answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
Polar aprotic solvents leave nucleophiles free to attack. | {
"A": "Polar protic",
"B": "Nonpolar only",
"C": "Polar aprotic",
"D": "Water only"
} | che_0551 | multiple_choice | en | Which solvent class speeds SN2 most? | [
"Free nucleophile needed.",
"Aprotic avoids caging.",
"Select option C."
] | chemistry | Answer: Polar aprotic.
Steps:
1. Free nucleophile needed.
2. Aprotic avoids caging.
3. Select option C.
Answer: (C) | organic_mechanisms |
HBr adds across ethene to give bromoethane. | null | che_0552 | worked_example | en | What is the product of HBr addition to ethene? | [
"H adds to one carbon.",
"Br adds to the other.",
"Product: bromoethane."
] | chemistry | Answer: Bromoethane.
Steps:
1. H adds to one carbon.
2. Br adds to the other.
3. Product: bromoethane. | organic_mechanisms |
Carbocations rearrange by hydride or alkyl shifts toward greater stability. | null | che_0553 | concept | en | How do carbocations rearrange? | [
"Neighbor H or alkyl shifts.",
"Charge moves to stabler carbon.",
"Tertiary preferred."
] | chemistry | Answer: Shifts toward stability.
Steps:
1. Neighbor H or alkyl shifts.
2. Charge moves to stabler carbon.
3. Tertiary preferred. | organic_mechanisms |
Hydridový přesmyk přesune H- na sousední karbokation. | null | che_0554 | concept | cs | Co je hydridový přesmyk? | [
"H- migruje ke kationu.",
"Náboj se přesune.",
"Stabilnější kation vznikne."
] | chemistry | Answer: Migrace H- ke kationu.
Steps:
1. H- migruje ke kationu.
2. Náboj se přesune.
3. Stabilnější kation vznikne. | organic_mechanisms |
A methyl shift relocates a CH3 group to the cationic center. | {
"A": "CH3 group",
"B": "H-",
"C": "Leaving group",
"D": "Proton"
} | che_0555 | multiple_choice | en | What moves in a methyl shift? | [
"Methyl migrates.",
"Carries bonding pair.",
"Select option A."
] | chemistry | Answer: CH3 group.
Steps:
1. Methyl migrates.
2. Carries bonding pair.
3. Select option A.
Answer: (A) | organic_mechanisms |
The product is 2-chloropropane, with Cl on the middle carbon. | null | che_0556 | worked_example | en | What is the HCl addition product of propene? | [
"H adds to C1.",
"Secondary cation forms.",
"Cl captures it: 2-chloropropane."
] | chemistry | Answer: 2-chloropropane.
Steps:
1. H adds to C1.
2. Secondary cation forms.
3. Cl captures it: 2-chloropropane. | organic_mechanisms |
Backside attack in SN2 flips the stereocenter like an umbrella. | null | che_0557 | concept | en | Why does SN2 invert stereochemistry? | [
"Attack comes from behind.",
"Bonds flip over.",
"Configuration inverts."
] | chemistry | Answer: Backside attack flips center.
Steps:
1. Attack comes from behind.
2. Bonds flip over.
3. Configuration inverts. | organic_mechanisms |
The product is 2-bromobutane, via the secondary carbocation. | null | che_0558 | worked_example | en | What is the HBr addition product of 1-butene? | [
"H adds to C1.",
"Cation forms on C2.",
"Br captures it: 2-bromobutane."
] | chemistry | Answer: 2-bromobutane.
Steps:
1. H adds to C1.
2. Cation forms on C2.
3. Br captures it: 2-bromobutane. | organic_mechanisms |
SN2 always inverts the configuration at the attacked carbon. | {
"A": "Retention",
"B": "Racemization",
"C": "No change",
"D": "Inversion"
} | che_0559 | multiple_choice | en | What stereochemistry results from SN2? | [
"Backside attack occurs.",
"Center flips.",
"Select option D."
] | chemistry | Answer: Inversion.
Steps:
1. Backside attack occurs.
2. Center flips.
3. Select option D.
Answer: (D) | organic_mechanisms |
Eliminací 2-brompropanu vzniká propen. | null | che_0560 | worked_example | cs | Jaký produkt vznikne eliminací 2-brompropanu se silnou bází? | [
"Odstoupí Br a sousední H.",
"Vznikne dvojná vazba.",
"Produkt: propen."
] | chemistry | Answer: Propen.
Steps:
1. Odstoupí Br a sousední H.
2. Vznikne dvojná vazba.
3. Produkt: propen. | organic_mechanisms |
Halogenation prefers tertiary H over secondary over primary. | null | che_0561 | concept | en | What is the selectivity of radical halogenation? | [
"Weaker C-H breaks first.",
"Order: 3° > 2° > 1°.",
"Bromine most selective."
] | chemistry | Answer: Tertiary > secondary > primary.
Steps:
1. Weaker C-H breaks first.
2. Order: 3° > 2° > 1°.
3. Bromine most selective. | organic_mechanisms |
Propane gives 1-chloropropane and 2-chloropropane as a mixture. | null | che_0562 | worked_example | en | What products come from propane monochlorination? | [
"Primary H gives 1-chloro.",
"Secondary H gives 2-chloro.",
"Both form as a mixture."
] | chemistry | Answer: 1-chloro and 2-chloropropane.
Steps:
1. Primary H gives 1-chloro.
2. Secondary H gives 2-chloro.
3. Both form as a mixture. | organic_mechanisms |
Tertiary hydrogens abstract fastest in radical halogenation. | {
"A": "Primary",
"B": "Secondary",
"C": "Tertiary",
"D": "Aromatic"
} | che_0563 | multiple_choice | en | Which hydrogen reacts fastest in halogenation? | [
"Weakest C-H wins.",
"Tertiary is weakest.",
"Select option C."
] | chemistry | Answer: Tertiary.
Steps:
1. Weakest C-H wins.
2. Tertiary is weakest.
3. Select option C.
Answer: (C) | organic_mechanisms |
Elektrofil je částice chudá na elektrony, která je přijímá. | null | che_0564 | concept | cs | Co je elektrofil? | [
"Chudý na elektrony.",
"Napadá nukleofil.",
"Příklad: H+, NO2+."
] | chemistry | Answer: Akceptor elektronového páru.
Steps:
1. Chudý na elektrony.
2. Napadá nukleofil.
3. Příklad: H+, NO2+. | organic_mechanisms |
At equilibrium ΔG is zero and forward and reverse rates match. | null | che_0565 | concept | en | How does Gibbs energy relate to equilibrium? | [
"ΔG drives change.",
"Zero means no drive.",
"Rates are equal there."
] | chemistry | Answer: ΔG = 0 at equilibrium.
Steps:
1. ΔG drives change.
2. Zero means no drive.
3. Rates are equal there. | equilibrium |
Dynamic equilibrium has equal forward and reverse rates. | {
"A": "Equal",
"B": "Zero both",
"C": "Forward faster",
"D": "Reverse faster"
} | che_0566 | multiple_choice | en | How do forward and reverse rates compare at equilibrium? | [
"Both still run.",
"Rates match.",
"Select option A."
] | chemistry | Answer: Equal.
Steps:
1. Both still run.
2. Rates match.
3. Select option A.
Answer: (A) | equilibrium |
The overall K is 5, from 10 times 0.5. | null | che_0567 | worked_example | en | Step 1 has K1 = 10 and step 2 has K2 = 0.5. Find overall K. | [
"Multiply step constants.",
"K = 10 x 0.5.",
"K = 5."
] | chemistry | Answer: 5.
Steps:
1. Multiply step constants.
2. K = 10 x 0.5.
3. K = 5. | equilibrium |
Rovnovážná směs obsahuje reaktanty i produkty ve stálých koncentracích. | null | che_0568 | concept | cs | Co je rovnovážná směs? | [
"Oba směry běží.",
"Koncentrace stálé.",
"Poměr určuje K."
] | chemistry | Answer: Směs reaktantů a produktů.
Steps:
1. Oba směry běží.
2. Koncentrace stálé.
3. Poměr určuje K. | equilibrium |
Halving all coefficients takes the square root of K. | {
"A": "K squared",
"B": "2K",
"C": "Square root of K",
"D": "K halved"
} | che_0569 | multiple_choice | en | What happens to K when the equation is halved? | [
"Exponents halve.",
"That roots the value.",
"Select option C."
] | chemistry | Answer: Square root of K.
Steps:
1. Exponents halve.
2. That roots the value.
3. Select option C.
Answer: (C) | equilibrium |
The system shifts left because Q = 20 exceeds K = 5. | null | che_0570 | worked_example | en | Q = 20 and K = 5. Which way does the system shift? | [
"Compare: 20 > 5.",
"Too many products.",
"Shifts left."
] | chemistry | Answer: Shifts left.
Steps:
1. Compare: 20 > 5.
2. Too many products.
3. Shifts left. | equilibrium |
A catalyst at equilibrium only helps reach it faster, changing nothing else. | null | che_0571 | concept | en | What does adding a catalyst at equilibrium do? | [
"Both rates rise.",
"K unchanged.",
"No shift occurs."
] | chemistry | Answer: Nothing but faster balance.
Steps:
1. Both rates rise.
2. K unchanged.
3. No shift occurs. | equilibrium |
The catalyst leaves the equilibrium position untouched. | {
"A": "Shifts right",
"B": "No shift",
"C": "Shifts left",
"D": "Destroys K"
} | che_0572 | multiple_choice | en | A catalyst is added at equilibrium. What shift follows? | [
"Rates rise equally.",
"Position unchanged.",
"Select option B."
] | chemistry | Answer: No shift.
Steps:
1. Rates rise equally.
2. Position unchanged.
3. Select option B.
Answer: (B) | equilibrium |
Doubling volume lowers pressure, shifting toward more gas moles. | null | che_0573 | worked_example | en | Volume doubles for N2O4 <-> 2NO2. Which way is the shift? | [
"Pressure drops.",
"System favors more moles.",
"Shifts right to NO2."
] | chemistry | Answer: Shifts right.
Steps:
1. Pressure drops.
2. System favors more moles.
3. Shifts right to NO2. | equilibrium |
Při stejném počtu molů plynu tlak rovnováhu neposune. | null | che_0574 | worked_example | cs | Jak ovlivní tlak rovnováhu se stejnými moly plynu na obou stranách? | [
"Moly se rovnají.",
"Žádná strana nezvýhodněna.",
"Žádný posun."
] | chemistry | Answer: Žádný posun.
Steps:
1. Moly se rovnají.
2. Žádná strana nezvýhodněna.
3. Žádný posun. | equilibrium |
Pressure cannot favor either side when gas moles are equal. | {
"A": "Shifts right",
"B": "Shifts left",
"C": "K changes",
"D": "No shift"
} | che_0575 | multiple_choice | en | Equal gas moles on both sides: what does pressure do? | [
"No mole advantage.",
"Q unchanged.",
"Select option D."
] | chemistry | Answer: No shift.
Steps:
1. No mole advantage.
2. Q unchanged.
3. Select option D.
Answer: (D) | equilibrium |
Heating an endothermic reaction raises K as products gain favor. | null | che_0576 | concept | en | How does temperature change K for endothermic reactions? | [
"Heat acts as reactant.",
"Warming shifts right.",
"K grows."
] | chemistry | Answer: K rises with temperature.
Steps:
1. Heat acts as reactant.
2. Warming shifts right.
3. K grows. | equilibrium |
Rising K with temperature marks the reaction as endothermic. | null | che_0577 | worked_example | en | K doubles when heated. Is the reaction endo- or exothermic? | [
"K rises with T.",
"Heat favors products.",
"Endothermic."
] | chemistry | Answer: Endothermic.
Steps:
1. K rises with T.
2. Heat favors products.
3. Endothermic. | equilibrium |
A growing K on heating means heat is consumed: endothermic. | {
"A": "Endothermic",
"B": "Exothermic",
"C": "Athermal",
"D": "Impossible"
} | che_0578 | multiple_choice | en | Heating increases K. What type of reaction is it? | [
"K up means products favored.",
"Heat drives it.",
"Select option A."
] | chemistry | Answer: Endothermic.
Steps:
1. K up means products favored.
2. Heat drives it.
3. Select option A.
Answer: (A) | equilibrium |
Compare the ion product Q to Ksp: above it, solid precipitates. | null | che_0579 | concept | en | How do Q and Ksp predict precipitation? | [
"Compute ion product Q.",
"Compare to Ksp.",
"Q > Ksp means precipitate."
] | chemistry | Answer: Q above Ksp precipitates.
Steps:
1. Compute ion product Q.
2. Compare to Ksp.
3. Q > Ksp means precipitate. | equilibrium |
Srážení je vznik pevné látky z roztoku při překročení rozpustnosti. | null | che_0580 | concept | cs | Co je srážení? | [
"Ionty překročí Ksp.",
"Vznikne sraženina.",
"Příklad: AgCl z Ag+ a Cl-."
] | chemistry | Answer: Vznik pevné látky z roztoku.
Steps:
1. Ionty překročí Ksp.
2. Vznikne sraženina.
3. Příklad: AgCl z Ag+ a Cl-. | equilibrium |
Excess ions beyond saturation fall out as solid. | {
"A": "More dissolves",
"B": "Precipitate forms",
"C": "Nothing happens",
"D": "Gas evolves"
} | che_0581 | multiple_choice | en | What happens when Q exceeds Ksp? | [
"Solution is supersaturated.",
"Excess must leave.",
"Select option B."
] | chemistry | Answer: Precipitate forms.
Steps:
1. Solution is supersaturated.
2. Excess must leave.
3. Select option B.
Answer: (B) | equilibrium |
No precipitate forms because Q = 1.0 x 10^-10 is below Ksp = 1.8 x 10^-10. | null | che_0582 | worked_example | en | [Ag+] = 1.0e-5 M and [Cl-] = 1.0e-5 M. Does AgCl precipitate (Ksp = 1.8e-10)? | [
"Q = 1.0e-5 x 1.0e-5 = 1.0e-10.",
"Compare: 1.0e-10 < 1.8e-10.",
"No precipitate."
] | chemistry | Answer: No precipitate.
Steps:
1. Q = 1.0e-5 x 1.0e-5 = 1.0e-10.
2. Compare: 1.0e-10 < 1.8e-10.
3. No precipitate. | equilibrium |
The Daniell cell pairs zinc and copper half-cells for a steady 1.10 V. | null | che_0583 | concept | en | What is a Daniell cell? | [
"Zn anode dissolves.",
"Cu cathode plates.",
"Salt bridge joins halves."
] | chemistry | Answer: Zn/Cu galvanic cell.
Steps:
1. Zn anode dissolves.
2. Cu cathode plates.
3. Salt bridge joins halves. | electrochemistry |
The Daniell cell uses zinc and copper electrodes. | {
"A": "Zn and Cu",
"B": "Pb and PbO2",
"C": "Ag and Au",
"D": "Fe and Ni"
} | che_0584 | multiple_choice | en | Which electrodes does a Daniell cell use? | [
"Classic Zn/Cu pair.",
"Eliminate others.",
"Select option A."
] | chemistry | Answer: Zn and Cu.
Steps:
1. Classic Zn/Cu pair.
2. Eliminate others.
3. Select option A.
Answer: (A) | electrochemistry |
The cell potential is 0.59 V, from 0.34 - (-0.25). | null | che_0585 | worked_example | en | Compute E cell for Ni/Cu with E(Ni) = -0.25 V and E(Cu) = 0.34 V. | [
"Cathode is Cu at 0.34 V.",
"Anode is Ni at -0.25 V.",
"E = 0.34 - (-0.25) = 0.59 V."
] | chemistry | Answer: 0.59 V.
Steps:
1. Cathode is Cu at 0.34 V.
2. Anode is Ni at -0.25 V.
3. E = 0.34 - (-0.25) = 0.59 V. | electrochemistry |
Daniellův článek spojuje zinkovou a měděnou polovinu o napětí 1,10 V. | null | che_0586 | concept | cs | Co je Daniellův článek? | [
"Zinková anoda.",
"Měděná katoda.",
"Solný můstek spojuje."
] | chemistry | Answer: Zn/Cu článek 1,10 V.
Steps:
1. Zinková anoda.
2. Měděná katoda.
3. Solný můstek spojuje. | electrochemistry |
Conventional current flows opposite to electrons: cathode to anode externally. | {
"A": "Anode to cathode",
"B": "Cathode to anode",
"C": "No flow",
"D": "Alternating"
} | che_0587 | multiple_choice | en | Which way does conventional current flow in the external wire? | [
"Electrons go anode to cathode.",
"Current is defined opposite.",
"Select option B."
] | chemistry | Answer: Cathode to anode.
Steps:
1. Electrons go anode to cathode.
2. Current is defined opposite.
3. Select option B.
Answer: (B) | electrochemistry |
Overpotential is the extra voltage beyond E needed to drive electrolysis at a rate. | null | che_0588 | concept | en | What is overpotential? | [
"Kinetics resist reaction.",
"Extra voltage overcomes it.",
"Varies by electrode."
] | chemistry | Answer: Extra voltage for kinetics.
Steps:
1. Kinetics resist reaction.
2. Extra voltage overcomes it.
3. Varies by electrode. | electrochemistry |
The potential is 1.16 V, from 1.10 - 0.0296 x log(0.01). | null | che_0589 | worked_example | en | E° = 1.10 V, n = 2, Q = 0.01 at 25 C. Compute E. | [
"Write: E = 1.10 - 0.0296 x log(0.01).",
"log(0.01) = -2.",
"E = 1.10 + 0.0592 = 1.16 V."
] | chemistry | Answer: 1.16 V.
Steps:
1. Write: E = 1.10 - 0.0296 x log(0.01).
2. log(0.01) = -2.
3. E = 1.10 + 0.0592 = 1.16 V. | electrochemistry |
K is about 3.5 x 10^15, from log K = 2 x 0.46 / 0.0592 = 15.54. | null | che_0590 | worked_example | en | A cell has n = 2 and E° = 0.46 V. Estimate K at 25 C. | [
"Write: log K = nE°/0.0592.",
"Substitute: 2 x 0.46/0.0592 = 15.54.",
"K = 10^15.54 = 3.5 x 10^15."
] | chemistry | Answer: 3.5 x 10^15.
Steps:
1. Write: log K = nE°/0.0592.
2. Substitute: 2 x 0.46/0.0592 = 15.54.
3. K = 10^15.54 = 3.5 x 10^15. | electrochemistry |
A large positive E° means a huge equilibrium constant favoring products. | {
"A": "Tiny K",
"B": "K is zero",
"C": "Very large K",
"D": "Negative K"
} | che_0591 | multiple_choice | en | What K matches a large positive E°? | [
"log K scales with E°.",
"Large E° gives huge K.",
"Select option C."
] | chemistry | Answer: Very large K.
Steps:
1. log K scales with E°.
2. Large E° gives huge K.
3. Select option C.
Answer: (C) | electrochemistry |
Fuel cells convert fuel and oxygen directly into electricity and water. | null | che_0592 | concept | en | How does a fuel cell work? | [
"Fuel oxidizes at anode.",
"Oxygen reduces at cathode.",
"Steady feed sustains power."
] | chemistry | Answer: Fuel plus O2 give power.
Steps:
1. Fuel oxidizes at anode.
2. Oxygen reduces at cathode.
3. Steady feed sustains power. | electrochemistry |
Náboj je 7200 C, protože 2 A krát 3600 s se rovná 7200 C. | null | che_0593 | worked_example | cs | Jaký náboj proteče při 2 A za 1 hodinu? | [
"Převeď čas: 3600 s.",
"Q = 2 x 3600.",
"Q = 7200 C."
] | chemistry | Answer: 7200 C.
Steps:
1. Převeď čas: 3600 s.
2. Q = 2 x 3600.
3. Q = 7200 C. | electrochemistry |
The plated zinc mass is 1.22 g, from 3600 C giving 0.01866 mol Zn. | null | che_0594 | worked_example | en | What mass of Zn plates at 2.00 A for 30.0 min (M = 65.38 g/mol)? | [
"Q = 2.00 x 1800 = 3600 C.",
"Mol e- = 0.03731; Zn = 0.01866 mol.",
"Mass = 0.01866 x 65.38 = 1.22 g."
] | chemistry | Answer: 1.22 g.
Steps:
1. Q = 2.00 x 1800 = 3600 C.
2. Mol e- = 0.03731; Zn = 0.01866 mol.
3. Mass = 0.01866 x 65.38 = 1.22 g. | electrochemistry |
The anode supplies the plating metal by dissolving into the bath. | {
"A": "Plating metal",
"B": "Object",
"C": "Inert rod",
"D": "Plastic"
} | che_0595 | multiple_choice | en | In electroplating, what is the anode made of? | [
"Anode dissolves to feed ions.",
"Made of plating metal.",
"Select option A."
] | chemistry | Answer: Plating metal.
Steps:
1. Anode dissolves to feed ions.
2. Made of plating metal.
3. Select option A.
Answer: (A) | electrochemistry |
Pipelines gain cathodic protection from buried sacrificial anodes. | null | che_0596 | concept | en | How are pipelines protected from corrosion? | [
"Bury Mg or Zn anodes.",
"They corrode instead.",
"Pipe stays cathodic."
] | chemistry | Answer: Sacrificial anodes.
Steps:
1. Bury Mg or Zn anodes.
2. They corrode instead.
3. Pipe stays cathodic. | electrochemistry |
The oxygen volume is 22.4 L, from 1.00 mol times 22.414 L/mol. | null | che_0597 | worked_example | en | What volume of O2 at STP comes from 4.00 mol of electrons? | [
"4 e- give 1 O2: 1.00 mol.",
"Multiply: 1.00 x 22.414.",
"V = 22.4 L."
] | chemistry | Answer: 22.4 L.
Steps:
1. 4 e- give 1 O2: 1.00 mol.
2. Multiply: 1.00 x 22.414.
3. V = 22.4 L. | electrochemistry |
Palivový článek mění palivo a kyslík přímo na elektřinu a vodu. | null | che_0598 | concept | cs | Co je palivový článek? | [
"Vodík na anodě.",
"Kyslík na katodě.",
"Trvalý přívod = výkon."
] | chemistry | Answer: Přímá přeměna paliva na proud.
Steps:
1. Vodík na anodě.
2. Kyslík na katodě.
3. Trvalý přívod = výkon. | electrochemistry |
Hydrogen fuel cells combine H2 and O2 into water and power. | {
"A": "Coal",
"B": "Hydrogen",
"C": "Uranium",
"D": "Gasoline"
} | che_0599 | multiple_choice | en | What fuel do common fuel cells use? | [
"H2 oxidizes cleanly.",
"Only water forms.",
"Select option B."
] | chemistry | Answer: Hydrogen.
Steps:
1. H2 oxidizes cleanly.
2. Only water forms.
3. Select option B.
Answer: (B) | electrochemistry |
The charge is 9.57 x 10^4 C, from 0.992 mol of electrons. | null | che_0600 | worked_example | en | What charge makes 1.00 g of H2 by electrolysis (M = 2.016 g/mol)? | [
"Mol H2 = 1.00/2.016 = 0.496.",
"Mol e- = 0.992.",
"Q = 0.992 x 96485 = 9.57 x 10^4 C."
] | chemistry | Answer: 9.57 x 10^4 C.
Steps:
1. Mol H2 = 1.00/2.016 = 0.496.
2. Mol e- = 0.992.
3. Q = 0.992 x 96485 = 9.57 x 10^4 C. | electrochemistry |
Chemists count particles by weighing because moles link atomic mass to grams. | null | che_0601 | concept | en | Why do chemists use moles? | [
"Atoms are too small to count.",
"Moles group 6e23 particles.",
"Mass in grams equals atomic mass."
] | chemistry | Answer: Counting particles by weighing.
Steps:
1. Atoms are too small to count.
2. Moles group 6e23 particles.
3. Mass in grams equals atomic mass. | stoichiometry |
The mass is 15.89 g, from 0.250 mol times 63.55 g/mol. | {
"A": "7.94 g",
"B": "31.78 g",
"C": "15.89 g",
"D": "63.55 g"
} | che_0602 | multiple_choice | en | What mass is 0.250 mol of copper (M = 63.55 g/mol)? | [
"Write: m = 0.250 x 63.55.",
"Multiply: 15.89.",
"Select option C."
] | chemistry | Answer: 15.89 g.
Steps:
1. Write: m = 0.250 x 63.55.
2. Multiply: 15.89.
3. Select option C.
Answer: (C) | stoichiometry |
There are 0.185 mol, from 5.00 g divided by 26.98 g/mol. | null | che_0603 | worked_example | en | How many moles are in 5.00 g of aluminum (M = 26.98 g/mol)? | [
"Write: n = 5.00/26.98.",
"Divide.",
"n = 0.185 mol."
] | chemistry | Answer: 0.185 mol.
Steps:
1. Write: n = 5.00/26.98.
2. Divide.
3. n = 0.185 mol. | stoichiometry |
Molární objem plynu je objem 1 molu plynu, za STP 22,414 l/mol. | null | che_0604 | concept | cs | Co je molární objem plynu? | [
"Objem 1 molu plynu.",
"Za STP 22,414 l/mol.",
"Stejný pro všechny ideální plyny."
] | chemistry | Answer: Objem 1 molu plynu.
Steps:
1. Objem 1 molu plynu.
2. Za STP 22,414 l/mol.
3. Stejný pro všechny ideální plyny. | stoichiometry |
There are 5.97 x 10^23 molecules, from 0.992 mol times Avogadro's number. | null | che_0605 | worked_example | en | How many molecules are in 2.00 g of H2 (M = 2.016 g/mol)? | [
"Moles = 2.00/2.016 = 0.992.",
"Multiply: 0.992 x 6.022 x 10^23.",
"N = 5.97 x 10^23."
] | chemistry | Answer: 5.97 x 10^23.
Steps:
1. Moles = 2.00/2.016 = 0.992.
2. Multiply: 0.992 x 6.022 x 10^23.
3. N = 5.97 x 10^23. | stoichiometry |
Calcium hydroxide totals 74.10 g/mol from Ca, 2 O, and 2 H. | {
"A": "74.10 g/mol",
"B": "57.09 g/mol",
"C": "100.09 g/mol",
"D": "40.08 g/mol"
} | che_0606 | multiple_choice | en | What is the molar mass of Ca(OH)2? | [
"40.08 + 2x16.00 + 2x1.008.",
"Total 74.10.",
"Select option A."
] | chemistry | Answer: 74.10 g/mol.
Steps:
1. 40.08 + 2x16.00 + 2x1.008.
2. Total 74.10.
3. Select option A.
Answer: (A) | stoichiometry |
The gravimetric factor converts precipitate mass to analyte mass via molar mass ratio. | null | che_0607 | concept | en | What is a gravimetric factor? | [
"Weigh a precipitate.",
"Ratio of molar masses.",
"Multiply to get analyte."
] | chemistry | Answer: Precipitate-to-analyte ratio.
Steps:
1. Weigh a precipitate.
2. Ratio of molar masses.
3. Multiply to get analyte. | stoichiometry |
The chloride mass is 70.90 g, from 2.00 mol times 35.45 g/mol. | null | che_0608 | worked_example | en | 2.00 mol of AgCl precipitates. What mass of Cl does it hold? | [
"Each AgCl has 1 Cl.",
"Moles Cl = 2.00.",
"Mass = 2.00 x 35.45 = 70.90 g."
] | chemistry | Answer: 70.90 g.
Steps:
1. Each AgCl has 1 Cl.
2. Moles Cl = 2.00.
3. Mass = 2.00 x 35.45 = 70.90 g. | stoichiometry |
Hmotnost je 112,04 g, protože 4 krát 28,01 se rovná 112,04. | null | che_0609 | worked_example | cs | Jaká je hmotnost 4 molů CO (M = 28,01 g/mol)? | [
"Napiš vzorec: m = 4 x 28,01.",
"Vynásob.",
"m = 112,04 g."
] | chemistry | Answer: 112,04 g.
Steps:
1. Napiš vzorec: m = 4 x 28,01.
2. Vynásob.
3. m = 112,04 g. | stoichiometry |
The balanced equation is 2Fe + 3Cl2 -> 2FeCl3. | {
"A": "1, 3, 2",
"B": "2, 3, 2",
"C": "2, 1, 2",
"D": "1, 1, 1"
} | che_0610 | multiple_choice | en | What coefficients balance Fe + Cl2 -> FeCl3? | [
"Fe: 2 each side.",
"Cl: 6 each side.",
"Select option B."
] | chemistry | Answer: 2, 3, 2.
Steps:
1. Fe: 2 each side.
2. Cl: 6 each side.
3. Select option B.
Answer: (B) | stoichiometry |
One mole of O2 is needed, from 0.5 mol CH4 times the 2/1 ratio. | null | che_0611 | worked_example | en | How much O2 burns 0.50 mol of CH4 (CH4 + 2O2 -> CO2 + 2H2O)? | [
"Ratio O2:CH4 = 2:1.",
"Multiply: 0.50 x 2.",
"O2 = 1.00 mol."
] | chemistry | Answer: 1.00 mol.
Steps:
1. Ratio O2:CH4 = 2:1.
2. Multiply: 0.50 x 2.
3. O2 = 1.00 mol. | stoichiometry |
Side reactions consume reactants into unwanted products, cutting the main yield. | null | che_0612 | concept | en | Why do side reactions lower yield? | [
"Reactants split pathways.",
"Less goes to product.",
"Yield falls below theory."
] | chemistry | Answer: Reactants diverted elsewhere.
Steps:
1. Reactants split pathways.
2. Less goes to product.
3. Yield falls below theory. | stoichiometry |
Hydrogen limits because 2 mol H2 needs only 1 mol O2, leaving O2 over. | {
"A": "H2 limits",
"B": "O2 limits",
"C": "Neither limits",
"D": "Both limit"
} | che_0613 | multiple_choice | en | With 2 mol H2 and 2 mol O2 for 2H2 + O2 -> 2H2O, what limits? | [
"2 H2 needs 1 O2.",
"O2 is in excess.",
"Select option A."
] | chemistry | Answer: H2 limits.
Steps:
1. 2 H2 needs 1 O2.
2. O2 is in excess.
3. Select option A.
Answer: (A) | stoichiometry |
Teoretický výtěžek je maximální množství produktu ze stechiometrie. | null | che_0614 | concept | cs | Co je teoretický výtěžek? | [
"Výpočet z rovnice.",
"Předpoklad úplné reakce.",
"Skutečnost bývá nižší."
] | chemistry | Answer: Stechiometrické maximum.
Steps:
1. Výpočet z rovnice.
2. Předpoklad úplné reakce.
3. Skutečnost bývá nižší. | stoichiometry |
The actual yield is 30 g, from 0.75 times 40 g. | null | che_0615 | worked_example | en | Theoretical yield is 40 g at 75% yield. Find the actual yield. | [
"Write: actual = 0.75 x 40.",
"Multiply.",
"Actual = 30 g."
] | chemistry | Answer: 30 g.
Steps:
1. Write: actual = 0.75 x 40.
2. Multiply.
3. Actual = 30 g. | stoichiometry |
The molarity is 0.20 M, from 0.0500 mol divided by 0.250 L. | {
"A": "0.02 M",
"B": "0.50 M",
"C": "5.0 M",
"D": "0.20 M"
} | che_0616 | multiple_choice | en | What is the molarity of 0.0500 mol in 250 mL? | [
"Convert: 250 mL = 0.250 L.",
"Divide: 0.0500/0.250.",
"Select option D."
] | chemistry | Answer: 0.20 M.
Steps:
1. Convert: 250 mL = 0.250 L.
2. Divide: 0.0500/0.250.
3. Select option D.
Answer: (D) | stoichiometry |
Dilute to 400 mL total, since 50 x 4.0 / 0.50 = 400 mL. | null | che_0617 | worked_example | en | How to dilute 50 mL of 4.0 M stock to 0.50 M? | [
"Use M1V1 = M2V2.",
"V2 = 50 x 4.0/0.50 = 400 mL.",
"Add 350 mL solvent."
] | chemistry | Answer: Dilute to 400 mL.
Steps:
1. Use M1V1 = M2V2.
2. V2 = 50 x 4.0/0.50 = 400 mL.
3. Add 350 mL solvent. | stoichiometry |
Je to 0,2 molu, protože 0,1 krát 2 se rovná 0,2. | null | che_0618 | worked_example | cs | Kolik molů je ve 100 ml 2M roztoku? | [
"Převeď objem: 0,1 l.",
"Vynásob: 0,1 x 2.",
"n = 0,2 mol."
] | chemistry | Answer: 0,2 mol.
Steps:
1. Převeď objem: 0,1 l.
2. Vynásob: 0,1 x 2.
3. n = 0,2 mol. | stoichiometry |
Carbon dioxide density is 1.96 g/L, from 44.01 g divided by 22.414 L. | {
"A": "1.25 g/L",
"B": "1.96 g/L",
"C": "0.98 g/L",
"D": "2.50 g/L"
} | che_0619 | multiple_choice | en | What is the density of CO2 gas at STP? | [
"Mass of 1 mol = 44.01 g.",
"Divide: 44.01/22.414.",
"Select option B."
] | chemistry | Answer: 1.96 g/L.
Steps:
1. Mass of 1 mol = 44.01 g.
2. Divide: 44.01/22.414.
3. Select option B.
Answer: (B) | stoichiometry |
Ammonia and water form two conjugate pairs: NH4+/NH3 and H2O/OH-. | null | che_0620 | concept | en | What conjugate pairs appear in NH3 + H2O <-> NH4+ + OH-? | [
"NH4+ pairs with NH3.",
"H2O pairs with OH-.",
"Each differs by one H+."
] | chemistry | Answer: NH4+/NH3 and H2O/OH-.
Steps:
1. NH4+ pairs with NH3.
2. H2O pairs with OH-.
3. Each differs by one H+. | acids_bases |
Removing a proton from water gives hydroxide, OH-. | {
"A": "OH(-)",
"B": "H3O(+)",
"C": "O2(-)",
"D": "H2"
} | che_0621 | multiple_choice | en | What is the conjugate base of H2O? | [
"Remove one H+.",
"Left with OH-.",
"Select option A."
] | chemistry | Answer: OH-.
Steps:
1. Remove one H+.
2. Left with OH-.
3. Select option A.
Answer: (A) | acids_bases |
The pH is 1.40, from -log(0.040). | null | che_0622 | worked_example | en | Compute the pH of 0.040 M HBr. | [
"HBr dissociates fully.",
"pH = -log(0.040).",
"pH = 1.40."
] | chemistry | Answer: 1.40.
Steps:
1. HBr dissociates fully.
2. pH = -log(0.040).
3. pH = 1.40. | acids_bases |
Konjugovaný pár tvoří kyselina a zásada lišící se o jeden proton. | null | che_0623 | concept | cs | Co je konjugovaný pár? | [
"Dvě částice, rozdíl H+.",
"Příklad: HCl/Cl-.",
"Silná kyselina = slabá báze."
] | chemistry | Answer: Dvojice lišící se o H+.
Steps:
1. Dvě částice, rozdíl H+.
2. Příklad: HCl/Cl-.
3. Silná kyselina = slabá báze. | acids_bases |
Adding a proton to carbonate gives hydrogen carbonate, HCO3-. | {
"A": "CO2",
"B": "HCO3(-)",
"C": "H2CO3",
"D": "OH(-)"
} | che_0624 | multiple_choice | en | What is the conjugate acid of CO3(2-)? | [
"Add one H+.",
"Result is HCO3-.",
"Select option B."
] | chemistry | Answer: HCO3-.
Steps:
1. Add one H+.
2. Result is HCO3-.
3. Select option B.
Answer: (B) | acids_bases |
Hydroxide concentration is 3.16 x 10^-3 M, from pOH = 14 - 11.5 = 2.5. | null | che_0625 | worked_example | en | A solution has pH 11.50. Compute [OH-]. | [
"pOH = 14 - 11.50 = 2.50.",
"Write: [OH-] = 10^(-2.50).",
"[OH-] = 3.16 x 10^-3 M."
] | chemistry | Answer: 3.16 x 10^-3 M.
Steps:
1. pOH = 14 - 11.50 = 2.50.
2. Write: [OH-] = 10^(-2.50).
3. [OH-] = 3.16 x 10^-3 M. | acids_bases |
Group 1 hydroxides and heavy Group 2 hydroxides are strong bases. | null | che_0626 | concept | en | Which hydroxides are strong bases? | [
"Group 1: NaOH, KOH.",
"Heavy Group 2: Ca, Sr, Ba.",
"They dissociate fully."
] | chemistry | Answer: Group 1 and heavy Group 2.
Steps:
1. Group 1: NaOH, KOH.
2. Heavy Group 2: Ca, Sr, Ba.
3. They dissociate fully. | acids_bases |
The pH is 12.30, from [OH-] = 0.020 M giving pOH 1.70. | null | che_0627 | worked_example | en | Compute the pH of 0.010 M Ca(OH)2. | [
"[OH-] = 2 x 0.010 = 0.020 M.",
"pOH = -log(0.020) = 1.70.",
"pH = 14 - 1.70 = 12.30."
] | chemistry | Answer: 12.30.
Steps:
1. [OH-] = 2 x 0.010 = 0.020 M.
2. pOH = -log(0.020) = 1.70.
3. pH = 14 - 1.70 = 12.30. | acids_bases |
Each Ca(OH)2 releases two hydroxide ions. | {
"A": "1",
"B": "3",
"C": "2",
"D": "0"
} | che_0628 | multiple_choice | en | How many OH- does one Ca(OH)2 release? | [
"Formula has two OH.",
"Both release.",
"Select option C."
] | chemistry | Answer: 2.
Steps:
1. Formula has two OH.
2. Both release.
3. Select option C.
Answer: (C) | acids_bases |
pOH je 1,30, protože -log(0,05) = 1,30. | null | che_0629 | worked_example | cs | Vypočítej pOH roztoku 0,05 M NaOH. | [
"NaOH disociuje plně.",
"pOH = -log(0,05).",
"pOH = 1,30."
] | chemistry | Answer: 1,30.
Steps:
1. NaOH disociuje plně.
2. pOH = -log(0,05).
3. pOH = 1,30. | acids_bases |
Acid rain forms when SO2 and NOx dissolve into sulfuric and nitric acids. | null | che_0630 | concept | en | What chemistry causes acid rain? | [
"Burning releases SO2, NOx.",
"They oxidize and dissolve.",
"Rain pH drops below 5.6."
] | chemistry | Answer: SO2 and NOx acids.
Steps:
1. Burning releases SO2, NOx.
2. They oxidize and dissolve.
3. Rain pH drops below 5.6. | acids_bases |
Rain below pH 5.6 counts as acid rain. | {
"A": "7.0",
"B": "8.0",
"C": "6.5",
"D": "5.6"
} | che_0631 | multiple_choice | en | Below what pH is rain called acid rain? | [
"Natural rain is ~5.6.",
"Lower means polluted.",
"Select option D."
] | chemistry | Answer: 5.6.
Steps:
1. Natural rain is ~5.6.
2. Lower means polluted.
3. Select option D.
Answer: (D) | acids_bases |
Exactly 40.0 mL of NaOH is needed, matching 4.00 mmol of H+. | null | che_0632 | worked_example | en | What volume of 0.100 M NaOH neutralizes 10.0 mL of 0.200 M H2SO4? | [
"H+ = 2 x 10.0 x 0.200 = 4.00 mmol.",
"Same mmol NaOH needed.",
"V = 4.00/0.100 = 40.0 mL."
] | chemistry | Answer: 40.0 mL.
Steps:
1. H+ = 2 x 10.0 x 0.200 = 4.00 mmol.
2. Same mmol NaOH needed.
3. V = 4.00/0.100 = 40.0 mL. | acids_bases |
Diprotic titrations show two endpoints, one per proton. | null | che_0633 | concept | en | Why do diprotic titrations have two endpoints? | [
"First proton neutralized.",
"Then the second.",
"Two jumps appear."
] | chemistry | Answer: One endpoint per proton.
Steps:
1. First proton neutralized.
2. Then the second.
3. Two jumps appear. | acids_bases |
Kyselý déšť je déšť s pH pod 5,6 kvůli oxidům síry a dusíku. | null | che_0634 | concept | cs | Co je kyselý déšť? | [
"Emise SO2 a NOx.",
"Vznik kyselin v atmosféře.",
"pH pod 5,6."
] | chemistry | Answer: Déšť s pH pod 5,6.
Steps:
1. Emise SO2 a NOx.
2. Vznik kyselin v atmosféře.
3. pH pod 5,6. | acids_bases |
Phenol red is yellow in acid and red in base. | {
"A": "Yellow acid, red base",
"B": "Red acid, yellow base",
"C": "Blue both",
"D": "Pink both"
} | che_0635 | multiple_choice | en | How does phenol red change color? | [
"Acid form yellow.",
"Base form red.",
"Select option A."
] | chemistry | Answer: Yellow acid, red base.
Steps:
1. Acid form yellow.
2. Base form red.
3. Select option A.
Answer: (A) | acids_bases |
The concentration is 5.01 x 10^-7 M, from 10 raised to -6.30. | null | che_0636 | worked_example | en | A solution has pH 6.30. Compute [H+]. | [
"Write: [H+] = 10^(-6.30).",
"Evaluate.",
"[H+] = 5.01 x 10^-7 M."
] | chemistry | Answer: 5.01 x 10^-7 M.
Steps:
1. Write: [H+] = 10^(-6.30).
2. Evaluate.
3. [H+] = 5.01 x 10^-7 M. | acids_bases |
pH je 3, protože -log(0,001) = 3. | null | che_0637 | worked_example | cs | Vypočítej pH roztoku 0,001 M HNO3. | [
"HNO3 disociuje plně.",
"pH = -log(0,001).",
"pH = 3."
] | chemistry | Answer: 3.
Steps:
1. HNO3 disociuje plně.
2. pH = -log(0,001).
3. pH = 3. | acids_bases |
Carbonate hydrolyzes to make OH-, so the solution is basic. | {
"A": "Acidic",
"B": "Neutral",
"C": "Basic",
"D": "Cannot tell"
} | che_0638 | multiple_choice | en | Is Na2CO3 solution acidic or basic? | [
"CO3(2-) is base of weak acid.",
"It makes OH-.",
"Select option C."
] | chemistry | Answer: Basic.
Steps:
1. CO3(2-) is base of weak acid.
2. It makes OH-.
3. Select option C.
Answer: (C) | acids_bases |
Halogens are -1 except with oxygen or more electronegative halogens. | null | che_0639 | concept | en | What oxidation states do halogens take? | [
"Normal rule: -1.",
"With O: positive states.",
"Example: Cl in ClO3- is +5."
] | chemistry | Answer: Usually -1, positive with O.
Steps:
1. Normal rule: -1.
2. With O: positive states.
3. Example: Cl in ClO3- is +5. | redox |
Chlorine is +1, since +1 + Cl + (-2) = 0. | {
"A": "-1",
"B": "0",
"C": "+1",
"D": "+3"
} | che_0640 | multiple_choice | en | What is the oxidation state of Cl in HClO? | [
"H +1, O -2.",
"Solve: 1 + Cl - 2 = 0.",
"Cl = +1; select C."
] | chemistry | Answer: +1.
Steps:
1. H +1, O -2.
2. Solve: 1 + Cl - 2 = 0.
3. Cl = +1; select C.
Answer: (C) | redox |
Silicon is +4, since Si + 2(-2) = 0 gives Si = +4. | null | che_0641 | worked_example | en | Find the oxidation number of Si in SiO2. | [
"Assign O -2.",
"Equation: Si - 4 = 0.",
"Si = +4."
] | chemistry | Answer: +4.
Steps:
1. Assign O -2.
2. Equation: Si - 4 = 0.
3. Si = +4. | redox |
Oxidovadlo je látka, která přijímá elektrony a sama se redukuje. | null | che_0642 | concept | cs | Co je oxidovadlo? | [
"Přijímá elektrony.",
"Samo se redukuje.",
"Příklad: O2, Cl2."
] | chemistry | Answer: Příjemce elektronů.
Steps:
1. Přijímá elektrony.
2. Samo se redukuje.
3. Příklad: O2, Cl2. | redox |
Hydrogen goes from 0 to +1, so it is oxidized. | {
"A": "CuO",
"B": "H2",
"C": "Cu",
"D": "O"
} | che_0643 | multiple_choice | en | In CuO + H2 -> Cu + H2O, what is oxidized? | [
"H goes 0 to +1.",
"That is oxidation.",
"Select option B."
] | chemistry | Answer: H2.
Steps:
1. H goes 0 to +1.
2. That is oxidation.
3. Select option B.
Answer: (B) | redox |
The balanced equation is 4P + 5O2 -> 2P2O5. | null | che_0644 | worked_example | en | Balance: P + O2 -> P2O5. | [
"Balance P with 2 P2O5 and 4 P.",
"Count O: 10 each side.",
"Use 5 O2: 4P + 5O2 -> 2P2O5."
] | chemistry | Answer: 4P + 5O2 -> 2P2O5.
Steps:
1. Balance P with 2 P2O5 and 4 P.
2. Count O: 10 each side.
3. Use 5 O2: 4P + 5O2 -> 2P2O5. | redox |
Permanganate titrations use purple MnO4- that turns colorless as Mn2+ at the endpoint. | null | che_0645 | concept | en | How does permanganate titration work? | [
"MnO4- oxidizes analyte.",
"Purple fades to colorless.",
"First pink marks endpoint."
] | chemistry | Answer: Purple to colorless endpoint.
Steps:
1. MnO4- oxidizes analyte.
2. Purple fades to colorless.
3. First pink marks endpoint. | redox |
Three electrons are gained as chromium drops from +6 to +3. | null | che_0646 | worked_example | en | How many electrons are gained in Cr(+6) -> Cr(+3)? | [
"Change: +6 to +3.",
"Difference = 3 down.",
"Electrons gained = 3."
] | chemistry | Answer: 3 electrons.
Steps:
1. Change: +6 to +3.
2. Difference = 3 down.
3. Electrons gained = 3. | redox |
Iodide gives up electrons most easily among the halides. | {
"A": "F(-)",
"B": "Cl(-)",
"C": "Br(-)",
"D": "I(-)"
} | che_0647 | multiple_choice | en | Which halide is the strongest reducing agent? | [
"Largest ion oxidizes easiest.",
"I- tops the series.",
"Select option D."
] | chemistry | Answer: I-.
Steps:
1. Largest ion oxidizes easiest.
2. I- tops the series.
3. Select option D.
Answer: (D) | redox |
Mangan má oxidační číslo +4, protože Mn + 2(-2) = 0. | null | che_0648 | worked_example | cs | Urči oxidační číslo manganu v MnO2. | [
"Přiřaď O -2.",
"Rovnice: Mn - 4 = 0.",
"Mn = +4."
] | chemistry | Answer: +4.
Steps:
1. Přiřaď O -2.
2. Rovnice: Mn - 4 = 0.
3. Mn = +4. | redox |
Aluminum forms a tight oxide skin that blocks further attack. | null | che_0649 | concept | en | Why does aluminum resist corrosion? | [
"Surface oxidizes fast.",
"Dense Al2O3 layer forms.",
"Layer seals the metal."
] | chemistry | Answer: Protective oxide layer.
Steps:
1. Surface oxidizes fast.
2. Dense Al2O3 layer forms.
3. Layer seals the metal. | redox |
A sealing oxide layer stops aluminum corrosion. | {
"A": "Oxide layer",
"B": "Paint only",
"C": "Oil coating",
"D": "Noble metal"
} | che_0650 | multiple_choice | en | What protects aluminum from corrosion? | [
"Al2O3 seals surface.",
"Eliminate paint, oil, noble.",
"Select option A."
] | chemistry | Answer: Oxide layer.
Steps:
1. Al2O3 seals surface.
2. Eliminate paint, oil, noble.
3. Select option A.
Answer: (A) | redox |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.