 pathway."
] | chemistry | Answer: Nu replaces leaving group.
Steps:
1. Nucleophile attacks carbon.
2. Leaving group departs.
3. SN1 or SN2 pathway. | organic_mechanisms |
SN2 rate depends on both substrate and nucleophile concentrations. | {
"A": "k[RX] only",
"B": "k[RX][Nu]",
"C": "k[Nu] only",
"D": "k always"
} | che_0547 | multiple_choice | en | What is the SN2 rate law? | [
"One step with both.",
"Rate = k[RX][Nu].",
"Select option B."
] | chemistry | Answer: k[RX][Nu].
Steps:
1. One step with both.
2. Rate = k[RX][Nu].
3. Select option B.
Answer: (B) | organic_mechanisms |
The SN2 rate triples to 3x because rate scales with [Nu]. | null | che_0548 | worked_example | en | SN2 rate = k[RX][Nu]. What happens if [Nu] triples? | [
"Write: rate = k[RX] x 3[Nu].",
"Factor: 3 x k[RX][Nu].",
"Rate becomes 3x."
] | chemistry | Answer: Rate triples.
Steps:
1. Write: rate = k[RX] x 3[Nu].
2. Factor: 3 x k[RX][Nu].
3. Rate becomes 3x. | organic_mechanisms |
Nukleofilní substituce nahradí odstupující skupinu nukleofilem. | null | che_0549 | concept | cs | Co je nukleofilní substituce? | [
"Nukleofil napadne uhlík.",
"Skupina odstoupí.",
"Mechanismus SN1 nebo SN2."
] | chemistry | Answer: Náhrada skupiny nukleofilem.
Steps:
1. Nukleofil napadne uhlík.
2. Skupina odstoupí.
3. Mechanismus SN1 nebo SN2. | organic_mechanisms |
Polar aprotic solvents free nucleophiles and speed SN2. | null | che_0550 | concept | en | Which solvents favor SN2? | [
"Nucleophile must stay free.",
"Aprotic avoids H-bond cage.",
"Examples: acetone, DMSO."
] | chemistry | Answer: Polar aprotic solvents.
Steps:
1. Nucleophile must stay free.
2. Aprotic avoids H-bond cage.
3. Examples: acetone, DMSO. | organic_mechanisms |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.