answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
Balance atoms first, then add H+, H2O, and electrons to balance charge in acidic solution. | null | che_0051 | concept | en | How is redox balanced in acidic solution? | [
"Split into half-reactions.",
"Balance atoms with H2O and H+.",
"Balance charge with electrons."
] | chemistry | Answer: Half-reactions with H+, H2O, electrons.
Steps:
1. Split into half-reactions.
2. Balance atoms with H2O and H+.
3. Balance charge with electrons. | redox |
Redoxní reakce je přenos elektronů mezi oxidovadlem a redukovadlem. | null | che_0052 | concept | cs | Co je redoxní reakce? | [
"Definuj přenos elektronů.",
"Oxidovadlo se redukuje.",
"Redukovadlo se oxiduje."
] | chemistry | Answer: Přenos elektronů.
Steps:
1. Definuj přenos elektronů.
2. Oxidovadlo se redukuje.
3. Redukovadlo se oxiduje. | redox |
Carbon is -4 in methane, since C + 4(+1) = 0 gives C = -4. | {
"A": "-4",
"B": "0",
"C": "+2",
"D": "+4"
} | che_0053 | multiple_choice | en | What is the oxidation state of carbon in CH4? | [
"Assign H +1.",
"Solve: C + 4 = 0.",
"C = -4; select A."
] | chemistry | Answer: -4.
Steps:
1. Assign H +1.
2. Solve: C + 4 = 0.
3. C = -4; select A.
Answer: (A) | redox |
The balanced half-reaction is MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O. | null | che_0054 | worked_example | en | Balance the half-reaction MnO4- -> Mn2+ in acid. | [
"Balance O with 4 H2O.",
"Balance H with 8 H+.",
"Balance charge: left +7, add 5 e-."
] | chemistry | Answer: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O.
Steps:
1. Balance O with 4 H2O.
2. Balance H with 8 H+.
3. Balance charge: left +7, add 5 e-. | redox |
Spectator ions appear unchanged on both sides and are omitted from the net ionic equation. | null | che_0055 | concept | en | What are spectator ions? | [
"Identify ions unchanged by reaction.",
"Example: Na+ in precipitation.",
"Cancel them for net equation."
] | chemistry | Answer: Ions unchanged, omitted from net equation.
Steps:
1. Identify ions unchanged by reaction.
2. Example: Na+ in precipitation.
3. Cancel them for net equation. | redox |
Aluminum is the reducing agent because it loses electrons going from 0 to +3. | {
"A": "Fe2O3",
"B": "Al",
"C": "Fe",
"D": "Al2O3"
} | che_0056 | multiple_choice | en | In 2Al + Fe2O3 -> Al2O3 + 2Fe, what is the reducing agent? | [
"Find species oxidized: Al 0 -> +3.",
"Reducer is the oxidized species.",
"Select option B."
] | chemistry | Answer: Al.
Steps:
1. Find species oxidized: Al 0 -> +3.
2. Reducer is the oxidized species.
3. Select option B.
Answer: (B) | redox |
Vyčíslená rovnice je 2H2 + O2 -> 2H2O. | null | che_0057 | worked_example | cs | Vyčísli: H2 + O2 -> H2O. | [
"Vodík: 2 H2O a 2 H2.",
"Kyslík: 1 O2 na 2 O.",
"Výsledek: 2H2 + O2 -> 2H2O."
] | chemistry | Answer: 2H2 + O2 -> 2H2O.
Steps:
1. Vodík: 2 H2O a 2 H2.
2. Kyslík: 1 O2 na 2 O.
3. Výsledek: 2H2 + O2 -> 2H2O. | redox |
Exothermic reactions release heat with negative ΔH, while endothermic reactions absorb heat with positive ΔH. | null | che_0058 | concept | en | How do exothermic and endothermic reactions differ? | [
"Exothermic: heat out, ΔH negative.",
"Endothermic: heat in, ΔH positive.",
"Example: combustion versus ice melting."
] | chemistry | Answer: Heat released versus absorbed.
Steps:
1. Exothermic: heat out, ΔH negative.
2. Endothermic: heat in, ΔH positive.
3. Example: combustion versus ice melting. | thermochemistry |
An exothermic reaction has negative ΔH because the system loses heat to the surroundings. | {
"A": "Positive ΔH",
"B": "Negative ΔH",
"C": "Zero ΔH",
"D": "Infinite ΔH"
} | che_0059 | multiple_choice | en | What is the sign of ΔH for an exothermic reaction? | [
"Recall exo means heat exits.",
"Enthalpy of system drops.",
"Select option B."
] | chemistry | Answer: Negative ΔH.
Steps:
1. Recall exo means heat exits.
2. Enthalpy of system drops.
3. Select option B.
Answer: (B) | thermochemistry |
Entalpie je tepelný obsah soustavy; její změna ΔH udává teplo reakce při stálém tlaku. | null | che_0060 | concept | cs | Co je entalpie? | [
"Definuj tepelný obsah soustavy.",
"ΔH je teplo při stálém tlaku.",
"Záporná = exotermická."
] | chemistry | Answer: Tepelný obsah soustavy.
Steps:
1. Definuj tepelný obsah soustavy.
2. ΔH je teplo při stálém tlaku.
3. Záporná = exotermická. | thermochemistry |
The heat required is 50.2 kJ, from 200 g x 4.184 J/gC x 60 C. | null | che_0061 | worked_example | en | How much heat warms 200 g of water from 20 C to 80 C (c = 4.184 J/gC)? | [
"Write: q = m x c x ΔT.",
"Substitute: 200 x 4.184 x 60.",
"q = 50,208 J = 50.2 kJ."
] | chemistry | Answer: 50.2 kJ.
Steps:
1. Write: q = m x c x ΔT.
2. Substitute: 200 x 4.184 x 60.
3. q = 50,208 J = 50.2 kJ. | thermochemistry |
Liquid water has a specific heat of 4.184 J/gC, unusually high due to hydrogen bonding. | {
"A": "1.00 J/gC",
"B": "2.05 J/gC",
"C": "4.184 J/gC",
"D": "8.314 J/gC"
} | che_0062 | multiple_choice | en | What is the specific heat capacity of liquid water? | [
"Recall 4.184 J/gC.",
"Eliminate other values.",
"Select option C."
] | chemistry | Answer: 4.184 J/gC.
Steps:
1. Recall 4.184 J/gC.
2. Eliminate other values.
3. Select option C.
Answer: (C) | thermochemistry |
Hess's law states the total enthalpy change is the sum of step changes, independent of path. | null | che_0063 | concept | en | What is Hess's law? | [
"Enthalpy is a state function.",
"Sum ΔH of steps.",
"Path independence holds."
] | chemistry | Answer: Total ΔH equals sum of steps.
Steps:
1. Enthalpy is a state function.
2. Sum ΔH of steps.
3. Path independence holds. | thermochemistry |
The overall ΔH is -15 kJ, from +10 plus -25 kJ. | null | che_0064 | worked_example | en | A->B has ΔH = +10 kJ and B->C has ΔH = -25 kJ. Find ΔH for A->C. | [
"Apply Hess's law: sum steps.",
"Add: 10 + (-25).",
"ΔH = -15 kJ."
] | chemistry | Answer: -15 kJ.
Steps:
1. Apply Hess's law: sum steps.
2. Add: 10 + (-25).
3. ΔH = -15 kJ. | thermochemistry |
Standard enthalpy of formation uses 1 bar pressure and 298 K with elements in reference states. | {
"A": "1 bar and 298 K",
"B": "0 bar and 0 K",
"C": "10 bar and 500 K",
"D": "STP only"
} | che_0065 | multiple_choice | en | What are standard conditions for ΔHf? | [
"Recall 1 bar pressure.",
"Recall 298 K temperature.",
"Select option A."
] | chemistry | Answer: 1 bar and 298 K.
Steps:
1. Recall 1 bar pressure.
2. Recall 298 K temperature.
3. Select option A.
Answer: (A) | thermochemistry |
Potřebné teplo je 4184 J, protože 100 x 4,184 x 10 se rovná 4184. | null | che_0066 | worked_example | cs | Kolik tepla ohřeje 100 g vody o 10 °C (c = 4,184 J/g°C)? | [
"Napiš vzorec: q = m x c x ΔT.",
"Dosaď: 100 x 4,184 x 10.",
"q = 4184 J."
] | chemistry | Answer: 4184 J.
Steps:
1. Napiš vzorec: q = m x c x ΔT.
2. Dosaď: 100 x 4,184 x 10.
3. q = 4184 J. | thermochemistry |
Reaction ΔH is estimated as bonds broken minus bonds formed, since breaking costs and forming releases energy. | null | che_0067 | concept | en | How do bond energies estimate reaction enthalpy? | [
"Sum energy of bonds broken.",
"Subtract energy of bonds formed.",
"Positive means endothermic."
] | chemistry | Answer: Broken minus formed.
Steps:
1. Sum energy of bonds broken.
2. Subtract energy of bonds formed.
3. Positive means endothermic. | thermochemistry |
The ΔH is -184 kJ, from 678 kJ broken minus 862 kJ formed. | null | che_0068 | worked_example | en | Estimate ΔH for H2 + Cl2 -> 2HCl using H-H 436, Cl-Cl 242, H-Cl 431 kJ/mol. | [
"Broken: 436 + 242 = 678 kJ.",
"Formed: 2 x 431 = 862 kJ.",
"ΔH = 678 - 862 = -184 kJ."
] | chemistry | Answer: -184 kJ.
Steps:
1. Broken: 436 + 242 = 678 kJ.
2. Formed: 2 x 431 = 862 kJ.
3. ΔH = 678 - 862 = -184 kJ. | thermochemistry |
Entropy rises when a solid vaporizes because gas particles have far more disorder. | {
"A": "Gas condenses",
"B": "Liquid freezes",
"C": "Gas dissolves",
"D": "Solid vaporizes"
} | che_0069 | multiple_choice | en | In which change does entropy increase most? | [
"Recall gas has highest entropy.",
"Only vaporization creates gas.",
"Select option D."
] | chemistry | Answer: Solid vaporizes.
Steps:
1. Recall gas has highest entropy.
2. Only vaporization creates gas.
3. Select option D.
Answer: (D) | thermochemistry |
Exotermická reakce uvolňuje teplo do okolí a má zápornou změnu entalpie. | null | che_0070 | concept | cs | Co je exotermická reakce? | [
"Definuj uvolnění tepla.",
"ΔH je záporné.",
"Příklad: hoření metanu."
] | chemistry | Answer: Reakce uvolňující teplo.
Steps:
1. Definuj uvolnění tepla.
2. ΔH je záporné.
3. Příklad: hoření metanu. | thermochemistry |
ΔG is -40.4 kJ, so the reaction is spontaneous at 298 K. | null | che_0071 | worked_example | en | Given ΔH = -100 kJ and ΔS = -0.200 kJ/K at 298 K, compute ΔG. | [
"Write: ΔG = ΔH - TΔS.",
"Substitute: -100 - 298 x (-0.200) = -100 + 59.6.",
"ΔG = -40.4 kJ, spontaneous."
] | chemistry | Answer: -40.4 kJ.
Steps:
1. Write: ΔG = ΔH - TΔS.
2. Substitute: -100 - 298 x (-0.200) = -100 + 59.6.
3. ΔG = -40.4 kJ, spontaneous. | thermochemistry |
A negative Gibbs free energy change means the process is spontaneous under the given conditions. | {
"A": "Positive ΔG",
"B": "Negative ΔG",
"C": "Zero ΔG always",
"D": "ΔG is irrelevant"
} | che_0072 | multiple_choice | en | What ΔG sign indicates a spontaneous process? | [
"Recall spontaneity needs ΔG < 0.",
"Eliminate positive, zero, irrelevant.",
"Select option B."
] | chemistry | Answer: Negative ΔG.
Steps:
1. Recall spontaneity needs ΔG < 0.
2. Eliminate positive, zero, irrelevant.
3. Select option B.
Answer: (B) | thermochemistry |
Calorimetry measures heat flow from temperature change in an insulated vessel of known heat capacity. | null | che_0073 | concept | en | What is calorimetry? | [
"Run reaction in insulated vessel.",
"Measure temperature change.",
"Compute q from mass and heat capacity."
] | chemistry | Answer: Measuring heat via temperature change.
Steps:
1. Run reaction in insulated vessel.
2. Measure temperature change.
3. Compute q from mass and heat capacity. | thermochemistry |
Hessův zákon říká, že celková změna entalpie je součtem změn dílčích kroků. | null | che_0074 | concept | cs | Co je Hessův zákon? | [
"Entalpie je stavová funkce.",
"Sečti ΔH kroků.",
"Cesta nehraje roli."
] | chemistry | Answer: Celkové ΔH je součet kroků.
Steps:
1. Entalpie je stavová funkce.
2. Sečti ΔH kroků.
3. Cesta nehraje roli. | thermochemistry |
A spontaneous process increases the entropy of the universe, per the second law. | {
"A": "ΔS universe < 0",
"B": "ΔS universe = 0",
"C": "ΔS universe > 0",
"D": "ΔS universe is undefined"
} | che_0075 | multiple_choice | en | What is ΔS of the universe for a spontaneous process? | [
"Recall second law of thermodynamics.",
"Spontaneous means universe entropy rises.",
"Select option C."
] | chemistry | Answer: ΔS universe > 0.
Steps:
1. Recall second law of thermodynamics.
2. Spontaneous means universe entropy rises.
3. Select option C.
Answer: (C) | thermochemistry |
Burning 32.0 g of methane releases 1780 kJ, from 2.00 mol times 890 kJ/mol. | null | che_0076 | worked_example | en | Methane's heat of combustion is 890 kJ/mol. How much heat from 32.0 g (M = 16.04 g/mol)? | [
"Moles = 32.0 / 16.04 = 2.00 mol.",
"Multiply: 2.00 x 890.",
"Heat = 1780 kJ released."
] | chemistry | Answer: 1780 kJ.
Steps:
1. Moles = 32.0 / 16.04 = 2.00 mol.
2. Multiply: 2.00 x 890.
3. Heat = 1780 kJ released. | thermochemistry |
Alkanes are named by longest carbon chain with the -ane suffix: methane, ethane, propane, butane. | null | che_0077 | concept | en | How are simple alkanes named? | [
"Find the longest carbon chain.",
"Use the -ane suffix.",
"Count: 1 methane, 2 ethane, 3 propane."
] | chemistry | Answer: Longest chain plus -ane.
Steps:
1. Find the longest carbon chain.
2. Use the -ane suffix.
3. Count: 1 methane, 2 ethane, 3 propane. | organic_naming |
A four-carbon straight chain is butane. | {
"A": "Propane",
"B": "Butane",
"C": "Pentane",
"D": "Ethene"
} | che_0078 | multiple_choice | en | What is the name of CH3CH2CH2CH3? | [
"Count carbons: four.",
"Single bonds mean alkane.",
"Select option B."
] | chemistry | Answer: Butane.
Steps:
1. Count carbons: four.
2. Single bonds mean alkane.
3. Select option B.
Answer: (B) | organic_naming |
Alkany se pojmenují podle nejdelšího řetězce koncovkou -an: methan, ethan, propan, butan. | null | che_0079 | concept | cs | Jak se pojmenovávají alkany? | [
"Najdi nejdelší řetězec.",
"Přidej koncovku -an.",
"Počítej: 1 methan, 2 ethan."
] | chemistry | Answer: Řetězec plus -an.
Steps:
1. Najdi nejdelší řetězec.
2. Přidej koncovku -an.
3. Počítej: 1 methan, 2 ethan. | organic_naming |
The name is 2-methylpropane: a three-carbon chain with a methyl branch on carbon 2. | null | che_0080 | worked_example | en | Name CH3CH(CH3)CH3. | [
"Longest chain: 3 carbons = propane.",
"Methyl branch on carbon 2.",
"Name: 2-methylpropane."
] | chemistry | Answer: 2-methylpropane.
Steps:
1. Longest chain: 3 carbons = propane.
2. Methyl branch on carbon 2.
3. Name: 2-methylpropane. | organic_naming |
The -OH group marks an alcohol, named with the -ol suffix. | {
"A": "Alcohol",
"B": "Ketone",
"C": "Ether",
"D": "Ester"
} | che_0081 | multiple_choice | en | Which functional group contains -OH on carbon? | [
"Recall -OH means alcohol.",
"Eliminate ketone, ether, ester.",
"Select option A."
] | chemistry | Answer: Alcohol.
Steps:
1. Recall -OH means alcohol.
2. Eliminate ketone, ether, ester.
3. Select option A.
Answer: (A) | organic_naming |
Alkenes use the -ene suffix for double bonds and alkynes use -yne for triple bonds. | null | che_0082 | concept | en | How do alkene and alkyne names differ? | [
"Double bond gives -ene.",
"Triple bond gives -yne.",
"Example: ethene versus ethyne."
] | chemistry | Answer: -ene for double, -yne for triple.
Steps:
1. Double bond gives -ene.
2. Triple bond gives -yne.
3. Example: ethene versus ethyne. | organic_naming |
The name is propene: three carbons with a double bond starting at carbon 1. | null | che_0083 | worked_example | en | Name CH2=CHCH3. | [
"Count 3 carbons.",
"Double bond gives -ene.",
"Name: propene."
] | chemistry | Answer: Propene.
Steps:
1. Count 3 carbons.
2. Double bond gives -ene.
3. Name: propene. | organic_naming |
CH3COOH je kyselina ethanová, běžně zvaná kyselina octová. | null | che_0084 | worked_example | cs | Pojmenuj sloučeninu CH3COOH. | [
"Skupina -COOH je karboxylová kyselina.",
"Dva uhlíky = ethanová.",
"Triviálně kyselina octová."
] | chemistry | Answer: Kyselina ethanová (octová).
Steps:
1. Skupina -COOH je karboxylová kyselina.
2. Dva uhlíky = ethanová.
3. Triviálně kyselina octová. | organic_naming |
Benzene is aromatic with a delocalized six-electron ring, unlike the non-aromatic choices. | {
"A": "Cyclohexane",
"B": "Hexene",
"C": "Ethanol",
"D": "Benzene"
} | che_0085 | multiple_choice | en | Which compound is aromatic? | [
"Recall aromatic means benzene-like ring.",
"Eliminate cyclohexane, hexene, ethanol.",
"Select option D."
] | chemistry | Answer: Benzene.
Steps:
1. Recall aromatic means benzene-like ring.
2. Eliminate cyclohexane, hexene, ethanol.
3. Select option D.
Answer: (D) | organic_naming |
Methyl is -CH3 and ethyl is -CH2CH3, named as substituents with position numbers. | null | che_0086 | concept | en | What are methyl and ethyl substituents? | [
"Methyl = one-carbon branch.",
"Ethyl = two-carbon branch.",
"Number their positions on the chain."
] | chemistry | Answer: One- and two-carbon branches.
Steps:
1. Methyl = one-carbon branch.
2. Ethyl = two-carbon branch.
3. Number their positions on the chain. | organic_naming |
The name is ethanol: two carbons with an -OH group. | null | che_0087 | worked_example | en | Name CH3CH2OH. | [
"Count 2 carbons.",
"-OH gives -ol suffix.",
"Name: ethanol."
] | chemistry | Answer: Ethanol.
Steps:
1. Count 2 carbons.
2. -OH gives -ol suffix.
3. Name: ethanol. | organic_naming |
The -COOH group defines a carboxylic acid. | {
"A": "Alcohol",
"B": "Aldehyde",
"C": "Carboxylic acid",
"D": "Amine"
} | che_0088 | multiple_choice | en | Which functional group is -COOH? | [
"Recall C(=O)OH is carboxyl.",
"Eliminate alcohol, aldehyde, amine.",
"Select option C."
] | chemistry | Answer: Carboxylic acid.
Steps:
1. Recall C(=O)OH is carboxyl.
2. Eliminate alcohol, aldehyde, amine.
3. Select option C.
Answer: (C) | organic_naming |
Structural isomers share a molecular formula but differ in atom connectivity. | null | che_0089 | concept | en | What are structural isomers? | [
"Same formula, different connections.",
"Example: butane vs methylpropane.",
"Properties differ."
] | chemistry | Answer: Same formula, different connectivity.
Steps:
1. Same formula, different connections.
2. Example: butane vs methylpropane.
3. Properties differ. | organic_naming |
Izomery mají stejný vzorec, ale různé uspořádání atomů. | null | che_0090 | concept | cs | Co jsou izomery? | [
"Stejný molekulový vzorec.",
"Různé uspořádání.",
"Příklad: butan a methylpropan."
] | chemistry | Answer: Stejný vzorec, jiné uspořádání.
Steps:
1. Stejný molekulový vzorec.
2. Různé uspořádání.
3. Příklad: butan a methylpropan. | organic_naming |
The name is butan-2-ol: four carbons with -OH on carbon 2. | null | che_0091 | worked_example | en | Name CH3CH(OH)CH2CH3. | [
"Longest chain: 4 = butane.",
"-OH on carbon 2.",
"Name: butan-2-ol."
] | chemistry | Answer: Butan-2-ol.
Steps:
1. Longest chain: 4 = butane.
2. -OH on carbon 2.
3. Name: butan-2-ol. | organic_naming |
A mid-chain C=O is a ketone, while a terminal C=O with H is an aldehyde. | {
"A": "Aldehyde",
"B": "Ketone",
"C": "Acid",
"D": "Ester"
} | che_0092 | multiple_choice | en | A carbonyl in the middle of a chain is what? | [
"Mid-chain C=O means ketone.",
"Eliminate aldehyde, acid, ester.",
"Select option B."
] | chemistry | Answer: Ketone.
Steps:
1. Mid-chain C=O means ketone.
2. Eliminate aldehyde, acid, ester.
3. Select option B.
Answer: (B) | organic_naming |
Esters combine an acid part and an alcohol part, named as alkyl alkanoate. | null | che_0093 | concept | en | How are esters named? | [
"Alcohol part gives alkyl prefix.",
"Acid part gives alkanoate.",
"Example: methyl ethanoate."
] | chemistry | Answer: Alkyl alkanoate.
Steps:
1. Alcohol part gives alkyl prefix.
2. Acid part gives alkanoate.
3. Example: methyl ethanoate. | organic_naming |
CH3COCH3 je propanon, triviálně aceton. | null | che_0094 | worked_example | cs | Pojmenuj sloučeninu CH3COCH3. | [
"Karbonyl uprostřed = keton.",
"Tři uhlíky = propanon.",
"Triviálně aceton."
] | chemistry | Answer: Propanon (aceton).
Steps:
1. Karbonyl uprostřed = keton.
2. Tři uhlíky = propanon.
3. Triviálně aceton. | organic_naming |
Carboxylic acids outrank aldehydes, ketones, alcohols, and amines for the suffix. | null | che_0095 | concept | en | Which functional group has naming priority? | [
"Carboxylic acid is highest.",
"Then aldehyde, ketone, alcohol.",
"Highest group sets the suffix."
] | chemistry | Answer: Carboxylic acid.
Steps:
1. Carboxylic acid is highest.
2. Then aldehyde, ketone, alcohol.
3. Highest group sets the suffix. | organic_naming |
Nucleophiles donate electron pairs to electron-poor electrophiles. | null | che_0096 | concept | en | What are nucleophiles and electrophiles? | [
"Nucleophile: electron-rich donor.",
"Electrophile: electron-poor acceptor.",
"They attract each other."
] | chemistry | Answer: Pair donors versus acceptors.
Steps:
1. Nucleophile: electron-rich donor.
2. Electrophile: electron-poor acceptor.
3. They attract each other. | organic_mechanisms |
Hydroxide is a nucleophile with lone pairs ready to donate. | {
"A": "OH(-)",
"B": "H(+)",
"C": "BF3",
"D": "Carbocation"
} | che_0097 | multiple_choice | en | Which species is a nucleophile? | [
"Need electron-rich donor.",
"Eliminate H+, BF3, carbocation.",
"Select option A."
] | chemistry | Answer: OH-.
Steps:
1. Need electron-rich donor.
2. Eliminate H+, BF3, carbocation.
3. Select option A.
Answer: (A) | organic_mechanisms |
SN1 is two-step via a carbocation with rate depending on substrate only, while SN2 is one-step with both reactants in the rate. | null | che_0098 | concept | en | How do SN1 and SN2 reactions differ? | [
"SN1: carbocation, rate = k[RX].",
"SN2: concerted, rate = k[RX][Nu].",
"SN1 favors tertiary, SN2 methyl."
] | chemistry | Answer: Two-step versus one-step substitution.
Steps:
1. SN1: carbocation, rate = k[RX].
2. SN2: concerted, rate = k[RX][Nu].
3. SN1 favors tertiary, SN2 methyl. | organic_mechanisms |
Nukleofil je částice bohatá na elektrony, která je donuje elektrofilu. | null | che_0099 | concept | cs | Co je nukleofil? | [
"Bohatý na elektrony.",
"Napadá elektrofil.",
"Příklad: OH-, CN-."
] | chemistry | Answer: Donor elektronového páru.
Steps:
1. Bohatý na elektrony.
2. Napadá elektrofil.
3. Příklad: OH-, CN-. | organic_mechanisms |
The SN1 rate doubles to 2x because rate depends only on substrate concentration. | null | che_0100 | worked_example | en | In an SN1 reaction with rate = k[RX], what happens to the rate if [RX] doubles? | [
"Write: rate = k[RX].",
"Double [RX]: rate = k x 2[RX].",
"Rate doubles to 2x."
] | chemistry | Answer: Rate doubles.
Steps:
1. Write: rate = k[RX].
2. Double [RX]: rate = k x 2[RX].
3. Rate doubles to 2x. | organic_mechanisms |
Tertiary halides form the most stable carbocations, so they react fastest by SN1. | {
"A": "Methyl",
"B": "Primary",
"C": "Tertiary",
"D": "Aryl"
} | che_0101 | multiple_choice | en | Which substrate reacts fastest by SN1? | [
"SN1 needs stable carbocation.",
"Tertiary is most stable.",
"Select option C."
] | chemistry | Answer: Tertiary.
Steps:
1. SN1 needs stable carbocation.
2. Tertiary is most stable.
3. Select option C.
Answer: (C) | organic_mechanisms |
Carbocation stability rises with substitution: tertiary beats secondary beats primary beats methyl. | null | che_0102 | concept | en | What is the order of carbocation stability? | [
"Alkyl groups donate electron density.",
"More substitution means more stable.",
"Order: 3 > 2 > 1 > methyl."
] | chemistry | Answer: Tertiary > secondary > primary > methyl.
Steps:
1. Alkyl groups donate electron density.
2. More substitution means more stable.
3. Order: 3 > 2 > 1 > methyl. | organic_mechanisms |
The major product is 2-bromopropane, as H adds to carbon 1 and Br to carbon 2. | null | che_0103 | worked_example | en | Predict the major product of HBr addition to propene. | [
"Apply Markovnikov: H to less substituted C.",
"H adds to C1, Br to C2.",
"Product: 2-bromopropane."
] | chemistry | Answer: 2-bromopropane.
Steps:
1. Apply Markovnikov: H to less substituted C.
2. H adds to C1, Br to C2.
3. Product: 2-bromopropane. | organic_mechanisms |
Karbokation je kladně nabitý uhlík se šesti elektrony, stabilizovaný alkyly. | null | che_0104 | concept | cs | Co je karbokation? | [
"Kladný uhlík, sextet elektronů.",
"Planární geometrie.",
"Stabilita: terciární nejvyšší."
] | chemistry | Answer: Kladně nabitý uhlík.
Steps:
1. Kladný uhlík, sextet elektronů.
2. Planární geometrie.
3. Stabilita: terciární nejvyšší. | organic_mechanisms |
E2 elimination needs a strong base to remove a proton as the leaving group departs. | {
"A": "Weak base, polar protic",
"B": "Strong base, concerted step",
"C": "No base at all",
"D": "Acid catalyst only"
} | che_0105 | multiple_choice | en | Which conditions favor E2 elimination? | [
"E2 is concerted with base.",
"Strong base removes proton.",
"Select option B."
] | chemistry | Answer: Strong base, concerted step.
Steps:
1. E2 is concerted with base.
2. Strong base removes proton.
3. Select option B.
Answer: (B) | organic_mechanisms |
Electrophilic addition breaks the pi bond of an alkene as two groups add across it. | null | che_0106 | concept | en | What is electrophilic addition to alkenes? | [
"Pi bond attacks electrophile.",
"Carbocation forms.",
"Nucleophile captures it."
] | chemistry | Answer: Addition across the double bond.
Steps:
1. Pi bond attacks electrophile.
2. Carbocation forms.
3. Nucleophile captures it. | organic_mechanisms |
Ethene plus H2 over nickel gives ethane, saturating the double bond. | null | che_0107 | worked_example | en | What is the product of ethene hydrogenation (H2, Ni)? | [
"H2 adds across C=C.",
"Each carbon gains one H.",
"Product: ethane."
] | chemistry | Answer: Ethane.
Steps:
1. H2 adds across C=C.
2. Each carbon gains one H.
3. Product: ethane. | organic_mechanisms |
Radical chains run through initiation forming radicals, propagation cycling them, and termination combining them. | null | che_0108 | concept | en | What are the steps of a radical chain reaction? | [
"Initiation creates radicals.",
"Propagation sustains the cycle.",
"Termination ends it by pairing."
] | chemistry | Answer: Initiation, propagation, termination.
Steps:
1. Initiation creates radicals.
2. Propagation sustains the cycle.
3. Termination ends it by pairing. | organic_mechanisms |
Initiation splits a bond homolytically to create the first radicals. | {
"A": "Joins two radicals",
"B": "Transfers a radical",
"C": "Quenches all radicals",
"D": "Creates the first radicals"
} | che_0109 | multiple_choice | en | What happens in chain initiation? | [
"Recall initiation starts the chain.",
"Homolysis creates radicals.",
"Select option D."
] | chemistry | Answer: Creates the first radicals.
Steps:
1. Recall initiation starts the chain.
2. Homolysis creates radicals.
3. Select option D.
Answer: (D) | organic_mechanisms |
Adicí bromu na ethen vzniká 1,2-dibromethan. | null | che_0110 | worked_example | cs | Jaký produkt vznikne adicí Br2 na ethen? | [
"Dvojná vazba se rozštěpí.",
"Každý uhlík přijme jeden brom.",
"Produkt: 1,2-dibromethan."
] | chemistry | Answer: 1,2-dibromethan.
Steps:
1. Dvojná vazba se rozštěpí.
2. Každý uhlík přijme jeden brom.
3. Produkt: 1,2-dibromethan. | organic_mechanisms |
Electrophilic aromatic substitution replaces a ring H with an electrophile while restoring aromaticity. | null | che_0111 | concept | en | What is electrophilic aromatic substitution? | [
"Electrophile attacks the ring.",
"Sigma complex forms.",
"Base removes H, aromaticity returns."
] | chemistry | Answer: Ring H replaced by electrophile.
Steps:
1. Electrophile attacks the ring.
2. Sigma complex forms.
3. Base removes H, aromaticity returns. | organic_mechanisms |
Nitration of benzene with HNO3/H2SO4 gives nitrobenzene plus water. | null | che_0112 | worked_example | en | What is the product of benzene nitration (HNO3, H2SO4)? | [
"NO2+ electrophile forms.",
"It substitutes one ring H.",
"Product: nitrobenzene."
] | chemistry | Answer: Nitrobenzene.
Steps:
1. NO2+ electrophile forms.
2. It substitutes one ring H.
3. Product: nitrobenzene. | organic_mechanisms |
A catalyst speeds reaction by lowering activation energy without being consumed. | {
"A": "Lowers activation energy",
"B": "Raises activation energy",
"C": "Changes equilibrium",
"D": "Is always consumed"
} | che_0113 | multiple_choice | en | How does a catalyst work? | [
"Recall alternative lower path.",
"Activation energy drops.",
"Select option A."
] | chemistry | Answer: Lowers activation energy.
Steps:
1. Recall alternative lower path.
2. Activation energy drops.
3. Select option A.
Answer: (A) | organic_mechanisms |
Aktivační energie je minimální energie nutná k dosažení přechodového stavu. | null | che_0114 | concept | cs | Co je aktivační energie? | [
"Energie přechodového stavu.",
"Vyšší bariéra = pomalejší reakce.",
"Katalyzátor ji snižuje."
] | chemistry | Answer: Energie nutná k reakci.
Steps:
1. Energie přechodového stavu.
2. Vyšší bariéra = pomalejší reakce.
3. Katalyzátor ji snižuje. | organic_mechanisms |
Dynamic equilibrium means forward and reverse rates are equal with constant concentrations. | null | che_0115 | concept | en | What is dynamic equilibrium? | [
"Both directions still react.",
"Rates become equal.",
"Concentrations stay constant."
] | chemistry | Answer: Equal rates, constant concentrations.
Steps:
1. Both directions still react.
2. Rates become equal.
3. Concentrations stay constant. | equilibrium |
Kc equals [NH3]^2 divided by [N2][H2]^3, with coefficients as exponents. | {
"A": "[N2][H2]/[NH3]",
"B": "[NH3]/([N2][H2])",
"C": "[NH3]^2/([N2][H2]^3)",
"D": "[N2][H2]^3/[NH3]^2"
} | che_0116 | multiple_choice | en | What is Kc for N2 + 3H2 <-> 2NH3? | [
"Products over reactants.",
"Apply exponents 2, 1, 3.",
"Select option C."
] | chemistry | Answer: [NH3]^2/([N2][H2]^3).
Steps:
1. Products over reactants.
2. Apply exponents 2, 1, 3.
3. Select option C.
Answer: (C) | equilibrium |
Kc is 8.0, from 2.0 divided by (0.50 x 0.50). | null | che_0117 | worked_example | en | For A + B <-> C with [A] = 0.50 M, [B] = 0.50 M, [C] = 2.0 M, compute Kc. | [
"Write: Kc = [C]/([A][B]).",
"Substitute: 2.0 / (0.50 x 0.50).",
"Kc = 8.0."
] | chemistry | Answer: 8.0.
Steps:
1. Write: Kc = [C]/([A][B]).
2. Substitute: 2.0 / (0.50 x 0.50).
3. Kc = 8.0. | equilibrium |
Rovnovážná konstanta je poměr koncentrací produktů a reaktantů v rovnováze. | null | che_0118 | concept | cs | Co je rovnovážná konstanta? | [
"Poměr produkty/reaktanty.",
"Koeficienty jako exponenty.",
"Velká K = převaha produktů."
] | chemistry | Answer: Poměr koncentrací v rovnováze.
Steps:
1. Poměr produkty/reaktanty.
2. Koeficienty jako exponenty.
3. Velká K = převaha produktů. | equilibrium |
A very large K means products dominate at equilibrium. | {
"A": "Reactants dominate",
"B": "Products dominate",
"C": "Nothing reacts",
"D": "K is meaningless"
} | che_0119 | multiple_choice | en | What does a very large equilibrium constant mean? | [
"K = products/reactants.",
"Large K favors numerator.",
"Select option B."
] | chemistry | Answer: Products dominate.
Steps:
1. K = products/reactants.
2. Large K favors numerator.
3. Select option B.
Answer: (B) | equilibrium |
Le Chatelier's principle states a system shifts to counteract an imposed change. | null | che_0120 | concept | en | What is Le Chatelier's principle? | [
"Apply stress: concentration, pressure, heat.",
"System shifts to relieve it.",
"New equilibrium results."
] | chemistry | Answer: Systems counteract imposed changes.
Steps:
1. Apply stress: concentration, pressure, heat.
2. System shifts to relieve it.
3. New equilibrium results. | equilibrium |
The system shifts right toward products because Q = 2 is less than K = 10. | null | che_0121 | worked_example | en | For a reaction with Q = 2 and K = 10, which way does it shift? | [
"Compare Q to K: 2 < 10.",
"Too many reactants.",
"Shifts right to products."
] | chemistry | Answer: Shifts right.
Steps:
1. Compare Q to K: 2 < 10.
2. Too many reactants.
3. Shifts right to products. | equilibrium |
Higher pressure favors the side with fewer moles of gas. | {
"A": "More moles of gas",
"B": "No shift ever",
"C": "Only liquids",
"D": "Fewer moles of gas"
} | che_0122 | multiple_choice | en | How does increased pressure shift gas equilibrium? | [
"System relieves pressure.",
"Fewer gas moles reduce pressure.",
"Select option D."
] | chemistry | Answer: Fewer moles of gas.
Steps:
1. System relieves pressure.
2. Fewer gas moles reduce pressure.
3. Select option D.
Answer: (D) | equilibrium |
Catalysts speed both directions equally, so equilibrium position and K stay unchanged. | null | che_0123 | concept | en | Do catalysts shift equilibrium? | [
"Both rates rise equally.",
"K stays the same.",
"Equilibrium reached faster only."
] | chemistry | Answer: No, only faster to reach it.
Steps:
1. Both rates rise equally.
2. K stays the same.
3. Equilibrium reached faster only. | equilibrium |
Kc je 2,0, protože 3,0 děleno 1,5 se rovná 2,0. | null | che_0124 | worked_example | cs | Pro A <-> B je [A] = 1,5 M a [B] = 3,0 M. Vypočítej Kc. | [
"Napiš vzorec: Kc = [B]/[A].",
"Dosaď: 3,0 / 1,5.",
"Kc = 2,0."
] | chemistry | Answer: 2,0.
Steps:
1. Napiš vzorec: Kc = [B]/[A].
2. Dosaď: 3,0 / 1,5.
3. Kc = 2,0. | equilibrium |
Adding heat to an endothermic reaction shifts it right, toward products. | {
"A": "Shifts right",
"B": "Shifts left",
"C": "No shift",
"D": "K becomes zero"
} | che_0125 | multiple_choice | en | Heating an endothermic reaction shifts it which way? | [
"Heat acts as reactant.",
"Adding reactant shifts right.",
"Select option A."
] | chemistry | Answer: Shifts right.
Steps:
1. Heat acts as reactant.
2. Adding reactant shifts right.
3. Select option A.
Answer: (A) | equilibrium |
ICE tables track initial, change, and equilibrium concentrations to solve for unknowns. | null | che_0126 | concept | en | What is an ICE table? | [
"List initial concentrations.",
"Define change as x.",
"Solve equilibrium row with K."
] | chemistry | Answer: Initial-change-equilibrium tracker.
Steps:
1. List initial concentrations.
2. Define change as x.
3. Solve equilibrium row with K. | equilibrium |
Equilibrium [B] is 0.80 M, solving x/(1.0 - x) = 4. | null | che_0127 | worked_example | en | For A <-> B with K = 4 and [A]0 = 1.0 M, find [B] at equilibrium. | [
"Write: x / (1.0 - x) = 4.",
"Solve: x = 4 - 4x, 5x = 4.",
"[B] = x = 0.80 M."
] | chemistry | Answer: 0.80 M.
Steps:
1. Write: x / (1.0 - x) = 4.
2. Solve: x = 4 - 4x, 5x = 4.
3. [B] = x = 0.80 M. | equilibrium |
Kp is written with partial pressures instead of molar concentrations. | {
"A": "Molar masses",
"B": "Mole counts",
"C": "Partial pressures",
"D": "Volumes"
} | che_0128 | multiple_choice | en | What does Kp use in its expression? | [
"Recall p means pressure.",
"Use partial pressures.",
"Select option C."
] | chemistry | Answer: Partial pressures.
Steps:
1. Recall p means pressure.
2. Use partial pressures.
3. Select option C.
Answer: (C) | equilibrium |
Homogeneous equilibria have one phase while heterogeneous equilibria mix phases, omitting pure solids and liquids from K. | null | che_0129 | concept | en | How do homogeneous and heterogeneous equilibria differ? | [
"Homogeneous: single phase.",
"Heterogeneous: multiple phases.",
"Omit solids and liquids from K."
] | chemistry | Answer: One phase versus mixed phases.
Steps:
1. Homogeneous: single phase.
2. Heterogeneous: multiple phases.
3. Omit solids and liquids from K. | equilibrium |
Le Chatelierův princip říká, že soustava se posune proti vnější změně. | null | che_0130 | concept | cs | Co říká Le Chatelierův princip? | [
"Změna koncentrace, tlaku či teploty.",
"Soustava působí proti ní.",
"Ustálí se nová rovnováha."
] | chemistry | Answer: Soustava působí proti změně.
Steps:
1. Změna koncentrace, tlaku či teploty.
2. Soustava působí proti ní.
3. Ustálí se nová rovnováha. | equilibrium |
Adding reactant drives the system toward products to consume the excess. | {
"A": "Shifts left",
"B": "Shifts right",
"C": "No shift",
"D": "K triples"
} | che_0131 | multiple_choice | en | What happens when a reactant is added at equilibrium? | [
"Excess reactant present.",
"System consumes it.",
"Select option B."
] | chemistry | Answer: Shifts right.
Steps:
1. Excess reactant present.
2. System consumes it.
3. Select option B.
Answer: (B) | equilibrium |
Kp is 12.2, from 0.50 x 0.08206 x 298. | null | che_0132 | worked_example | en | For N2O4 <-> 2NO2 with Kc = 0.50 at 298 K, compute Kp (R = 0.08206 L atm/mol K). | [
"Δn = 2 - 1 = 1.",
"Kp = 0.50 x 0.08206 x 298.",
"Kp = 12.2."
] | chemistry | Answer: 12.2.
Steps:
1. Δn = 2 - 1 = 1.
2. Kp = 0.50 x 0.08206 x 298.
3. Kp = 12.2. | equilibrium |
Galvanic cells release energy spontaneously while electrolytic cells consume energy to force reaction. | null | che_0133 | concept | en | How do galvanic and electrolytic cells differ? | [
"Galvanic: spontaneous, E positive.",
"Electrolytic: forced, E negative.",
"Example: battery versus plating."
] | chemistry | Answer: Spontaneous versus driven cells.
Steps:
1. Galvanic: spontaneous, E positive.
2. Electrolytic: forced, E negative.
3. Example: battery versus plating. | electrochemistry |
Oxidation occurs at the anode, remembered as AN OX. | {
"A": "Anode is oxidation",
"B": "Anode is reduction",
"C": "Anode is neutral",
"D": "Anode stores ions"
} | che_0134 | multiple_choice | en | What happens at the anode? | [
"Recall AN OX mnemonic.",
"Anode loses electrons.",
"Select option A."
] | chemistry | Answer: Anode is oxidation.
Steps:
1. Recall AN OX mnemonic.
2. Anode loses electrons.
3. Select option A.
Answer: (A) | electrochemistry |
The standard cell potential is 1.10 V, from 0.34 - (-0.76). | null | che_0135 | worked_example | en | Compute E cell for Zn/Cu with E(Cu) = 0.34 V and E(Zn) = -0.76 V. | [
"Cathode is Cu at 0.34 V.",
"Anode is Zn at -0.76 V.",
"E = 0.34 - (-0.76) = 1.10 V."
] | chemistry | Answer: 1.10 V.
Steps:
1. Cathode is Cu at 0.34 V.
2. Anode is Zn at -0.76 V.
3. E = 0.34 - (-0.76) = 1.10 V. | electrochemistry |
Galvanický článek mění chemickou energii samovolné reakce na elektrickou. | null | che_0136 | concept | cs | Co je galvanický článek? | [
"Samovolná redoxní reakce.",
"Elektrony tečou vnějším obvodem.",
"Příklad: Daniellův článek."
] | chemistry | Answer: Zdroj proudu z chemie.
Steps:
1. Samovolná redoxní reakce.
2. Elektrony tečou vnějším obvodem.
3. Příklad: Daniellův článek. | electrochemistry |
Electrons flow from anode to cathode through the external wire. | {
"A": "Cathode to anode",
"B": "Anode to cathode",
"C": "Both directions",
"D": "Through the salt bridge"
} | che_0137 | multiple_choice | en | Which way do electrons flow in a galvanic cell? | [
"Anode releases electrons.",
"Wire carries them to cathode.",
"Select option B."
] | chemistry | Answer: Anode to cathode.
Steps:
1. Anode releases electrons.
2. Wire carries them to cathode.
3. Select option B.
Answer: (B) | electrochemistry |
The standard hydrogen electrode is assigned exactly 0.00 V as the reference. | null | che_0138 | concept | en | What is the standard hydrogen electrode? | [
"H2 gas over platinum.",
"1 M H+ at 1 atm.",
"Defined as 0.00 V."
] | chemistry | Answer: Reference electrode at 0.00 V.
Steps:
1. H2 gas over platinum.
2. 1 M H+ at 1 atm.
3. Defined as 0.00 V. | electrochemistry |
ΔG is -212 kJ, from -2 x 96485 x 1.10 J. | null | che_0139 | worked_example | en | Compute ΔG for a cell with n = 2 and E = 1.10 V (F = 96485 C/mol). | [
"Write: ΔG = -nFE.",
"Substitute: -2 x 96485 x 1.10.",
"ΔG = -212,267 J = -212 kJ."
] | chemistry | Answer: -212 kJ.
Steps:
1. Write: ΔG = -nFE.
2. Substitute: -2 x 96485 x 1.10.
3. ΔG = -212,267 J = -212 kJ. | electrochemistry |
The Nernst equation adjusts cell potential for nonstandard concentrations via Q. | null | che_0140 | concept | en | What is the Nernst equation? | [
"E = E° - (RT/nF) ln Q.",
"At 25 C: E° - (0.0592/n) log Q.",
"Q reflects concentrations."
] | chemistry | Answer: E from E° and reaction quotient.
Steps:
1. E = E° - (RT/nF) ln Q.
2. At 25 C: E° - (0.0592/n) log Q.
3. Q reflects concentrations. | electrochemistry |
In electrolysis the cathode is negative, attracting cations. | {
"A": "Positive",
"B": "Neutral",
"C": "Negative",
"D": "Absent"
} | che_0141 | multiple_choice | en | What is the cathode's sign in an electrolytic cell? | [
"External supply drives electrons in.",
"Cathode receives electrons.",
"Select option C."
] | chemistry | Answer: Negative.
Steps:
1. External supply drives electrons in.
2. Cathode receives electrons.
3. Select option C.
Answer: (C) | electrochemistry |
The cell potential is 1.07 V, from 1.10 - (0.0592/2) x log(10). | null | che_0142 | worked_example | en | E° = 1.10 V, n = 2, Q = 10 at 25 C. Compute E. | [
"Write: E = 1.10 - (0.0592/2) log Q.",
"Substitute: 1.10 - 0.0296 x 1.",
"E = 1.07 V."
] | chemistry | Answer: 1.07 V.
Steps:
1. Write: E = 1.10 - (0.0592/2) log Q.
2. Substitute: 1.10 - 0.0296 x 1.
3. E = 1.07 V. | electrochemistry |
Napětí Daniellova článku je 1,10 V, protože 0,34 - (-0,76) = 1,10. | null | che_0143 | worked_example | cs | Vypočítej napětí článku Zn/Cu: E(Cu) = 0,34 V, E(Zn) = -0,76 V. | [
"Katoda Cu: 0,34 V.",
"Anoda Zn: -0,76 V.",
"E = 0,34 - (-0,76) = 1,10 V."
] | chemistry | Answer: 1,10 V.
Steps:
1. Katoda Cu: 0,34 V.
2. Anoda Zn: -0,76 V.
3. E = 0,34 - (-0,76) = 1,10 V. | electrochemistry |
Faraday's laws state deposited mass is proportional to charge passed and equivalent weight. | null | che_0144 | concept | en | What are Faraday's laws of electrolysis? | [
"Charge: Q = I x t.",
"Moles electrons = Q / F.",
"Mass from moles and molar mass."
] | chemistry | Answer: Mass proportional to charge.
Steps:
1. Charge: Q = I x t.
2. Moles electrons = Q / F.
3. Mass from moles and molar mass. | electrochemistry |
One mole of electrons carries 96,485 coulombs, the Faraday constant. | {
"A": "6,022 C/mol",
"B": "8,314 C/mol",
"C": "1,602 C/mol",
"D": "96,485 C/mol"
} | che_0145 | multiple_choice | en | What is the Faraday constant? | [
"Recall charge per mole electrons.",
"Value 96,485 C/mol.",
"Select option D."
] | chemistry | Answer: 96,485 C/mol.
Steps:
1. Recall charge per mole electrons.
2. Value 96,485 C/mol.
3. Select option D.
Answer: (D) | electrochemistry |
The plated copper mass is 2.37 g, from 7200 C giving 0.0373 mol Cu. | null | che_0146 | worked_example | en | What mass of Cu plates at 2.00 A for 1.00 hour (M = 63.55 g/mol)? | [
"Q = 2.00 x 3600 = 7200 C.",
"Mol e- = 7200/96485 = 0.0746; Cu = 0.0373 mol.",
"Mass = 0.0373 x 63.55 = 2.37 g."
] | chemistry | Answer: 2.37 g.
Steps:
1. Q = 2.00 x 3600 = 7200 C.
2. Mol e- = 7200/96485 = 0.0746; Cu = 0.0373 mol.
3. Mass = 0.0373 x 63.55 = 2.37 g. | electrochemistry |
Corrosion is spontaneous oxidation of metals like iron forming rust in air and water. | null | che_0147 | concept | en | What is corrosion? | [
"Metal oxidizes at anodic sites.",
"Oxygen reduces at cathodic sites.",
"Rust is hydrated iron oxide."
] | chemistry | Answer: Spontaneous metal oxidation.
Steps:
1. Metal oxidizes at anodic sites.
2. Oxygen reduces at cathodic sites.
3. Rust is hydrated iron oxide. | electrochemistry |
Elektrolýza je rozklad látky stejnosměrným proudem na elektrodách. | null | che_0148 | concept | cs | Co je elektrolýza? | [
"Proud protéká elektrolytem.",
"Na katodě redukce, na anodě oxidace.",
"Příklad: rozklad vody."
] | chemistry | Answer: Rozklad proudem.
Steps:
1. Proud protéká elektrolytem.
2. Na katodě redukce, na anodě oxidace.
3. Příklad: rozklad vody. | electrochemistry |
The salt bridge carries ions to keep each half-cell electrically neutral. | {
"A": "Keeps charge neutrality",
"B": "Carries electrons",
"C": "Adds reactants",
"D": "Cools the cell"
} | che_0149 | multiple_choice | en | What is the purpose of a salt bridge? | [
"Cells would charge up otherwise.",
"Ions migrate to balance.",
"Select option A."
] | chemistry | Answer: Keeps charge neutrality.
Steps:
1. Cells would charge up otherwise.
2. Ions migrate to balance.
3. Select option A.
Answer: (A) | electrochemistry |
Plating 6.35 g of Cu at 2.00 A takes 2.68 hours. | null | che_0150 | worked_example | en | How long at 2.00 A to plate 6.35 g of Cu (M = 63.55 g/mol)? | [
"Mol Cu = 6.35/63.55 = 0.100; e- = 0.200 mol.",
"Q = 0.200 x 96485 = 19,297 C.",
"t = 19,297/2.00 = 9,649 s = 2.68 h."
] | chemistry | Answer: 2.68 h.
Steps:
1. Mol Cu = 6.35/63.55 = 0.100; e- = 0.200 mol.
2. Q = 0.200 x 96485 = 19,297 C.
3. t = 19,297/2.00 = 9,649 s = 2.68 h. | electrochemistry |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.