answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
Two electrons are gained as Cl2 becomes 2Cl-. | null | che_0351 | worked_example | en | How many electrons are gained in Cl2 -> 2Cl-? | [
"Each Cl goes 0 to -1.",
"Two Cl gain 1 each.",
"Total gained = 2."
] | chemistry | Answer: 2 electrons.
Steps:
1. Each Cl goes 0 to -1.
2. Two Cl gain 1 each.
3. Total gained = 2. | redox |
Řada reaktivity řadí kovy podle ochoty oxidovat se. | null | che_0352 | concept | cs | Co je řada reaktivity kovů? | [
"Reaktivnější kov výše.",
"Vytěsní méně reaktivní.",
"Příklad: Zn vytěsní Cu."
] | chemistry | Answer: Žebříček reaktivity kovů.
Steps:
1. Reaktivnější kov výše.
2. Vytěsní méně reaktivní.
3. Příklad: Zn vytěsní Cu. | redox |
Magnesium sits above hydrogen and dissolves in acid with H2 release. | {
"A": "Cu",
"B": "Mg",
"C": "Pt",
"D": "Au"
} | che_0353 | multiple_choice | en | Which metal releases H2 from dilute acid? | [
"Need metal above H.",
"Eliminate Cu, Pt, Au.",
"Select option B."
] | chemistry | Answer: Mg.
Steps:
1. Need metal above H.
2. Eliminate Cu, Pt, Au.
3. Select option B.
Answer: (B) | redox |
The balanced equation is 2S + 3O2 -> 2SO3. | null | che_0354 | worked_example | en | Balance: S + O2 -> SO3. | [
"Balance S with 2 SO3 and 2 S.",
"Count O: 6 each side.",
"Use 3 O2: 2S + 3O2 -> 2SO3."
] | chemistry | Answer: 2S + 3O2 -> 2SO3.
Steps:
1. Balance S with 2 SO3 and 2 S.
2. Count O: 6 each side.
3. Use 3 O2: 2S + 3O2 -> 2SO3. | redox |
Vyčíslená rovnice je N2 + 3H2 -> 2NH3. | null | che_0355 | worked_example | cs | Vyčísli: N2 + H2 -> NH3. | [
"Dusík: 2 NH3 a 1 N2.",
"Vodík: 3 H2 na 6 H.",
"Výsledek: N2 + 3H2 -> 2NH3."
] | chemistry | Answer: N2 + 3H2 -> 2NH3.
Steps:
1. Dusík: 2 NH3 a 1 N2.
2. Vodík: 3 H2 na 6 H.
3. Výsledek: N2 + 3H2 -> 2NH3. | redox |
Chromium is +6, since Cr + 4(-2) = -2 gives Cr = +6. | {
"A": "+2",
"B": "+4",
"C": "+6",
"D": "+8"
} | che_0356 | multiple_choice | en | What is the oxidation state of Cr in CrO4(2-)? | [
"Assign O -2.",
"Solve: Cr - 8 = -2.",
"Cr = +6; select C."
] | chemistry | Answer: +6.
Steps:
1. Assign O -2.
2. Solve: Cr - 8 = -2.
3. Cr = +6; select C.
Answer: (C) | redox |
Synthesis joins parts, decomposition splits them, and combustion burns fuel in oxygen. | null | che_0357 | concept | en | How are synthesis, decomposition, and combustion classified? | [
"Synthesis: A + B -> AB.",
"Decomposition: AB -> A + B.",
"Combustion: fuel + O2."
] | chemistry | Answer: Joining, splitting, burning.
Steps:
1. Synthesis: A + B -> AB.
2. Decomposition: AB -> A + B.
3. Combustion: fuel + O2. | redox |
The system is the studied part while everything else is the surroundings. | null | che_0358 | concept | en | What are system and surroundings? | [
"System: reaction studied.",
"Surroundings: all else.",
"Boundary separates them."
] | chemistry | Answer: Studied part versus rest.
Steps:
1. System: reaction studied.
2. Surroundings: all else.
3. Boundary separates them. | thermochemistry |
An isolated system exchanges neither matter nor energy with surroundings. | {
"A": "Both matter and energy",
"B": "Energy only",
"C": "Neither matter nor energy",
"D": "Matter only"
} | che_0359 | multiple_choice | en | What does an isolated system exchange? | [
"Isolated means fully sealed.",
"No matter crosses.",
"No energy crosses; select C."
] | chemistry | Answer: Neither matter nor energy.
Steps:
1. Isolated means fully sealed.
2. No matter crosses.
3. No energy crosses; select C.
Answer: (C) | thermochemistry |
The expansion work is -1000 J, from -100,000 Pa times 0.010 m3. | null | che_0360 | worked_example | en | Gas expands by 0.010 m3 against 100 kPa. Compute the work. | [
"Write: w = -PΔV.",
"Substitute: -100,000 x 0.010.",
"w = -1000 J."
] | chemistry | Answer: -1000 J.
Steps:
1. Write: w = -PΔV.
2. Substitute: -100,000 x 0.010.
3. w = -1000 J. | thermochemistry |
Teplo je přenos energie rozdílem teplot, práce je přenos silou na dráze. | null | che_0361 | concept | cs | Co je teplo a práce? | [
"Teplo teče z teplejšího.",
"Práce je síla krát dráha.",
"Obě mění energii soustavy."
] | chemistry | Answer: Dva přenosy energie.
Steps:
1. Teplo teče z teplejšího.
2. Práce je síla krát dráha.
3. Obě mění energii soustavy. | thermochemistry |
Work done by the system on surroundings is negative. | {
"A": "Positive",
"B": "Negative",
"C": "Zero",
"D": "Undefined"
} | che_0362 | multiple_choice | en | What sign does work done BY the system have? | [
"System loses energy.",
"Loss is negative.",
"Select option B."
] | chemistry | Answer: Negative.
Steps:
1. System loses energy.
2. Loss is negative.
3. Select option B.
Answer: (B) | thermochemistry |
The heat is 4485 J, from 100 x 0.897 x 50. | null | che_0363 | worked_example | en | How much heat warms 100 g of Al by 50 C (c = 0.897 J/gC)? | [
"Write: q = 100 x 0.897 x 50.",
"Multiply.",
"q = 4485 J."
] | chemistry | Answer: 4485 J.
Steps:
1. Write: q = 100 x 0.897 x 50.
2. Multiply.
3. q = 4485 J. | thermochemistry |
Calorimeter heat capacity links its temperature rise to absorbed heat. | null | che_0364 | concept | en | What is calorimeter heat capacity? | [
"Vessel absorbs some heat.",
"Capacity times ΔT gives q.",
"Calibrate by burning standard."
] | chemistry | Answer: Heat per degree of vessel.
Steps:
1. Vessel absorbs some heat.
2. Capacity times ΔT gives q.
3. Calibrate by burning standard. | thermochemistry |
The heat is 2000 J, from 500 J/C times 4 C. | null | che_0365 | worked_example | en | A calorimeter (C = 500 J/C) warms 4.0 C. How much heat did it absorb? | [
"Write: q = C x ΔT.",
"Substitute: 500 x 4.0.",
"q = 2000 J."
] | chemistry | Answer: 2000 J.
Steps:
1. Write: q = C x ΔT.
2. Substitute: 500 x 4.0.
3. q = 2000 J. | thermochemistry |
A bomb calorimeter holds volume fixed, so it measures internal energy change. | {
"A": "Constant volume",
"B": "Constant pressure",
"C": "Constant temperature",
"D": "Constant mass"
} | che_0366 | multiple_choice | en | What does a bomb calorimeter hold constant? | [
"Steel bomb is rigid.",
"Volume cannot change.",
"Select option A."
] | chemistry | Answer: Constant volume.
Steps:
1. Steel bomb is rigid.
2. Volume cannot change.
3. Select option A.
Answer: (A) | thermochemistry |
Potřebné teplo je 4184 J, protože 50 x 4,184 x 20 se rovná 4184. | null | che_0367 | worked_example | cs | Kolik tepla ohřeje 50 g vody o 20 °C? | [
"Napiš vzorec: q = 50 x 4,184 x 20.",
"Vynásob.",
"q = 4184 J."
] | chemistry | Answer: 4184 J.
Steps:
1. Napiš vzorec: q = 50 x 4,184 x 20.
2. Vynásob.
3. q = 4184 J. | thermochemistry |
State functions like H, U, S, and G depend only on current state, not path. | null | che_0368 | concept | en | What is a state function? | [
"Depends on state only.",
"Path does not matter.",
"Examples: H, U, S, G."
] | chemistry | Answer: Path-independent property.
Steps:
1. Depends on state only.
2. Path does not matter.
3. Examples: H, U, S, G. | thermochemistry |
Enthalpy change is a state function, unlike path-dependent heat and work. | {
"A": "Heat q",
"B": "Work w",
"C": "Enthalpy change ΔH",
"D": "Path length"
} | che_0369 | multiple_choice | en | Which is a state function? | [
"q and w depend on path.",
"ΔH depends on states.",
"Select option C."
] | chemistry | Answer: Enthalpy change ΔH.
Steps:
1. q and w depend on path.
2. ΔH depends on states.
3. Select option C.
Answer: (C) | thermochemistry |
The ΔH is -285.8 kJ per mole of water formed. | null | che_0370 | worked_example | en | H2 + 1/2O2 -> H2O has ΔH = -285.8 kJ. How much heat per mole of water? | [
"One mole water forms.",
"Heat equals ΔH.",
"-285.8 kJ per mole."
] | chemistry | Answer: -285.8 kJ.
Steps:
1. One mole water forms.
2. Heat equals ΔH.
3. -285.8 kJ per mole. | thermochemistry |
Kalorimetr je izolovaná nádoba měřící teplo ze změny teploty. | null | che_0371 | concept | cs | Co je kalorimetr? | [
"Izolovaná nádoba.",
"Měří změnu teploty.",
"Teplo z kapacity a ΔT."
] | chemistry | Answer: Nádoba na měření tepla.
Steps:
1. Izolovaná nádoba.
2. Měří změnu teploty.
3. Teplo z kapacity a ΔT. | thermochemistry |
Going from solid to gas raises disorder most, giving positive ΔS. | {
"A": "Gas to solid",
"B": "Fewer gas moles",
"C": "Liquid to solid",
"D": "Solid to gas"
} | che_0372 | multiple_choice | en | Which change gives positive ΔS? | [
"Disorder must rise.",
"Only solid to gas qualifies.",
"Select option D."
] | chemistry | Answer: Solid to gas.
Steps:
1. Disorder must rise.
2. Only solid to gas qualifies.
3. Select option D.
Answer: (D) | thermochemistry |
ΔG is -30 kJ, from -80 minus 500 x (-0.100). | null | che_0373 | worked_example | en | ΔH = -80 kJ and ΔS = -0.100 kJ/K at 500 K. Compute ΔG. | [
"Write: ΔG = -80 - 500 x (-0.100).",
"Simplify: -80 + 50.",
"ΔG = -30 kJ."
] | chemistry | Answer: -30 kJ.
Steps:
1. Write: ΔG = -80 - 500 x (-0.100).
2. Simplify: -80 + 50.
3. ΔG = -30 kJ. | thermochemistry |
Gibbsova energie určuje samovolnost děje: záporná změna znamená samovolný děj. | null | che_0374 | concept | cs | Co je Gibbsova energie? | [
"G = H - TS.",
"ΔG < 0 samovolné.",
"ΔG = 0 rovnováha."
] | chemistry | Answer: Energie samovolnosti.
Steps:
1. G = H - TS.
2. ΔG < 0 samovolné.
3. ΔG = 0 rovnováha. | thermochemistry |
Negative ΔH and positive ΔS make ΔG negative at every temperature. | {
"A": "Always spontaneous",
"B": "Never spontaneous",
"C": "Only at high T",
"D": "Only at low T"
} | che_0375 | multiple_choice | en | Exothermic with entropy increase is spontaneous when? | [
"ΔG = ΔH - TΔS.",
"Both terms negative.",
"Select option A."
] | chemistry | Answer: Always spontaneous.
Steps:
1. ΔG = ΔH - TΔS.
2. Both terms negative.
3. Select option A.
Answer: (A) | thermochemistry |
Boiling needs 40.7 kJ, from 1.00 mol times 40.7 kJ/mol. | null | che_0376 | worked_example | en | How much heat boils 18.0 g of water (ΔHvap = 40.7 kJ/mol)? | [
"Moles = 18.0/18.015 = 1.00.",
"Multiply: 1.00 x 40.7.",
"Heat = 40.7 kJ."
] | chemistry | Answer: 40.7 kJ.
Steps:
1. Moles = 18.0/18.015 = 1.00.
2. Multiply: 1.00 x 40.7.
3. Heat = 40.7 kJ. | thermochemistry |
Haloalkanes name the halogen as a fluoro-, chloro-, bromo-, or iodo- prefix. | null | che_0377 | concept | en | How are haloalkanes named? | [
"Halogen becomes prefix.",
"Number its position.",
"Example: 2-bromopropane."
] | chemistry | Answer: Halogen prefix plus alkane.
Steps:
1. Halogen becomes prefix.
2. Number its position.
3. Example: 2-bromopropane. | organic_naming |
CH3Cl is chloromethane: one carbon with chlorine. | {
"A": "Chloromethane",
"B": "Chloroethane",
"C": "Dichloromethane",
"D": "Methanol"
} | che_0378 | multiple_choice | en | What is the name of CH3Cl? | [
"One carbon present.",
"Chlorine substituent.",
"Select option A."
] | chemistry | Answer: Chloromethane.
Steps:
1. One carbon present.
2. Chlorine substituent.
3. Select option A.
Answer: (A) | organic_naming |
The name is bromoethane: two carbons with bromine. | null | che_0379 | worked_example | en | Name CH3CH2Br. | [
"Count 2 carbons.",
"Bromine substituent.",
"Name: bromoethane."
] | chemistry | Answer: Bromoethane.
Steps:
1. Count 2 carbons.
2. Bromine substituent.
3. Name: bromoethane. | organic_naming |
Halogenalkany se pojmenují předponou fluor-, chlor-, brom- či jod- před alkanem. | null | che_0380 | concept | cs | Jak se pojmenovávají halogenalkany? | [
"Halogen jako předpona.",
"Očísluj polohu.",
"Příklad: 2-brompropan."
] | chemistry | Answer: Předpona halogenu + alkan.
Steps:
1. Halogen jako předpona.
2. Očísluj polohu.
3. Příklad: 2-brompropan. | organic_naming |
Two -OH groups use the -diol suffix with position numbers. | {
"A": "-ol",
"B": "-diol",
"C": "-al",
"D": "-one"
} | che_0381 | multiple_choice | en | What suffix marks two alcohol groups? | [
"One -OH gives -ol.",
"Two give -diol.",
"Select option B."
] | chemistry | Answer: -diol.
Steps:
1. One -OH gives -ol.
2. Two give -diol.
3. Select option B.
Answer: (B) | organic_naming |
The name is ethane-1,2-diol: -OH on both carbons. | null | che_0382 | worked_example | en | Name HOCH2CH2OH. | [
"Count 2 carbons.",
"-OH on carbons 1 and 2.",
"Name: ethane-1,2-diol."
] | chemistry | Answer: Ethane-1,2-diol.
Steps:
1. Count 2 carbons.
2. -OH on carbons 1 and 2.
3. Name: ethane-1,2-diol. | organic_naming |
The carboxylic acid group outranks all others and sets the -oic acid suffix. | null | che_0383 | concept | en | Why do carboxylic acids set the suffix? | [
"Highest naming priority.",
"Others become prefixes.",
"Suffix is -oic acid."
] | chemistry | Answer: Highest priority group.
Steps:
1. Highest naming priority.
2. Others become prefixes.
3. Suffix is -oic acid. | organic_naming |
The name is methanoic acid: one carbon with -COOH. | null | che_0384 | worked_example | en | Name HCOOH. | [
"One carbon present.",
"-COOH gives -oic acid.",
"Name: methanoic acid."
] | chemistry | Answer: Methanoic acid.
Steps:
1. One carbon present.
2. -COOH gives -oic acid.
3. Name: methanoic acid. | organic_naming |
Acetic acid is systematically ethanoic acid. | {
"A": "Methanoic acid",
"B": "Propanoic acid",
"C": "Ethanoic acid",
"D": "Butanoic acid"
} | che_0385 | multiple_choice | en | What is the IUPAC name of acetic acid? | [
"Two carbons present.",
"Acid suffix applies.",
"Select option C."
] | chemistry | Answer: Ethanoic acid.
Steps:
1. Two carbons present.
2. Acid suffix applies.
3. Select option C.
Answer: (C) | organic_naming |
CH3CHO je ethanal. | null | che_0386 | worked_example | cs | Pojmenuj sloučeninu CH3CHO. | [
"Koncová skupina -CHO.",
"Dva uhlíky.",
"Název: ethanal."
] | chemistry | Answer: Ethanal.
Steps:
1. Koncová skupina -CHO.
2. Dva uhlíky.
3. Název: ethanal. | organic_naming |
Ketones number the chain to give the carbonyl the lowest locant with -one. | null | che_0387 | concept | en | How are ketones numbered? | [
"Find C=O carbon.",
"Number chain lowest there.",
"Suffix -one with locant."
] | chemistry | Answer: Lowest locant for C=O.
Steps:
1. Find C=O carbon.
2. Number chain lowest there.
3. Suffix -one with locant. | organic_naming |
The carbonyl sits on carbon 2 of four, giving butan-2-one. | {
"A": "Butan-1-one",
"B": "Butan-2-one",
"C": "Butanal",
"D": "Butanoic acid"
} | che_0388 | multiple_choice | en | What is the name of CH3CH2COCH3? | [
"Four carbons, mid C=O.",
"Carbonyl on carbon 2.",
"Select option B."
] | chemistry | Answer: Butan-2-one.
Steps:
1. Four carbons, mid C=O.
2. Carbonyl on carbon 2.
3. Select option B.
Answer: (B) | organic_naming |
The name is phenol: -OH attached directly to benzene. | null | che_0389 | worked_example | en | Name C6H5OH. | [
"Benzene ring present.",
"-OH on the ring.",
"Name: phenol."
] | chemistry | Answer: Phenol.
Steps:
1. Benzene ring present.
2. -OH on the ring.
3. Name: phenol. | organic_naming |
Phenols attach -OH directly to an aromatic ring, making them more acidic than alcohols. | null | che_0390 | concept | en | What are phenols? | [
"-OH on aromatic ring.",
"Ring stabilizes anion.",
"More acidic than alcohols."
] | chemistry | Answer: -OH on aromatic ring.
Steps:
1. -OH on aromatic ring.
2. Ring stabilizes anion.
3. More acidic than alcohols. | organic_naming |
Benzenové jádro je šestičlenný aromatický kruh s delokalizovanými elektrony. | null | che_0391 | concept | cs | Co je benzenové jádro? | [
"Šest uhlíků v kruhu.",
"Delokalizace elektronů.",
"Základ aromatických sloučenin."
] | chemistry | Answer: Aromatický šestičlenný kruh.
Steps:
1. Šest uhlíků v kruhu.
2. Delokalizace elektronů.
3. Základ aromatických sloučenin. | organic_naming |
Ortho means adjacent 1,2 positions on a benzene ring. | {
"A": "1,2 positions",
"B": "1,3 positions",
"C": "1,4 positions",
"D": "1,5 positions"
} | che_0392 | multiple_choice | en | What does ortho- mean on benzene? | [
"Ortho means adjacent.",
"That is 1,2.",
"Select option A."
] | chemistry | Answer: 1,2 positions.
Steps:
1. Ortho means adjacent.
2. That is 1,2.
3. Select option A.
Answer: (A) | organic_naming |
The name is methyl ethanoate: methyl alcohol part plus ethanoate. | null | che_0393 | worked_example | en | Name CH3COOCH3. | [
"Acid part: ethanoate.",
"Alcohol part: methyl.",
"Name: methyl ethanoate."
] | chemistry | Answer: Methyl ethanoate.
Steps:
1. Acid part: ethanoate.
2. Alcohol part: methyl.
3. Name: methyl ethanoate. | organic_naming |
CH3NH2 je methylamin. | null | che_0394 | worked_example | cs | Pojmenuj sloučeninu CH3NH2. | [
"Jeden uhlík.",
"Skupina -NH2 = amin.",
"Název: methylamin."
] | chemistry | Answer: Methylamin.
Steps:
1. Jeden uhlík.
2. Skupina -NH2 = amin.
3. Název: methylamin. | organic_naming |
Amides attach -NH2 to a carbonyl and use the -amide suffix. | null | che_0395 | concept | en | How are amides named? | [
"Find C(=O)N group.",
"Name chain plus -amide.",
"Example: ethanamide."
] | chemistry | Answer: Chain plus -amide.
Steps:
1. Find C(=O)N group.
2. Name chain plus -amide.
3. Example: ethanamide. | organic_naming |
SN1 first loses the leaving group to form a carbocation, then the nucleophile attacks. | null | che_0396 | concept | en | What are the SN1 mechanism steps? | [
"Leaving group departs.",
"Carbocation forms.",
"Nucleophile attacks."
] | chemistry | Answer: Ionize, then attack.
Steps:
1. Leaving group departs.
2. Carbocation forms.
3. Nucleophile attacks. | organic_mechanisms |
The SN1 intermediate is a planar carbocation. | {
"A": "Carbocation",
"B": "Carbanion",
"C": "Radical",
"D": "Epoxide"
} | che_0397 | multiple_choice | en | What intermediate forms in SN1? | [
"Leaving group takes pair.",
"Positive carbon remains.",
"Select option A."
] | chemistry | Answer: Carbocation.
Steps:
1. Leaving group takes pair.
2. Positive carbon remains.
3. Select option A.
Answer: (A) | organic_mechanisms |
The SN1 rate triples to 3x because rate equals k[RX]. | null | che_0398 | worked_example | en | SN1 rate = k[RX]. What happens if [RX] triples? | [
"Write: rate = k x 3[RX].",
"Factor: 3 x k[RX].",
"Rate becomes 3x."
] | chemistry | Answer: Rate triples.
Steps:
1. Write: rate = k x 3[RX].
2. Factor: 3 x k[RX].
3. Rate becomes 3x. | organic_mechanisms |
Přesmyk karbokationu přesune vodík či alkyl na stabilnější uhlík. | null | che_0399 | concept | cs | Co je přesmyk karbokationu? | [
"Nestabilní karbokation.",
"Přesun H nebo alkylu.",
"Vznikne stabilnější ion."
] | chemistry | Answer: Přesun na stabilnější uhlík.
Steps:
1. Nestabilní karbokation.
2. Přesun H nebo alkylu.
3. Vznikne stabilnější ion. | organic_mechanisms |
A hydride shift moves H- to the cationic carbon, relocating the charge. | null | che_0400 | concept | en | What is a hydride shift? | [
"H- migrates to cation.",
"Charge moves along.",
"More stable cation results."
] | chemistry | Answer: H- migration to cation.
Steps:
1. H- migrates to cation.
2. Charge moves along.
3. More stable cation results. | organic_mechanisms |
Tertiary carbocations are most stable through three donating alkyl groups. | {
"A": "Methyl",
"B": "Primary",
"C": "Tertiary",
"D": "Vinyl"
} | che_0401 | multiple_choice | en | Which carbocation is most stable? | [
"More alkyls mean more stable.",
"Tertiary has three.",
"Select option C."
] | chemistry | Answer: Tertiary.
Steps:
1. More alkyls mean more stable.
2. Tertiary has three.
3. Select option C.
Answer: (C) | organic_mechanisms |
The product is 2-bromo-2-methylpropane, via the tertiary carbocation. | null | che_0402 | worked_example | en | Predict the HBr addition product of 2-methylpropene. | [
"H adds to CH2 end.",
"Tertiary carbocation forms.",
"Br captures it: 2-bromo-2-methylpropane."
] | chemistry | Answer: 2-bromo-2-methylpropane.
Steps:
1. H adds to CH2 end.
2. Tertiary carbocation forms.
3. Br captures it: 2-bromo-2-methylpropane. | organic_mechanisms |
Regioselectivity is the preference for one positional outcome over others. | null | che_0403 | concept | en | What is regioselectivity? | [
"Multiple positions possible.",
"One dominates.",
"Example: Markovnikov addition."
] | chemistry | Answer: Positional preference.
Steps:
1. Multiple positions possible.
2. One dominates.
3. Example: Markovnikov addition. | organic_mechanisms |
Markovnikovovo pravidlo říká, že vodík se aduje na méně substituovaný uhlík. | null | che_0404 | concept | cs | Co je Markovnikovovo pravidlo? | [
"H na méně substituovaný C.",
"Stabilnější karbokation.",
"Příklad: propen + HBr."
] | chemistry | Answer: H na méně substituovaný uhlík.
Steps:
1. H na méně substituovaný C.
2. Stabilnější karbokation.
3. Příklad: propen + HBr. | organic_mechanisms |
Peroxides trigger radical addition that reverses the usual orientation. | {
"A": "Acid catalyst",
"B": "Peroxides",
"C": "Low temperature",
"D": "Polar solvent"
} | che_0405 | multiple_choice | en | What causes anti-Markovnikov HBr addition? | [
"Needs radical pathway.",
"Peroxides start radicals.",
"Select option B."
] | chemistry | Answer: Peroxides.
Steps:
1. Needs radical pathway.
2. Peroxides start radicals.
3. Select option B.
Answer: (B) | organic_mechanisms |
The E1 product is 2-methylpropene, the only possible alkene. | null | che_0406 | worked_example | en | What is the E1 product of tert-butyl bromide with weak base? | [
"Br leaves, tertiary cation forms.",
"Base removes a methyl H.",
"Product: 2-methylpropene."
] | chemistry | Answer: 2-methylpropene.
Steps:
1. Br leaves, tertiary cation forms.
2. Base removes a methyl H.
3. Product: 2-methylpropene. | organic_mechanisms |
E1 is two-step via a carbocation while E2 is one concerted step with base. | null | che_0407 | concept | en | How do E1 and E2 differ? | [
"E1: cation, rate = k[RX].",
"E2: concerted, needs strong base.",
"E1 with weak base, E2 with strong."
] | chemistry | Answer: Two-step versus concerted.
Steps:
1. E1: cation, rate = k[RX].
2. E2: concerted, needs strong base.
3. E1 with weak base, E2 with strong. | organic_mechanisms |
Dehydration of ethanol gives ethene plus water. | null | che_0408 | worked_example | en | What is the product of ethanol dehydration (conc. H2SO4, heat)? | [
"Remove -OH and adjacent H.",
"Double bond forms.",
"Product: ethene."
] | chemistry | Answer: Ethene.
Steps:
1. Remove -OH and adjacent H.
2. Double bond forms.
3. Product: ethene. | organic_mechanisms |
Concentrated sulfuric acid with heat dehydrates alcohols to alkenes. | {
"A": "Dilute NaOH",
"B": "Cold water",
"C": "H2 with Ni",
"D": "Conc. H2SO4, heat"
} | che_0409 | multiple_choice | en | Which reagent dehydrates alcohols? | [
"Need strong dehydrating acid.",
"Conc. H2SO4 with heat works.",
"Select option D."
] | chemistry | Answer: Conc. H2SO4, heat.
Steps:
1. Need strong dehydrating acid.
2. Conc. H2SO4 with heat works.
3. Select option D.
Answer: (D) | organic_mechanisms |
Adicí HCl na ethen vzniká chlorethan. | null | che_0410 | worked_example | cs | Jaký produkt vznikne adicí HCl na ethen? | [
"H se aduje na uhlík.",
"Cl na druhý uhlík.",
"Produkt: chlorethan."
] | chemistry | Answer: Chlorethan.
Steps:
1. H se aduje na uhlík.
2. Cl na druhý uhlík.
3. Produkt: chlorethan. | organic_mechanisms |
Common electrophiles include H+, NO2+, carbocations, and carbonyl carbons. | null | che_0411 | concept | en | What are common electrophiles? | [
"Electron-poor centers.",
"Examples: H+, NO2+.",
"Carbonyl C also qualifies."
] | chemistry | Answer: Electron-poor acceptors.
Steps:
1. Electron-poor centers.
2. Examples: H+, NO2+.
3. Carbonyl C also qualifies. | organic_mechanisms |
Chlorination gives chloromethane plus HCl via a radical chain. | null | che_0412 | worked_example | en | What are the products of methane chlorination (Cl2, light)? | [
"Cl radical abstracts H.",
"Methyl radical takes Cl.",
"Products: CH3Cl + HCl."
] | chemistry | Answer: CH3Cl + HCl.
Steps:
1. Cl radical abstracts H.
2. Methyl radical takes Cl.
3. Products: CH3Cl + HCl. | organic_mechanisms |
UV light splits Cl2 into radicals that start the chain. | {
"A": "UV light",
"B": "Darkness",
"C": "Water",
"D": "Salt"
} | che_0413 | multiple_choice | en | What initiates methane chlorination? | [
"Need Cl radicals.",
"Light splits Cl2.",
"Select option A."
] | chemistry | Answer: UV light.
Steps:
1. Need Cl radicals.
2. Light splits Cl2.
3. Select option A.
Answer: (A) | organic_mechanisms |
Radikál je částice s nepárovým elektronem, velmi reaktivní. | null | che_0414 | concept | cs | Co je radikál? | [
"Nepárový elektron.",
"Vzniká homolýzou.",
"Řetězové reakce."
] | chemistry | Answer: Částice s nepárovým elektronem.
Steps:
1. Nepárový elektron.
2. Vzniká homolýzou.
3. Řetězové reakce. | organic_mechanisms |
The law of mass action states reaction rate is proportional to reactant concentration products. | null | che_0415 | concept | en | What is the law of mass action? | [
"Rate scales with concentrations.",
"Exponents from stoichiometry.",
"Basis of K expressions."
] | chemistry | Answer: Rate from concentration products.
Steps:
1. Rate scales with concentrations.
2. Exponents from stoichiometry.
3. Basis of K expressions. | equilibrium |
Pure solids and liquids have constant activity and are left out of K. | {
"A": "Gases",
"B": "Solids and liquids",
"C": "Ions",
"D": "Nothing"
} | che_0416 | multiple_choice | en | What is omitted from K expressions? | [
"Activities of pure phases are 1.",
"Omit solids and liquids.",
"Select option B."
] | chemistry | Answer: Solids and liquids.
Steps:
1. Activities of pure phases are 1.
2. Omit solids and liquids.
3. Select option B.
Answer: (B) | equilibrium |
K is 20, from 0.80 divided by 0.20 squared. | null | che_0417 | worked_example | en | For 2A <-> B with [A] = 0.20 M and [B] = 0.80 M, compute K. | [
"Write: K = [B]/[A]^2.",
"Substitute: 0.80/(0.20^2) = 0.80/0.04.",
"K = 20."
] | chemistry | Answer: 20.
Steps:
1. Write: K = [B]/[A]^2.
2. Substitute: 0.80/(0.20^2) = 0.80/0.04.
3. K = 20. | equilibrium |
Zákon působení hmoty říká, že rychlost závisí na součinu koncentrací reaktantů. | null | che_0418 | concept | cs | Co je zákon působení hmoty? | [
"Rychlost ze součinu koncentrací.",
"Exponenty ze stechiometrie.",
"Základ pro K."
] | chemistry | Answer: Rychlost ze součinu koncentrací.
Steps:
1. Rychlost ze součinu koncentrací.
2. Exponenty ze stechiometrie.
3. Základ pro K. | equilibrium |
A tiny K means reactants dominate at equilibrium. | {
"A": "Reactants dominate",
"B": "Products dominate",
"C": "No equilibrium",
"D": "K is zero"
} | che_0419 | multiple_choice | en | What does a very small K mean? | [
"K = products/reactants.",
"Small K favors denominator.",
"Select option A."
] | chemistry | Answer: Reactants dominate.
Steps:
1. K = products/reactants.
2. Small K favors denominator.
3. Select option A.
Answer: (A) | equilibrium |
The system is already at equilibrium because Q equals K. | null | che_0420 | worked_example | en | Q = 5 and K = 5. Which way does the system shift? | [
"Compare: Q = K.",
"No excess either side.",
"No shift; at equilibrium."
] | chemistry | Answer: No shift.
Steps:
1. Compare: Q = K.
2. No excess either side.
3. No shift; at equilibrium. | equilibrium |
Adding a species shifts equilibrium away; removing one shifts toward it. | null | che_0421 | concept | en | How do concentration changes shift equilibrium? | [
"Add reactant: shift right.",
"Remove product: shift right.",
"Reverse for opposite changes."
] | chemistry | Answer: Away from added species.
Steps:
1. Add reactant: shift right.
2. Remove product: shift right.
3. Reverse for opposite changes. | equilibrium |
Extra reactant is consumed by shifting toward products. | {
"A": "Shifts left",
"B": "No shift",
"C": "Shifts right",
"D": "K changes"
} | che_0422 | multiple_choice | en | Adding more reactant shifts equilibrium where? | [
"Reactants in excess.",
"System consumes them.",
"Select option C."
] | chemistry | Answer: Shifts right.
Steps:
1. Reactants in excess.
2. System consumes them.
3. Select option C.
Answer: (C) | equilibrium |
High pressure and low temperature favor ammonia: fewer moles and exothermic reaction. | null | che_0423 | worked_example | en | Which conditions favor NH3 in N2 + 3H2 <-> 2NH3 (exothermic)? | [
"Fewer gas moles right: high P.",
"Exothermic: low T.",
"High P plus low T favor NH3."
] | chemistry | Answer: High pressure, low temperature.
Steps:
1. Fewer gas moles right: high P.
2. Exothermic: low T.
3. High P plus low T favor NH3. | equilibrium |
Ochlazení exotermické rovnováhy ji posune k produktům. | null | che_0424 | worked_example | cs | Kam posune ochlazení exotermickou rovnováhu? | [
"Teplo je produkt.",
"Odebrání tepla.",
"Posun k produktům."
] | chemistry | Answer: K produktům.
Steps:
1. Teplo je produkt.
2. Odebrání tepla.
3. Posun k produktům. | equilibrium |
Removing heat from an exothermic reaction pulls it toward products. | {
"A": "Shifts right",
"B": "Shifts left",
"C": "No shift",
"D": "K becomes 1"
} | che_0425 | multiple_choice | en | Cooling an exothermic reaction shifts it which way? | [
"Heat is a product.",
"Removing product shifts right.",
"Select option A."
] | chemistry | Answer: Shifts right.
Steps:
1. Heat is a product.
2. Removing product shifts right.
3. Select option A.
Answer: (A) | equilibrium |
Heterogeneous equilibria mix phases, and K includes only gases and dissolved species. | null | che_0426 | concept | en | How is K written for heterogeneous equilibria? | [
"Include gases.",
"Include aqueous ions.",
"Omit pure solids, liquids."
] | chemistry | Answer: Gases and ions only.
Steps:
1. Include gases.
2. Include aqueous ions.
3. Omit pure solids, liquids. | equilibrium |
K equals [CO2] = 0.050, since solids are omitted. | null | che_0427 | worked_example | en | For CaCO3(s) <-> CaO(s) + CO2(g) with [CO2] = 0.050 M, find K. | [
"Write: K = [CO2].",
"Omit both solids.",
"K = 0.050."
] | chemistry | Answer: 0.050.
Steps:
1. Write: K = [CO2].
2. Omit both solids.
3. K = 0.050. | equilibrium |
With no mole change, the RT term vanishes and Kp equals Kc. | {
"A": "Δn = 1",
"B": "Δn = 0",
"C": "Δn = -1",
"D": "Never"
} | che_0428 | multiple_choice | en | When does Kp equal Kc? | [
"Kp = Kc(RT)^Δn.",
"Need exponent zero.",
"Select option B."
] | chemistry | Answer: Δn = 0.
Steps:
1. Kp = Kc(RT)^Δn.
2. Need exponent zero.
3. Select option B.
Answer: (B) | equilibrium |
Δn is -2, from 2 product moles minus 4 reactant moles. | null | che_0429 | worked_example | en | Compute Δn(gas) for N2 + 3H2 <-> 2NH3. | [
"Products: 2 moles.",
"Reactants: 1 + 3 = 4.",
"Δn = 2 - 4 = -2."
] | chemistry | Answer: -2.
Steps:
1. Products: 2 moles.
2. Reactants: 1 + 3 = 4.
3. Δn = 2 - 4 = -2. | equilibrium |
Heterogenní rovnováha zahrnuje více skupenství, například pevnou látku a plyn. | null | che_0430 | concept | cs | Co je heterogenní rovnováha? | [
"Více fází najednou.",
"Pevné látky se vynechávají z K.",
"Příklad: CaCO3 <-> CaO + CO2."
] | chemistry | Answer: Rovnováha více fází.
Steps:
1. Více fází najednou.
2. Pevné látky se vynechávají z K.
3. Příklad: CaCO3 <-> CaO + CO2. | equilibrium |
Catalysts speed both directions equally, leaving K unchanged. | {
"A": "K doubles",
"B": "K halves",
"C": "K becomes 1",
"D": "K unchanged"
} | che_0431 | multiple_choice | en | What does a catalyst do to K? | [
"Both rates rise equally.",
"Ratio stays same.",
"Select option D."
] | chemistry | Answer: K unchanged.
Steps:
1. Both rates rise equally.
2. Ratio stays same.
3. Select option D.
Answer: (D) | equilibrium |
The solubility is 1.21 x 10^-3 M, from (7.1 x 10^-9 / 4)^(1/3). | null | che_0432 | worked_example | en | PbI2 has Ksp = 7.1 x 10^-9. Compute its molar solubility. | [
"Write: Ksp = 4s^3.",
"s^3 = 7.1 x 10^-9 / 4 = 1.775 x 10^-9.",
"s = 1.21 x 10^-3 M."
] | chemistry | Answer: 1.21 x 10^-3 M.
Steps:
1. Write: Ksp = 4s^3.
2. s^3 = 7.1 x 10^-9 / 4 = 1.775 x 10^-9.
3. s = 1.21 x 10^-3 M. | equilibrium |
Each half-cell pairs an electrode with its ion solution for one half-reaction. | null | che_0433 | concept | en | What is a half-cell? | [
"Electrode plus solution.",
"Hosts one half-reaction.",
"Two halves make a cell."
] | chemistry | Answer: Electrode plus half-reaction.
Steps:
1. Electrode plus solution.
2. Hosts one half-reaction.
3. Two halves make a cell. | electrochemistry |
Reduction gains electrons at the cathode, remembered as RED CAT. | {
"A": "Anode",
"B": "Cathode",
"C": "Salt bridge",
"D": "Wire"
} | che_0434 | multiple_choice | en | Where does reduction occur? | [
"Recall RED CAT mnemonic.",
"Cathode gains electrons.",
"Select option B."
] | chemistry | Answer: Cathode.
Steps:
1. Recall RED CAT mnemonic.
2. Cathode gains electrons.
3. Select option B.
Answer: (B) | electrochemistry |
The cell potential is 1.61 V, from -0.76 - (-2.37). | null | che_0435 | worked_example | en | Compute E cell for Mg/Zn with E(Mg) = -2.37 V and E(Zn) = -0.76 V. | [
"Cathode is Zn at -0.76 V.",
"Anode is Mg at -2.37 V.",
"E = -0.76 - (-2.37) = 1.61 V."
] | chemistry | Answer: 1.61 V.
Steps:
1. Cathode is Zn at -0.76 V.
2. Anode is Mg at -2.37 V.
3. E = -0.76 - (-2.37) = 1.61 V. | electrochemistry |
Anoda je elektroda, kde probíhá oxidace; katoda, kde probíhá redukce. | null | che_0436 | concept | cs | Co je anoda a katoda? | [
"Anoda: oxidace.",
"Katoda: redukce.",
"Pomůcka: AN OX, RED KAT."
] | chemistry | Answer: Oxidace versus redukce.
Steps:
1. Anoda: oxidace.
2. Katoda: redukce.
3. Pomůcka: AN OX, RED KAT. | electrochemistry |
In a galvanic cell the anode is negative as it releases electrons. | {
"A": "Negative",
"B": "Positive",
"C": "Neutral",
"D": "Grounded"
} | che_0437 | multiple_choice | en | What is the anode's sign in a galvanic cell? | [
"Anode emits electrons.",
"It becomes negative.",
"Select option A."
] | chemistry | Answer: Negative.
Steps:
1. Anode emits electrons.
2. It becomes negative.
3. Select option A.
Answer: (A) | electrochemistry |
Standard potentials use 1 M ions, 1 atm gases, and 25 C. | null | che_0438 | concept | en | What are standard conditions for E°? | [
"Ions at 1 M.",
"Gases at 1 atm.",
"Temperature 25 C."
] | chemistry | Answer: 1 M, 1 atm, 25 C.
Steps:
1. Ions at 1 M.
2. Gases at 1 atm.
3. Temperature 25 C. | electrochemistry |
ΔG is -77.2 kJ, from -1 x 96485 x 0.80 J. | null | che_0439 | worked_example | en | Compute ΔG for n = 1 and E = 0.80 V (F = 96485 C/mol). | [
"Write: ΔG = -1 x 96485 x 0.80.",
"Multiply: -77,188 J.",
"ΔG = -77.2 kJ."
] | chemistry | Answer: -77.2 kJ.
Steps:
1. Write: ΔG = -1 x 96485 x 0.80.
2. Multiply: -77,188 J.
3. ΔG = -77.2 kJ. | electrochemistry |
The potential is 0.68 V, from 0.80 - 0.0592 x log(100). | null | che_0440 | worked_example | en | E° = 0.80 V, n = 1, Q = 100 at 25 C. Compute E. | [
"Write: E = 0.80 - 0.0592 x log(100).",
"log(100) = 2.",
"E = 0.80 - 0.118 = 0.68 V."
] | chemistry | Answer: 0.68 V.
Steps:
1. Write: E = 0.80 - 0.0592 x log(100).
2. log(100) = 2.
3. E = 0.80 - 0.118 = 0.68 V. | electrochemistry |
At 25 C the Nernst slope is 0.0592 V per decade per electron. | {
"A": "0.0257 V",
"B": "0.0592 mV",
"C": "0.0592 V",
"D": "1.00 V"
} | che_0441 | multiple_choice | en | What is the Nernst slope at 25 C? | [
"2.303 RT/F at 298 K.",
"Equals 0.0592 V.",
"Select option C."
] | chemistry | Answer: 0.0592 V.
Steps:
1. 2.303 RT/F at 298 K.
2. Equals 0.0592 V.
3. Select option C.
Answer: (C) | electrochemistry |
Water electrolysis gives hydrogen at the cathode and oxygen at the anode. | null | che_0442 | concept | en | What are the products of water electrolysis? | [
"Cathode reduces H+ to H2.",
"Anode oxidizes water to O2.",
"Ratio H2:O2 is 2:1."
] | chemistry | Answer: H2 and O2.
Steps:
1. Cathode reduces H+ to H2.
2. Anode oxidizes water to O2.
3. Ratio H2:O2 is 2:1. | electrochemistry |
Vzniknou 2,02 g vodíku, protože 2 moly elektronů dají 1 mol H2. | null | che_0443 | worked_example | cs | Kolik vodíku vznikne ze 2 molů elektronů při elektrolýze? | [
"2 e- dají 1 H2.",
"Moly H2 = 1.",
"Hmotnost = 1 x 2,016 = 2,02 g."
] | chemistry | Answer: 2,02 g.
Steps:
1. 2 e- dají 1 H2.
2. Moly H2 = 1.
3. Hmotnost = 1 x 2,016 = 2,02 g. | electrochemistry |
The hydrogen volume is 11.2 L, from 0.50 mol times 22.414 L/mol. | null | che_0444 | worked_example | en | What volume of H2 at STP comes from 1.00 mol of electrons? | [
"2 e- give 1 H2: 0.50 mol.",
"Multiply: 0.50 x 22.414.",
"V = 11.2 L."
] | chemistry | Answer: 11.2 L.
Steps:
1. 2 e- give 1 H2: 0.50 mol.
2. Multiply: 0.50 x 22.414.
3. V = 11.2 L. | electrochemistry |
Chloride oxidizes to chlorine gas at the anode of brine electrolysis. | {
"A": "O2",
"B": "H2",
"C": "Na",
"D": "Cl2"
} | che_0445 | multiple_choice | en | What forms at the anode in brine electrolysis? | [
"Cl- oxidizes.",
"Product is Cl2 gas.",
"Select option D."
] | chemistry | Answer: Cl2.
Steps:
1. Cl- oxidizes.
2. Product is Cl2 gas.
3. Select option D.
Answer: (D) | electrochemistry |
The dry cell pairs a zinc anode with a carbon cathode in ammonium chloride paste. | null | che_0446 | concept | en | How does a dry cell work? | [
"Zn can is the anode.",
"Carbon rod is cathode.",
"Paste electrolyte conducts."
] | chemistry | Answer: Zn anode, carbon cathode.
Steps:
1. Zn can is the anode.
2. Carbon rod is cathode.
3. Paste electrolyte conducts. | electrochemistry |
The current is 13.4 A, from 48,243 C divided by 3600 s. | null | che_0447 | worked_example | en | What current passes 0.500 mol of electrons in 1.00 hour? | [
"Q = 0.500 x 96485 = 48,243 C.",
"t = 3600 s.",
"I = 48,243/3600 = 13.4 A."
] | chemistry | Answer: 13.4 A.
Steps:
1. Q = 0.500 x 96485 = 48,243 C.
2. t = 3600 s.
3. I = 48,243/3600 = 13.4 A. | electrochemistry |
Akumulátor je dobíjecí článek, jehož reakci lze proudem obrátit. | null | che_0448 | concept | cs | Co je akumulátor? | [
"Galvanický článek.",
"Reakce je vratná.",
"Nabíjení ji obrátí."
] | chemistry | Answer: Dobíjecí článek.
Steps:
1. Galvanický článek.
2. Reakce je vratná.
3. Nabíjení ji obrátí. | electrochemistry |
The lead-acid cell is rechargeable by driving current backward. | {
"A": "Leclanché dry cell",
"B": "Lead-acid cell",
"C": "Daniell cell",
"D": "Concentration cell"
} | che_0449 | multiple_choice | en | Which cell is rechargeable? | [
"Need reversible chemistry.",
"Lead-acid reverses.",
"Select option B."
] | chemistry | Answer: Lead-acid cell.
Steps:
1. Need reversible chemistry.
2. Lead-acid reverses.
3. Select option B.
Answer: (B) | electrochemistry |
The time is 32.2 min, from 9,649 C divided by 5.00 A. | null | che_0450 | worked_example | en | How long at 5.00 A to pass 0.100 mol of electrons? | [
"Q = 0.100 x 96485 = 9,649 C.",
"t = 9,649/5.00 = 1,930 s.",
"t = 32.2 min."
] | chemistry | Answer: 32.2 min.
Steps:
1. Q = 0.100 x 96485 = 9,649 C.
2. t = 9,649/5.00 = 1,930 s.
3. t = 32.2 min. | electrochemistry |
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