answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
The balanced equation is C2H4 + 3O2 -> 2CO2 + 2H2O. | null | che_0651 | worked_example | en | Balance: C2H4 + O2 -> CO2 + H2O. | [
"Balance C with 2 CO2.",
"Balance H with 2 H2O.",
"Balance O with 3 O2."
] | chemistry | Answer: C2H4 + 3O2 -> 2CO2 + 2H2O.
Steps:
1. Balance C with 2 CO2.
2. Balance H with 2 H2O.
3. Balance O with 3 O2. | redox |
Elektrodový děj je oxidace na anodě nebo redukce na katodě. | null | che_0652 | concept | cs | Co je elektrodový děj? | [
"Na anodě oxidace.",
"Na katodě redukce.",
"Elektrony tečou vnějškem."
] | chemistry | Answer: Oxidace či redukce na elektrodě.
Steps:
1. Na anodě oxidace.
2. Na katodě redukce.
3. Elektrony tečou vnějškem. | redox |
Sulfur is -2, since 2(+1) + S = 0 gives S = -2. | {
"A": "-2",
"B": "0",
"C": "+2",
"D": "+4"
} | che_0653 | multiple_choice | en | What is the oxidation state of sulfur in H2S? | [
"H is +1 each.",
"Solve: 2 + S = 0.",
"S = -2; select A."
] | chemistry | Answer: -2.
Steps:
1. H is +1 each.
2. Solve: 2 + S = 0.
3. S = -2; select A.
Answer: (A) | redox |
Combustion analysis burns a sample so CO2 and H2O reveal C and H content. | null | che_0654 | concept | en | How does combustion analysis find formulas? | [
"Burn sample in O2.",
"Weigh CO2 and H2O.",
"Back-calculate C and H."
] | chemistry | Answer: Burn, weigh, back-calculate.
Steps:
1. Burn sample in O2.
2. Weigh CO2 and H2O.
3. Back-calculate C and H. | redox |
Vyčíslená rovnice je 2HgO -> 2Hg + O2. | null | che_0655 | worked_example | cs | Vyčísli: HgO -> Hg + O2. | [
"Kyslík: 2 HgO a 1 O2.",
"Rtuť: 2 Hg.",
"Výsledek: 2HgO -> 2Hg + O2."
] | chemistry | Answer: 2HgO -> 2Hg + O2.
Steps:
1. Kyslík: 2 HgO a 1 O2.
2. Rtuť: 2 Hg.
3. Výsledek: 2HgO -> 2Hg + O2. | redox |
Each nitrogen averages +1, since 2N + (-2) = 0. | {
"A": "0",
"B": "+1",
"C": "+2",
"D": "-1"
} | che_0656 | multiple_choice | en | What is nitrogen's oxidation state in N2O? | [
"O is -2.",
"Solve: 2N - 2 = 0.",
"N = +1; select B."
] | chemistry | Answer: +1.
Steps:
1. O is -2.
2. Solve: 2N - 2 = 0.
3. N = +1; select B.
Answer: (B) | redox |
Two electrons transfer: each Na loses 1 and Cl2 gains 2. | null | che_0657 | worked_example | en | How many electrons transfer in 2Na + Cl2 -> 2NaCl? | [
"Each Na: 0 -> +1, loses 1.",
"Two Na lose 2 total.",
"Cl2 gains 2; transfer = 2 e-."
] | chemistry | Answer: 2 electrons.
Steps:
1. Each Na: 0 -> +1, loses 1.
2. Two Na lose 2 total.
3. Cl2 gains 2; transfer = 2 e-. | redox |
The zeroth law states bodies in thermal equilibrium with a third share a temperature. | null | che_0658 | concept | en | What is the zeroth law of thermodynamics? | [
"A matches C in temperature.",
"B matches C too.",
"Then A matches B."
] | chemistry | Answer: Shared temperature transitive.
Steps:
1. A matches C in temperature.
2. B matches C too.
3. Then A matches B. | thermochemistry |
The zeroth law founds the temperature concept and thermometry. | {
"A": "Temperature",
"B": "Pressure",
"C": "Entropy",
"D": "Enthalpy"
} | che_0659 | multiple_choice | en | What concept does the zeroth law define? | [
"Equilibrium transitivity.",
"That defines temperature.",
"Select option A."
] | chemistry | Answer: Temperature.
Steps:
1. Equilibrium transitivity.
2. That defines temperature.
3. Select option A.
Answer: (A) | thermochemistry |
The internal energy change is +1000 J, from +800 plus +200 J. | null | che_0660 | worked_example | en | A system absorbs 800 J of heat while 200 J of work is done on it. Find ΔU. | [
"Write: ΔU = 800 + 200.",
"Add.",
"ΔU = +1000 J."
] | chemistry | Answer: +1000 J.
Steps:
1. Write: ΔU = 800 + 200.
2. Add.
3. ΔU = +1000 J. | thermochemistry |
Tepelná rovnováha znamená stejné teploty a nulový čistý tok tepla. | null | che_0661 | concept | cs | Co je tepelná rovnováha? | [
"Tělesa v kontaktu.",
"Teploty se vyrovnají.",
"Tok tepla ustane."
] | chemistry | Answer: Stejné teploty, žádný tok.
Steps:
1. Tělesa v kontaktu.
2. Teploty se vyrovnají.
3. Tok tepla ustane. | thermochemistry |
Heat always flows spontaneously from hot bodies to cold ones. | {
"A": "Cold to hot",
"B": "Hot to cold",
"C": "Never flows",
"D": "Sideways only"
} | che_0662 | multiple_choice | en | Which way does heat flow spontaneously? | [
"Second law direction.",
"Hot loses, cold gains.",
"Select option B."
] | chemistry | Answer: Hot to cold.
Steps:
1. Second law direction.
2. Hot loses, cold gains.
3. Select option B.
Answer: (B) | thermochemistry |
The heat is 19,250 J, from 500 x 0.385 x 100. | null | che_0663 | worked_example | en | How much heat warms 500 g of copper by 100 C (c = 0.385 J/gC)? | [
"Write: q = 500 x 0.385 x 100.",
"Multiply.",
"q = 19,250 J."
] | chemistry | Answer: 19,250 J.
Steps:
1. Write: q = 500 x 0.385 x 100.
2. Multiply.
3. q = 19,250 J. | thermochemistry |
Heat moves by conduction through contact, convection by flow, and radiation by waves. | null | che_0664 | concept | en | What are the three heat transfer methods? | [
"Conduction via contact.",
"Convection via fluid flow.",
"Radiation via waves."
] | chemistry | Answer: Conduction, convection, radiation.
Steps:
1. Conduction via contact.
2. Convection via fluid flow.
3. Radiation via waves. | thermochemistry |
Melting needs 18.03 kJ, from 3.00 mol times 6.01 kJ/mol. | null | che_0665 | worked_example | en | How much heat melts 54.0 g of ice (ΔHfus = 6.01 kJ/mol)? | [
"Moles = 54.0/18.015 = 3.00.",
"Multiply: 3.00 x 6.01.",
"Heat = 18.03 kJ."
] | chemistry | Answer: 18.03 kJ.
Steps:
1. Moles = 54.0/18.015 = 3.00.
2. Multiply: 3.00 x 6.01.
3. Heat = 18.03 kJ. | thermochemistry |
Condensation forms liquid bonds and releases heat, so it is exothermic. | {
"A": "Exothermic",
"B": "Endothermic",
"C": "Athermal",
"D": "Isolated"
} | che_0666 | multiple_choice | en | Is condensation endothermic or exothermic? | [
"Bonds form going liquid.",
"Forming releases energy.",
"Select option A."
] | chemistry | Answer: Exothermic.
Steps:
1. Bonds form going liquid.
2. Forming releases energy.
3. Select option A.
Answer: (A) | thermochemistry |
K roztátí je třeba 30,05 kJ, protože 5 molů krát 6,01 se rovná 30,05. | null | che_0667 | worked_example | cs | Kolik tepla roztaví 90 g ledu (ΔH = 6,01 kJ/mol)? | [
"Moly = 90/18 = 5.",
"Vynásob: 5 x 6,01.",
"Teplo = 30,05 kJ."
] | chemistry | Answer: 30,05 kJ.
Steps:
1. Moly = 90/18 = 5.
2. Vynásob: 5 x 6,01.
3. Teplo = 30,05 kJ. | thermochemistry |
Enthalpy diagrams plot energy levels of reactants and products with ΔH as the gap. | null | che_0668 | concept | en | What is an enthalpy diagram? | [
"Levels for reactants.",
"Levels for products.",
"Gap equals ΔH."
] | chemistry | Answer: Energy levels with ΔH gap.
Steps:
1. Levels for reactants.
2. Levels for products.
3. Gap equals ΔH. | thermochemistry |
Endothermic products sit above reactants on the energy diagram. | {
"A": "Below reactants",
"B": "Equal to reactants",
"C": "Above reactants",
"D": "At zero"
} | che_0669 | multiple_choice | en | Where are products on an endothermic diagram? | [
"Endo absorbs energy.",
"Products hold more.",
"Select option C."
] | chemistry | Answer: Above reactants.
Steps:
1. Endo absorbs energy.
2. Products hold more.
3. Select option C.
Answer: (C) | thermochemistry |
Two moles release 200 kJ, double the per-mole ΔH of -100 kJ. | null | che_0670 | worked_example | en | A reaction releases 100 kJ per mole of fuel. How much for 2.00 mol? | [
"Scale: 2.00 x (-100).",
"Multiply.",
"Heat = -200 kJ."
] | chemistry | Answer: -200 kJ.
Steps:
1. Scale: 2.00 x (-100).
2. Multiply.
3. Heat = -200 kJ. | thermochemistry |
Endotermická reakce pohlcuje teplo z okolí a má kladnou změnu entalpie. | null | che_0671 | concept | cs | Co je endotermická reakce? | [
"Pohlcuje teplo.",
"ΔH je kladné.",
"Příklad: tání ledu."
] | chemistry | Answer: Reakce pohlcující teplo.
Steps:
1. Pohlcuje teplo.
2. ΔH je kladné.
3. Příklad: tání ledu. | thermochemistry |
Melting raises disorder, so ΔS is positive. | {
"A": "Negative",
"B": "Positive",
"C": "Zero",
"D": "Undefined"
} | che_0672 | multiple_choice | en | What is the sign of ΔS for melting ice? | [
"Solid to liquid.",
"Disorder rises.",
"Select option B."
] | chemistry | Answer: Positive.
Steps:
1. Solid to liquid.
2. Disorder rises.
3. Select option B.
Answer: (B) | thermochemistry |
ΔG is -19.6 kJ, from 40 minus 298 x 0.200. | null | che_0673 | worked_example | en | ΔH = +40 kJ and ΔS = +0.200 kJ/K at 298 K. Compute ΔG. | [
"Write: ΔG = 40 - 298 x 0.200.",
"Simplify: 40 - 59.6.",
"ΔG = -19.6 kJ."
] | chemistry | Answer: -19.6 kJ.
Steps:
1. Write: ΔG = 40 - 298 x 0.200.
2. Simplify: 40 - 59.6.
3. ΔG = -19.6 kJ. | thermochemistry |
Entropie vesmíru je součet entropie soustavy a okolí; u samovolných dějů roste. | null | che_0674 | concept | cs | Co je entropie vesmíru? | [
"Soustava plus okolí.",
"Samovolné děje ji zvyšují.",
"Druhý zákon ji řídí."
] | chemistry | Answer: Celková entropie děje.
Steps:
1. Soustava plus okolí.
2. Samovolné děje ji zvyšují.
3. Druhý zákon ji řídí. | thermochemistry |
Positive ΔH needs high T so the -TΔS term can win. | {
"A": "ΔH<0, ΔS>0",
"B": "ΔH<0, ΔS<0",
"C": "ΔH>0, ΔS<0",
"D": "ΔH>0, ΔS>0"
} | che_0675 | multiple_choice | en | Which signs give spontaneity only at high T? | [
"Need ΔG < 0 via -TΔS.",
"Requires ΔS > 0 with ΔH > 0.",
"Select option D."
] | chemistry | Answer: ΔH>0, ΔS>0.
Steps:
1. Need ΔG < 0 via -TΔS.
2. Requires ΔS > 0 with ΔH > 0.
3. Select option D.
Answer: (D) | thermochemistry |
Freezing releases 6.01 kJ per mole, the reverse of fusion. | null | che_0676 | worked_example | en | How much heat is released freezing 1.00 mol of water (ΔHfus = 6.01 kJ/mol)? | [
"Freezing reverses melting.",
"Heat = -6.01 kJ/mol.",
"Total = -6.01 kJ."
] | chemistry | Answer: -6.01 kJ.
Steps:
1. Freezing reverses melting.
2. Heat = -6.01 kJ/mol.
3. Total = -6.01 kJ. | thermochemistry |
Halogen substituents use fluoro-, chloro-, bromo-, and iodo- prefixes. | null | che_0677 | concept | en | What prefixes name halogen substituents? | [
"F gives fluoro.",
"Cl gives chloro.",
"Br gives bromo, I gives iodo."
] | chemistry | Answer: Fluoro, chloro, bromo, iodo.
Steps:
1. F gives fluoro.
2. Cl gives chloro.
3. Br gives bromo, I gives iodo. | organic_naming |
A fluorine substituent is named with the fluoro- prefix. | {
"A": "Fluorite",
"B": "Fluoro",
"C": "Fluoride",
"D": "Fluorine"
} | che_0678 | multiple_choice | en | What prefix names an F substituent? | [
"Substituent form needed.",
"That is fluoro.",
"Select option B."
] | chemistry | Answer: Fluoro.
Steps:
1. Substituent form needed.
2. That is fluoro.
3. Select option B.
Answer: (B) | organic_naming |
The name is 1-chloropropane: chlorine on carbon 1. | null | che_0679 | worked_example | en | Name CH3CH2CH2Cl. | [
"Count 3 carbons.",
"Cl on carbon 1.",
"Name: 1-chloropropane."
] | chemistry | Answer: 1-chloropropane.
Steps:
1. Count 3 carbons.
2. Cl on carbon 1.
3. Name: 1-chloropropane. | organic_naming |
Poloha substituentu se značí číslem uhlíku před názvem. | null | che_0680 | concept | cs | Jak se značí poloha substituentu? | [
"Očísluj řetězec.",
"Číslo před název.",
"Příklad: 2-chlorpropan."
] | chemistry | Answer: Číslem před názvem.
Steps:
1. Očísluj řetězec.
2. Číslo před název.
3. Příklad: 2-chlorpropan. | organic_naming |
1,2-dichloroethane has two carbons, each bearing one chlorine. | {
"A": "2",
"B": "1",
"C": "3",
"D": "4"
} | che_0681 | multiple_choice | en | How many carbons are in 1,2-dichloroethane? | [
"Ethane root = 2.",
"Dichloro adds none.",
"Select option A."
] | chemistry | Answer: 2.
Steps:
1. Ethane root = 2.
2. Dichloro adds none.
3. Select option A.
Answer: (A) | organic_naming |
The name is propan-2-ol: -OH on carbon 2 of three. | null | che_0682 | worked_example | en | Name CH3CH(OH)CH3. | [
"Count 3 carbons.",
"-OH on carbon 2.",
"Name: propan-2-ol."
] | chemistry | Answer: Propan-2-ol.
Steps:
1. Count 3 carbons.
2. -OH on carbon 2.
3. Name: propan-2-ol. | organic_naming |
Secondary alcohols have -OH on a carbon with two neighbors; tertiary have three. | null | che_0683 | concept | en | How do secondary and tertiary alcohols differ? | [
"Count carbon neighbors.",
"Two means secondary.",
"Three means tertiary."
] | chemistry | Answer: Two versus three neighbors.
Steps:
1. Count carbon neighbors.
2. Two means secondary.
3. Three means tertiary. | organic_naming |
The name is 2-methylpropan-2-ol: -OH and methyl on carbon 2. | null | che_0684 | worked_example | en | Name (CH3)3COH. | [
"Longest chain: 3 = propane.",
"-OH on carbon 2.",
"Methyl on carbon 2: 2-methylpropan-2-ol."
] | chemistry | Answer: 2-methylpropan-2-ol.
Steps:
1. Longest chain: 3 = propane.
2. -OH on carbon 2.
3. Methyl on carbon 2: 2-methylpropan-2-ol. | organic_naming |
The C-OH carbon has three carbon neighbors, so it is tertiary. | {
"A": "Primary",
"B": "Secondary",
"C": "Tertiary",
"D": "Quaternary"
} | che_0685 | multiple_choice | en | How is (CH3)3COH classified? | [
"Count neighbors: three.",
"Three means tertiary.",
"Select option C."
] | chemistry | Answer: Tertiary.
Steps:
1. Count neighbors: three.
2. Three means tertiary.
3. Select option C.
Answer: (C) | organic_naming |
CH3OH je methanol. | null | che_0686 | worked_example | cs | Pojmenuj sloučeninu CH3OH. | [
"Jeden uhlík.",
"Skupina -OH = -ol.",
"Název: methanol."
] | chemistry | Answer: Methanol.
Steps:
1. Jeden uhlík.
2. Skupina -OH = -ol.
3. Název: methanol. | organic_naming |
Alkene names combine locant, cis/trans tag, and the -ene root. | null | che_0687 | concept | en | How are cis/trans alkenes named fully? | [
"Number the double bond.",
"Assign cis or trans.",
"Write tag, locant, root."
] | chemistry | Answer: Tag plus locant plus root.
Steps:
1. Number the double bond.
2. Assign cis or trans.
3. Write tag, locant, root. | organic_naming |
Trans means the methyl groups sit on opposite sides. | {
"A": "Opposite sides",
"B": "Same side",
"C": "Ring form",
"D": "No geometry"
} | che_0688 | multiple_choice | en | What does trans-2-butene look like? | [
"Trans means across.",
"Groups oppose.",
"Select option A."
] | chemistry | Answer: Opposite sides.
Steps:
1. Trans means across.
2. Groups oppose.
3. Select option A.
Answer: (A) | organic_naming |
The name is but-1-ene: double bond starting at carbon 1. | null | che_0689 | worked_example | en | Name CH2=CHCH2CH3. | [
"Count 4 carbons.",
"Double bond at carbon 1.",
"Name: but-1-ene."
] | chemistry | Answer: But-1-ene.
Steps:
1. Count 4 carbons.
2. Double bond at carbon 1.
3. Name: but-1-ene. | organic_naming |
Alkyne locants mark where the triple bond starts with lowest numbers. | null | che_0690 | concept | en | How do alkyne locants work? | [
"Find triple bond.",
"Number lowest there.",
"Example: but-2-yne."
] | chemistry | Answer: Lowest number at triple bond.
Steps:
1. Find triple bond.
2. Number lowest there.
3. Example: but-2-yne. | organic_naming |
Trojná vazba se v názvu značí koncovkou -yn s číslem polohy. | null | che_0691 | concept | cs | Co značí trojná vazba v názvu? | [
"Koncovka -yn.",
"Číslo polohy.",
"Příklad: but-2-yn."
] | chemistry | Answer: Koncovku -yn s číslem.
Steps:
1. Koncovka -yn.
2. Číslo polohy.
3. Příklad: but-2-yn. | organic_naming |
The locant 2 marks the triple bond starting at carbon 2. | {
"A": "Carbon 1",
"B": "Carbon 2",
"C": "Carbon 3",
"D": "Carbon 4"
} | che_0692 | multiple_choice | en | Where is the triple bond in but-2-yne? | [
"Read the locant: 2.",
"Bond starts at carbon 2.",
"Select option B."
] | chemistry | Answer: Carbon 2.
Steps:
1. Read the locant: 2.
2. Bond starts at carbon 2.
3. Select option B.
Answer: (B) | organic_naming |
The name is ethylbenzene: an ethyl group on benzene. | null | che_0693 | worked_example | en | Name C6H5CH2CH3. | [
"Benzene ring present.",
"Ethyl substituent attached.",
"Name: ethylbenzene."
] | chemistry | Answer: Ethylbenzene.
Steps:
1. Benzene ring present.
2. Ethyl substituent attached.
3. Name: ethylbenzene. | organic_naming |
CH≡CH je ethyn. | null | che_0694 | worked_example | cs | Pojmenuj sloučeninu CH≡CH. | [
"Dva uhlíky.",
"Trojná vazba = -yn.",
"Název: ethyn."
] | chemistry | Answer: Ethyn.
Steps:
1. Dva uhlíky.
2. Trojná vazba = -yn.
3. Název: ethyn. | organic_naming |
Acid chlorides replace -OH with Cl and use the -oyl chloride suffix. | null | che_0695 | concept | en | How are acid chlorides named? | [
"Start from acid name.",
"Change -ic acid to -oyl.",
"Add chloride: ethanoyl chloride."
] | chemistry | Answer: -oyl chloride suffix.
Steps:
1. Start from acid name.
2. Change -ic acid to -oyl.
3. Add chloride: ethanoyl chloride. | organic_naming |
Polar protic solvents stabilize SN1 carbocations and leaving groups. | null | che_0696 | concept | en | Which solvents favor SN1? | [
"Carbocation needs solvation.",
"Protic solvents stabilize ions.",
"Examples: water, ethanol."
] | chemistry | Answer: Polar protic solvents.
Steps:
1. Carbocation needs solvation.
2. Protic solvents stabilize ions.
3. Examples: water, ethanol. | organic_mechanisms |
SN1 needs ion-stabilizing polar protic solvents. | {
"A": "Polar protic",
"B": "Polar aprotic",
"C": "Nonpolar",
"D": "No solvent"
} | che_0697 | multiple_choice | en | Which solvent class speeds SN1 most? | [
"Carbocation must be solvated.",
"Protic polar works best.",
"Select option A."
] | chemistry | Answer: Polar protic.
Steps:
1. Carbocation must be solvated.
2. Protic polar works best.
3. Select option A.
Answer: (A) | organic_mechanisms |
The SN1 rate halves to 0.5x because rate equals k[RX]. | null | che_0698 | worked_example | en | SN1 rate = k[RX]. What happens if [RX] is halved? | [
"Write: rate = k x 0.5[RX].",
"Factor: 0.5 x k[RX].",
"Rate becomes 0.5x."
] | chemistry | Answer: Rate halves.
Steps:
1. Write: rate = k x 0.5[RX].
2. Factor: 0.5 x k[RX].
3. Rate becomes 0.5x. | organic_mechanisms |
SN1 je dvoukroková substituce přes karbokation. | null | che_0699 | concept | cs | Co je substituce SN1? | [
"Odchod skupiny.",
"Vznik karbokationu.",
"Atak nukleofilu."
] | chemistry | Answer: Dvoukroková substituce.
Steps:
1. Odchod skupiny.
2. Vznik karbokationu.
3. Atak nukleofilu. | organic_mechanisms |
The cationic carbon is sp2 with three groups in one plane. | null | che_0700 | concept | en | What is carbocation geometry? | [
"sp2 hybridization.",
"Three planar bonds.",
"Empty p orbital above."
] | chemistry | Answer: Trigonal planar.
Steps:
1. sp2 hybridization.
2. Three planar bonds.
3. Empty p orbital above. | organic_mechanisms |
Three planar bonds around sp2 carbon make trigonal planar geometry. | {
"A": "Tetrahedral",
"B": "Linear",
"C": "Trigonal planar",
"D": "Bent"
} | che_0701 | multiple_choice | en | What shape does a carbocation have? | [
"sp2 carbon center.",
"Three groups flat.",
"Select option C."
] | chemistry | Answer: Trigonal planar.
Steps:
1. sp2 carbon center.
2. Three groups flat.
3. Select option C.
Answer: (C) | organic_mechanisms |
The product is propan-2-ol, with -OH on the middle carbon. | null | che_0702 | worked_example | en | What is the acid hydration product of propene? | [
"H adds to C1.",
"Water captures C2 cation.",
"Product: propan-2-ol."
] | chemistry | Answer: Propan-2-ol.
Steps:
1. H adds to C1.
2. Water captures C2 cation.
3. Product: propan-2-ol. | organic_mechanisms |
Primary alcohols oxidize to aldehydes then acids; secondary to ketones; tertiary resist. | null | che_0703 | concept | en | How do alcohol classes oxidize? | [
"Primary: aldehyde, then acid.",
"Secondary: ketone.",
"Tertiary: no reaction."
] | chemistry | Answer: Class sets oxidation path.
Steps:
1. Primary: aldehyde, then acid.
2. Secondary: ketone.
3. Tertiary: no reaction. | organic_mechanisms |
Oxidace alkoholů zvyšuje oxidační číslo uhlíku: primární na kyselinu, sekundární na keton. | null | che_0704 | concept | cs | Co je oxidace alkoholů? | [
"Primární: aldehyd, kyselina.",
"Sekundární: keton.",
"Terciární: nereaguje."
] | chemistry | Answer: Přeměna podle třídy.
Steps:
1. Primární: aldehyd, kyselina.
2. Sekundární: keton.
3. Terciární: nereaguje. | organic_mechanisms |
Primary alcohols first give aldehydes, which oxidize further to acids. | {
"A": "Ketone only",
"B": "Aldehyde then acid",
"C": "No reaction",
"D": "Alkene"
} | che_0705 | multiple_choice | en | What does primary alcohol oxidation give? | [
"First aldehyde forms.",
"Then acid forms.",
"Select option B."
] | chemistry | Answer: Aldehyde then acid.
Steps:
1. First aldehyde forms.
2. Then acid forms.
3. Select option B.
Answer: (B) | organic_mechanisms |
Chlorine adds across ethene to give 1,2-dichloroethane. | null | che_0706 | worked_example | en | What is the Cl2 addition product of ethene? | [
"Double bond opens.",
"Each carbon takes Cl.",
"Product: 1,2-dichloroethane."
] | chemistry | Answer: 1,2-dichloroethane.
Steps:
1. Double bond opens.
2. Each carbon takes Cl.
3. Product: 1,2-dichloroethane. | organic_mechanisms |
Bromine water loses its brown color when an alkene is present. | null | che_0707 | concept | en | How does the bromine test detect alkenes? | [
"Brown Br2 solution added.",
"Alkene consumes Br2.",
"Color vanishes."
] | chemistry | Answer: Brown color disappears.
Steps:
1. Brown Br2 solution added.
2. Alkene consumes Br2.
3. Color vanishes. | organic_mechanisms |
The balanced equation is C3H8 + 5O2 -> 3CO2 + 4H2O. | null | che_0708 | worked_example | en | Balance propane combustion: C3H8 + O2 -> CO2 + H2O. | [
"Balance C with 3 CO2.",
"Balance H with 4 H2O.",
"Balance O with 5 O2."
] | chemistry | Answer: C3H8 + 5O2 -> 3CO2 + 4H2O.
Steps:
1. Balance C with 3 CO2.
2. Balance H with 4 H2O.
3. Balance O with 5 O2. | organic_mechanisms |
Alkenes consume bromine, so the brown solution decolorizes. | {
"A": "Stays brown",
"B": "Forms precipitate",
"C": "Explodes",
"D": "Decolorizes"
} | che_0709 | multiple_choice | en | What does Br2 water do with an alkene? | [
"Br2 adds to C=C.",
"Brown color consumed.",
"Select option D."
] | chemistry | Answer: Decolorizes.
Steps:
1. Br2 adds to C=C.
2. Brown color consumed.
3. Select option D.
Answer: (D) | organic_mechanisms |
Adicí chloru na ethen vzniká 1,2-dichlorethan. | null | che_0710 | worked_example | cs | Jaký produkt vznikne adicí Cl2 na ethen? | [
"Dvojná vazba se otevře.",
"Každý uhlík přijme chlor.",
"Produkt: 1,2-dichlorethan."
] | chemistry | Answer: 1,2-dichlorethan.
Steps:
1. Dvojná vazba se otevře.
2. Každý uhlík přijme chlor.
3. Produkt: 1,2-dichlorethan. | organic_mechanisms |
Addition polymerization joins alkenes directly; condensation links monomers losing small molecules. | null | che_0711 | concept | en | How do addition and condensation polymers differ? | [
"Addition: C=C opens, no loss.",
"Condensation: water lost.",
"Examples: polyethene, nylon."
] | chemistry | Answer: Direct join versus loss of water.
Steps:
1. Addition: C=C opens, no loss.
2. Condensation: water lost.
3. Examples: polyethene, nylon. | organic_mechanisms |
The PVC monomer is chloroethene, CH2=CHCl. | null | che_0712 | worked_example | en | What monomer makes PVC? | [
"PVC repeats -CH2-CHCl-.",
"Monomer has C=C.",
"Monomer: chloroethene."
] | chemistry | Answer: Chloroethene.
Steps:
1. PVC repeats -CH2-CHCl-.
2. Monomer has C=C.
3. Monomer: chloroethene. | organic_mechanisms |
PVC polymerizes from chloroethene monomers. | {
"A": "Chloroethene",
"B": "Ethene",
"C": "Propene",
"D": "Styrene"
} | che_0713 | multiple_choice | en | Which monomer forms PVC? | [
"Repeat holds Cl.",
"Only chloroethene fits.",
"Select option A."
] | chemistry | Answer: Chloroethene.
Steps:
1. Repeat holds Cl.
2. Only chloroethene fits.
3. Select option A.
Answer: (A) | organic_mechanisms |
Polymerace spojuje monomery do dlouhých řetězců. | null | che_0714 | concept | cs | Co je polymerace? | [
"Monomery se spojují.",
"Vzniká polymer.",
"Adice nebo kondenzace."
] | chemistry | Answer: Tvorba polymerů z monomerů.
Steps:
1. Monomery se spojují.
2. Vzniká polymer.
3. Adice nebo kondenzace. | organic_mechanisms |
K expressions put products over reactants with coefficients as exponents. | null | che_0715 | concept | en | What are the rules for writing K? | [
"Products in numerator.",
"Reactants in denominator.",
"Coefficients become exponents."
] | chemistry | Answer: Products over reactants, powers.
Steps:
1. Products in numerator.
2. Reactants in denominator.
3. Coefficients become exponents. | equilibrium |
K equals [B] squared over [A] for A <-> 2B. | {
"A": "[A]/[B]",
"B": "[B]^2/[A]",
"C": "[A]/[B]^2",
"D": "[B]/[A]"
} | che_0716 | multiple_choice | en | What is K for A <-> 2B? | [
"Products over reactants.",
"Exponent 2 on B.",
"Select option B."
] | chemistry | Answer: [B]^2/[A].
Steps:
1. Products over reactants.
2. Exponent 2 on B.
3. Select option B.
Answer: (B) | equilibrium |
K is 0.90, from 0.30 squared divided by 0.10. | null | che_0717 | worked_example | en | For A <-> 2B with [A] = 0.10 M and [B] = 0.30 M, compute K. | [
"Write: K = [B]^2/[A].",
"Substitute: 0.09/0.10.",
"K = 0.90."
] | chemistry | Answer: 0.90.
Steps:
1. Write: K = [B]^2/[A].
2. Substitute: 0.09/0.10.
3. K = 0.90. | equilibrium |
Chemická rovnováha je stav stejných rychlostí obou směrů reakce. | null | che_0718 | concept | cs | Co je chemická rovnováha? | [
"Oba směry stejně rychlé.",
"Koncentrace stálé.",
"Dynamický stav."
] | chemistry | Answer: Stejné rychlosti obou směrů.
Steps:
1. Oba směry stejně rychlé.
2. Koncentrace stálé.
3. Dynamický stav. | equilibrium |
With K = 10^8, products vastly outnumber reactants at equilibrium. | {
"A": "Reactants dominate",
"B": "Nothing reacts",
"C": "Products dominate",
"D": "K is invalid"
} | che_0719 | multiple_choice | en | A reaction has K = 10^8. What dominates at equilibrium? | [
"Huge K favors products.",
"Reactants nearly gone.",
"Select option C."
] | chemistry | Answer: Products dominate.
Steps:
1. Huge K favors products.
2. Reactants nearly gone.
3. Select option C.
Answer: (C) | equilibrium |
The system shifts right because Q = 0.1 is less than K = 2. | null | che_0720 | worked_example | en | Q = 0.1 and K = 2. Which way does the system shift? | [
"Compare: 0.1 < 2.",
"Too few products.",
"Shifts right."
] | chemistry | Answer: Shifts right.
Steps:
1. Compare: 0.1 < 2.
2. Too few products.
3. Shifts right. | equilibrium |
Cooling an exothermic reaction shifts it toward products. | null | che_0721 | concept | en | How does temperature shift exothermic equilibrium? | [
"Heat is a product.",
"Cooling removes it.",
"Shift replaces it rightward."
] | chemistry | Answer: Cooling shifts right.
Steps:
1. Heat is a product.
2. Cooling removes it.
3. Shift replaces it rightward. | equilibrium |
At fixed pressure the volume grows, favoring more gas moles. | {
"A": "Fewer moles",
"B": "More moles",
"C": "No shift",
"D": "Solids"
} | che_0722 | multiple_choice | en | Adding inert gas at constant pressure shifts toward what? | [
"Volume expands.",
"Lower partial pressures.",
"Select option B."
] | chemistry | Answer: More moles.
Steps:
1. Volume expands.
2. Lower partial pressures.
3. Select option B.
Answer: (B) | equilibrium |
High temperature shifts the exothermic Haber reaction left, cutting NH3. | null | che_0723 | worked_example | en | How does high temperature affect NH3 yield (exothermic Haber)? | [
"Heat is a product.",
"Adding heat shifts left.",
"NH3 yield falls."
] | chemistry | Answer: Yield falls.
Steps:
1. Heat is a product.
2. Adding heat shifts left.
3. NH3 yield falls. | equilibrium |
Zahřátí endotermické rovnováhy ji posune k produktům. | null | che_0724 | worked_example | cs | Kam posune zahřátí endotermickou rovnováhu? | [
"Teplo je reaktant.",
"Přidání tepla.",
"Posun k produktům."
] | chemistry | Answer: K produktům.
Steps:
1. Teplo je reaktant.
2. Přidání tepla.
3. Posun k produktům. | equilibrium |
Cooling an endothermic reaction lowers K as reactants gain favor. | {
"A": "K rises",
"B": "No change",
"C": "Shifts right",
"D": "K falls"
} | che_0725 | multiple_choice | en | What does cooling do to an endothermic K? | [
"Heat is a reactant.",
"Removing it shifts left.",
"Select option D."
] | chemistry | Answer: K falls.
Steps:
1. Heat is a reactant.
2. Removing it shifts left.
3. Select option D.
Answer: (D) | equilibrium |
Coupled equilibria multiply their constants into one overall K. | null | che_0726 | concept | en | How do coupled equilibria combine? | [
"Add the equations.",
"Multiply the K values.",
"Intermediates cancel."
] | chemistry | Answer: K values multiply.
Steps:
1. Add the equations.
2. Multiply the K values.
3. Intermediates cancel. | equilibrium |
The reverse K is 5, the reciprocal of 0.2. | null | che_0727 | worked_example | en | Forward K = 0.20 for C <-> D. What is K for the reverse? | [
"Reverse inverts the ratio.",
"K_rev = 1/0.20.",
"K_rev = 5."
] | chemistry | Answer: 5.
Steps:
1. Reverse inverts the ratio.
2. K_rev = 1/0.20.
3. K_rev = 5. | equilibrium |
Reversing flips the ratio, so K becomes its reciprocal. | {
"A": "1/K",
"B": "K squared",
"C": "2K",
"D": "K"
} | che_0728 | multiple_choice | en | What is K of the reversed reaction? | [
"Products and reactants swap.",
"Ratio inverts.",
"Select option A."
] | chemistry | Answer: 1/K.
Steps:
1. Products and reactants swap.
2. Ratio inverts.
3. Select option A.
Answer: (A) | equilibrium |
The solubility is 1.05 x 10^-5 M, the square root of 1.1 x 10^-10. | null | che_0729 | worked_example | en | BaSO4 has Ksp = 1.1 x 10^-10. Compute its molar solubility. | [
"Write: s^2 = Ksp.",
"Substitute: s = sqrt(1.1 x 10^-10).",
"s = 1.05 x 10^-5 M."
] | chemistry | Answer: 1.05 x 10^-5 M.
Steps:
1. Write: s^2 = Ksp.
2. Substitute: s = sqrt(1.1 x 10^-10).
3. s = 1.05 x 10^-5 M. | equilibrium |
Nasycený roztok obsahuje maximální množství rozpuštěné látky při dané teplotě. | null | che_0730 | concept | cs | Co je nasycený roztok? | [
"Maximum rozpuštěné látky.",
"Rovnováha s pevnou fází.",
"Víc se nerozpustí."
] | chemistry | Answer: Roztok s maximem látky.
Steps:
1. Maximum rozpuštěné látky.
2. Rovnováha s pevnou fází.
3. Víc se nerozpustí. | equilibrium |
A saturated solution holds the maximum dissolved solute at that temperature. | {
"A": "No solute",
"B": "Max dissolved solute",
"C": "Precipitate only",
"D": "Pure solvent"
} | che_0731 | multiple_choice | en | What does a saturated solution hold? | [
"Dissolving at its limit.",
"Solid may coexist.",
"Select option B."
] | chemistry | Answer: Max dissolved solute.
Steps:
1. Dissolving at its limit.
2. Solid may coexist.
3. Select option B.
Answer: (B) | equilibrium |
A precipitate forms because Q = 5.0 x 10^-9 exceeds Ksp = 1.1 x 10^-10. | null | che_0732 | worked_example | en | Q = 5.0e-9 for BaSO4 (Ksp = 1.1e-10). Does it precipitate? | [
"Compare Q to Ksp.",
"5.0e-9 > 1.1e-10.",
"Yes, precipitate forms."
] | chemistry | Answer: Yes, precipitates.
Steps:
1. Compare Q to Ksp.
2. 5.0e-9 > 1.1e-10.
3. Yes, precipitate forms. | equilibrium |
The electrochemical series ranks half-cells by standard reduction potential. | null | che_0733 | concept | en | What is the electrochemical series? | [
"List half-reactions.",
"Order by E°.",
"Top oxidizes, bottom reduces."
] | chemistry | Answer: Half-cells ranked by E°.
Steps:
1. List half-reactions.
2. Order by E°.
3. Top oxidizes, bottom reduces. | electrochemistry |
The highest E° species gains electrons most eagerly. | {
"A": "Lowest E°",
"B": "Negative E°",
"C": "Highest E°",
"D": "Zero E°"
} | che_0734 | multiple_choice | en | Which species is the strongest oxidizing agent? | [
"Oxidizers want electrons.",
"Highest E° wants most.",
"Select option C."
] | chemistry | Answer: Highest E°.
Steps:
1. Oxidizers want electrons.
2. Highest E° wants most.
3. Select option C.
Answer: (C) | electrochemistry |
The cell potential is 0.78 V, from 0.34 - (-0.44). | null | che_0735 | worked_example | en | Compute E cell for Fe/Cu with E(Fe) = -0.44 V and E(Cu) = 0.34 V. | [
"Cathode is Cu at 0.34 V.",
"Anode is Fe at -0.44 V.",
"E = 0.34 - (-0.44) = 0.78 V."
] | chemistry | Answer: 0.78 V.
Steps:
1. Cathode is Cu at 0.34 V.
2. Anode is Fe at -0.44 V.
3. E = 0.34 - (-0.44) = 0.78 V. | electrochemistry |
Elektrochemická řada řadí kovy podle standardního potenciálu. | null | che_0736 | concept | cs | Co je elektrochemická řada? | [
"Seznam poloreakcí.",
"Seřazení podle E°.",
"Předpovídá reakce."
] | chemistry | Answer: Žebříček potenciálů.
Steps:
1. Seznam poloreakcí.
2. Seřazení podle E°.
3. Předpovídá reakce. | electrochemistry |
Metals above hydrogen reduce H+ to H2 gas in acids. | {
"A": "Dissolves in acids",
"B": "Never reacts",
"C": "Plates hydrogen",
"D": "Explodes"
} | che_0737 | multiple_choice | en | What does a metal above H in the series do? | [
"It reduces H+.",
"H2 gas evolves.",
"Select option A."
] | chemistry | Answer: Dissolves in acids.
Steps:
1. It reduces H+.
2. H2 gas evolves.
3. Select option A.
Answer: (A) | electrochemistry |
Positive E°cell predicts a spontaneous redox reaction. | null | che_0738 | concept | en | How does E° predict redox spontaneity? | [
"Compute E°cell.",
"Positive means ΔG negative.",
"Reaction runs forward."
] | chemistry | Answer: Positive E° means spontaneous.
Steps:
1. Compute E°cell.
2. Positive means ΔG negative.
3. Reaction runs forward. | electrochemistry |
ΔG is -114 kJ, from -2 x 96485 x 0.59 J. | null | che_0739 | worked_example | en | Compute ΔG for n = 2 and E = 0.59 V (F = 96485 C/mol). | [
"Write: ΔG = -2 x 96485 x 0.59.",
"Multiply: -113,852 J.",
"ΔG = -114 kJ."
] | chemistry | Answer: -114 kJ.
Steps:
1. Write: ΔG = -2 x 96485 x 0.59.
2. Multiply: -113,852 J.
3. ΔG = -114 kJ. | electrochemistry |
The potential is 0.53 V, from 0.59 - 0.0296 x log(100). | null | che_0740 | worked_example | en | E° = 0.59 V, n = 2, Q = 100 at 25 C. Compute E. | [
"Write: E = 0.59 - 0.0296 x 2.",
"Subtract: 0.59 - 0.0592.",
"E = 0.53 V."
] | chemistry | Answer: 0.53 V.
Steps:
1. Write: E = 0.59 - 0.0296 x 2.
2. Subtract: 0.59 - 0.0592.
3. E = 0.53 V. | electrochemistry |
With Q = 1, the log term vanishes and E equals E°. | {
"A": "E is zero",
"B": "E equals E°",
"C": "E doubles",
"D": "E reverses"
} | che_0741 | multiple_choice | en | What is E when Q = 1 in the Nernst equation? | [
"log(1) = 0.",
"Correction term zero.",
"Select option B."
] | chemistry | Answer: E equals E°.
Steps:
1. log(1) = 0.
2. Correction term zero.
3. Select option B.
Answer: (B) | electrochemistry |
Molten NaCl electrolysis yields sodium metal and chlorine gas. | null | che_0742 | concept | en | What are the products of molten NaCl electrolysis? | [
"Cathode: Na+ to Na.",
"Anode: Cl- to Cl2.",
"No water involved."
] | chemistry | Answer: Na and Cl2.
Steps:
1. Cathode: Na+ to Na.
2. Anode: Cl- to Cl2.
3. No water involved. | electrochemistry |
Vzniká sodík na katodě a chlor na anodě. | null | che_0743 | worked_example | cs | Jaké produkty dá elektrolýza taveniny NaCl? | [
"Katoda: Na+ na Na.",
"Anoda: Cl- na Cl2.",
"Produkty: Na a Cl2."
] | chemistry | Answer: Na a Cl2.
Steps:
1. Katoda: Na+ na Na.
2. Anoda: Cl- na Cl2.
3. Produkty: Na a Cl2. | electrochemistry |
The chlorine volume is 22.4 L, from 1.00 mol times 22.414 L/mol. | null | che_0744 | worked_example | en | What volume of Cl2 at STP comes from 2.00 mol of electrons? | [
"2 e- give 1 Cl2: 1.00 mol.",
"Multiply: 1.00 x 22.414.",
"V = 22.4 L."
] | chemistry | Answer: 22.4 L.
Steps:
1. 2 e- give 1 Cl2: 1.00 mol.
2. Multiply: 1.00 x 22.414.
3. V = 22.4 L. | electrochemistry |
Sodium ions reduce to sodium metal at the cathode. | {
"A": "Na metal",
"B": "H2 gas",
"C": "O2 gas",
"D": "NaOH"
} | che_0745 | multiple_choice | en | What forms at the cathode in molten NaCl electrolysis? | [
"Na+ gains electron.",
"Metal deposits.",
"Select option A."
] | chemistry | Answer: Na metal.
Steps:
1. Na+ gains electron.
2. Metal deposits.
3. Select option A.
Answer: (A) | electrochemistry |
Lithium-ion cells shuttle Li+ between graphite and metal oxide electrodes. | null | che_0746 | concept | en | How does a lithium-ion battery work? | [
"Li+ leaves graphite.",
"Li+ enters oxide.",
"Reverse on charge."
] | chemistry | Answer: Li+ shuttles between electrodes.
Steps:
1. Li+ leaves graphite.
2. Li+ enters oxide.
3. Reverse on charge. | electrochemistry |
The charge is 24,121 C, from 0.250 mol times 96,485 C/mol. | null | che_0747 | worked_example | en | What charge do 0.250 mol of electrons carry? | [
"Write: Q = 0.250 x 96485.",
"Multiply.",
"Q = 24,121 C."
] | chemistry | Answer: 24,121 C.
Steps:
1. Write: Q = 0.250 x 96485.
2. Multiply.
3. Q = 24,121 C. | electrochemistry |
Lithiový akumulátor přenáší ionty Li+ mezi elektrodami. | null | che_0748 | concept | cs | Co je lithiový akumulátor? | [
"Li+ putují mezi elektrodami.",
"Grafit a oxid kovu.",
"Dobíjecí chemie."
] | chemistry | Answer: Článek s Li+ ionty.
Steps:
1. Li+ putují mezi elektrodami.
2. Grafit a oxid kovu.
3. Dobíjecí chemie. | electrochemistry |
Lithium ions carry the charge inside a Li-ion cell. | {
"A": "Electrons only",
"B": "H+",
"C": "Li+",
"D": "O2-"
} | che_0749 | multiple_choice | en | What is the charge carrier in a Li-ion cell? | [
"Ions cross electrolyte.",
"Those ions are Li+.",
"Select option C."
] | chemistry | Answer: Li+.
Steps:
1. Ions cross electrolyte.
2. Those ions are Li+.
3. Select option C.
Answer: (C) | electrochemistry |
The plated iron mass is 3.13 g, from 10,800 C giving 0.05597 mol Fe. | null | che_0750 | worked_example | en | What mass of Fe plates from Fe2+ at 3.00 A for 1.00 h (M = 55.845 g/mol)? | [
"Q = 3.00 x 3600 = 10,800 C.",
"Mol e- = 0.11193; Fe = 0.05597 mol.",
"Mass = 0.05597 x 55.845 = 3.13 g."
] | chemistry | Answer: 3.13 g.
Steps:
1. Q = 3.00 x 3600 = 10,800 C.
2. Mol e- = 0.11193; Fe = 0.05597 mol.
3. Mass = 0.05597 x 55.845 = 3.13 g. | electrochemistry |
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