answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
The E1 intermediate is a carbocation formed after leaving group loss. | {
"A": "Carbocation",
"B": "Carbanion",
"C": "Radical",
"D": "Epoxide"
} | che_0851 | multiple_choice | en | What intermediate forms in E1? | [
"Leaving group leaves first.",
"Positive carbon remains.",
"Select option A."
] | chemistry | Answer: Carbocation.
Steps:
1. Leaving group leaves first.
2. Positive carbon remains.
3. Select option A.
Answer: (A) | organic_mechanisms |
HI adds across ethene to give iodoethane. | null | che_0852 | worked_example | en | What is the HI addition product of ethene? | [
"H adds to one carbon.",
"I adds to the other.",
"Product: iodoethane."
] | chemistry | Answer: Iodoethane.
Steps:
1. H adds to one carbon.
2. I adds to the other.
3. Product: iodoethane. | organic_mechanisms |
Bromine adds to opposite faces of the double bond via a bromonium ion. | null | che_0853 | concept | en | Why is Br2 addition anti? | [
"Bromonium ion forms.",
"Backside attack opens it.",
"Groups land opposite."
] | chemistry | Answer: Bromonium forces backside attack.
Steps:
1. Bromonium ion forms.
2. Backside attack opens it.
3. Groups land opposite. | organic_mechanisms |
Anti-adice přidává skupiny na opačné strany dvojné vazby. | null | che_0854 | concept | cs | Co je anti-adice? | [
"Skupiny na opačné strany.",
"Přes bromoniový ion.",
"Opak syn-adice."
] | chemistry | Answer: Adice na opačné strany.
Steps:
1. Skupiny na opačné strany.
2. Přes bromoniový ion.
3. Opak syn-adice. | organic_mechanisms |
Bromine adds anti across the double bond. | {
"A": "Syn",
"B": "Random",
"C": "Anti",
"D": "No addition"
} | che_0855 | multiple_choice | en | How does Br2 add to alkenes? | [
"Bromonium intermediate.",
"Backside opening.",
"Select option C."
] | chemistry | Answer: Anti.
Steps:
1. Bromonium intermediate.
2. Backside opening.
3. Select option C.
Answer: (C) | organic_mechanisms |
Full hydrogenation of propyne gives propane. | null | che_0856 | worked_example | en | What is the product of propyne + excess H2 (Pt)? | [
"Triple bond saturates fully.",
"Each carbon gains 2 H.",
"Product: propane."
] | chemistry | Answer: Propane.
Steps:
1. Triple bond saturates fully.
2. Each carbon gains 2 H.
3. Product: propane. | organic_mechanisms |
Both hydrogens add from the catalyst surface to the same face. | null | che_0857 | concept | en | Why is catalytic hydrogenation syn? | [
"Alkene adsorbs on metal.",
"H atoms add together.",
"Same face results."
] | chemistry | Answer: Same-face surface addition.
Steps:
1. Alkene adsorbs on metal.
2. H atoms add together.
3. Same face results. | organic_mechanisms |
The product is 2-methylpropan-2-ol, via the tertiary carbocation. | null | che_0858 | worked_example | en | What is the acid hydration product of 2-methylpropene? | [
"H adds to CH2 end.",
"Tertiary cation forms.",
"Water captures it: 2-methylpropan-2-ol."
] | chemistry | Answer: 2-methylpropan-2-ol.
Steps:
1. H adds to CH2 end.
2. Tertiary cation forms.
3. Water captures it: 2-methylpropan-2-ol. | organic_mechanisms |
Hydration places -OH on the more substituted carbon per Markovnikov. | {
"A": "Markovnikov",
"B": "Anti-Markovnikov",
"C": "Random",
"D": "Zaitsev"
} | che_0859 | multiple_choice | en | Which rule governs acid hydration? | [
"Stable cation forms.",
"OH lands on it.",
"Select option A."
] | chemistry | Answer: Markovnikov.
Steps:
1. Stable cation forms.
2. OH lands on it.
3. Select option A.
Answer: (A) | organic_mechanisms |
Úplnou hydrogenací ethynu vzniká ethan. | null | che_0860 | worked_example | cs | Jaký produkt vznikne hydrogenací ethynu s nadbytkem H2? | [
"Trojná vazba se nasytí.",
"Každý uhlík získá 2 H.",
"Produkt: ethan."
] | chemistry | Answer: Ethan.
Steps:
1. Trojná vazba se nasytí.
2. Každý uhlík získá 2 H.
3. Produkt: ethan. | organic_mechanisms |
Condensation polymers link monomers while expelling water or HCl. | null | che_0861 | concept | en | What is condensation polymerization? | [
"Bifunctional monomers join.",
"Small molecule leaves.",
"Examples: nylon, polyester."
] | chemistry | Answer: Joining with small loss.
Steps:
1. Bifunctional monomers join.
2. Small molecule leaves.
3. Examples: nylon, polyester. | organic_mechanisms |
Nylon forms from a diamine plus a diacid with water loss. | null | che_0862 | worked_example | en | What monomers make nylon? | [
"Need two functional ends each.",
"Diamine plus diacid.",
"Water splits out per link."
] | chemistry | Answer: Diamine plus diacid.
Steps:
1. Need two functional ends each.
2. Diamine plus diacid.
3. Water splits out per link. | organic_mechanisms |
Each nylon link expels one water molecule. | {
"A": "HCl",
"B": "Water",
"C": "H2",
"D": "CO2"
} | che_0863 | multiple_choice | en | What small molecule does nylon formation release? | [
"Amide bond forms.",
"H and OH leave.",
"Select option B."
] | chemistry | Answer: Water.
Steps:
1. Amide bond forms.
2. H and OH leave.
3. Select option B.
Answer: (B) | organic_mechanisms |
Kondenzační polymer vzniká spojováním monomerů za odštěpení malé molekuly. | null | che_0864 | concept | cs | Co je kondenzační polymer? | [
"Monomery se spojují.",
"Odchází voda či HCl.",
"Příklad: nylon."
] | chemistry | Answer: Polymer se ztrátou molekuly.
Steps:
1. Monomery se spojují.
2. Odchází voda či HCl.
3. Příklad: nylon. | organic_mechanisms |
The same equilibrium state results whether starting from reactants or products. | null | che_0865 | concept | en | Can equilibrium be reached from either side? | [
"Forward reaction proceeds.",
"Reverse also proceeds.",
"Both meet at same state."
] | chemistry | Answer: Yes, both sides meet.
Steps:
1. Forward reaction proceeds.
2. Reverse also proceeds.
3. Both meet at same state. | equilibrium |
Starting materials on either side converge to the same equilibrium. | {
"A": "Reactants only",
"B": "Products only",
"C": "Either side",
"D": "Never"
} | che_0866 | multiple_choice | en | From which side can equilibrium be reached? | [
"Both directions run.",
"Same state results.",
"Select option C."
] | chemistry | Answer: Either side.
Steps:
1. Both directions run.
2. Same state results.
3. Select option C.
Answer: (C) | equilibrium |
K is 8.0, from 1.0 squared divided by 0.5 cubed. | null | che_0867 | worked_example | en | For 3A <-> 2B with [A] = 0.50 M and [B] = 1.00 M, compute K. | [
"Write: K = [B]^2/[A]^3.",
"Substitute: 1.00/0.125.",
"K = 8.0."
] | chemistry | Answer: 8.0.
Steps:
1. Write: K = [B]^2/[A]^3.
2. Substitute: 1.00/0.125.
3. K = 8.0. | equilibrium |
Kc se vypočítá dosazením rovnovážných koncentrací do poměru produkty/reaktanty. | null | che_0868 | concept | cs | Jak se vypočítá Kc z koncentrací? | [
"Zjisti rovnovážné koncentrace.",
"Dosaď do vzorce.",
"Koeficienty jako exponenty."
] | chemistry | Answer: Dosazením do poměru.
Steps:
1. Zjisti rovnovážné koncentrace.
2. Dosaď do vzorce.
3. Koeficienty jako exponenty. | equilibrium |
Kc units depend on the reaction and are often omitted. | {
"A": "Always M",
"B": "Always atm",
"C": "Always mol",
"D": "Varies or omitted"
} | che_0869 | multiple_choice | en | What are the units of Kc? | [
"Depend on stoichiometry.",
"Often left out.",
"Select option D."
] | chemistry | Answer: Varies or omitted.
Steps:
1. Depend on stoichiometry.
2. Often left out.
3. Select option D.
Answer: (D) | equilibrium |
The system is at equilibrium with no shift. | null | che_0870 | worked_example | en | Q = 1.0 and K = 1.0. Which way does the system shift? | [
"Compare: Q equals K.",
"No excess either side.",
"No shift."
] | chemistry | Answer: No shift.
Steps:
1. Compare: Q equals K.
2. No excess either side.
3. No shift. | equilibrium |
Pressure shifts gas equilibrium toward the side with fewer moles. | null | che_0871 | concept | en | How does pressure shift gas equilibrium? | [
"Count gas moles each side.",
"High P favors fewer.",
"Low P favors more."
] | chemistry | Answer: Toward fewer moles at high P.
Steps:
1. Count gas moles each side.
2. High P favors fewer.
3. Low P favors more. | equilibrium |
Higher pressure favors the single product mole over two reactant moles. | {
"A": "Right",
"B": "Left",
"C": "None",
"D": "Stops"
} | che_0872 | multiple_choice | en | A(g) + B(g) <-> C(g) under pressure shifts where? | [
"Two moles left, one right.",
"Pressure favors fewer.",
"Select option A."
] | chemistry | Answer: Right.
Steps:
1. Two moles left, one right.
2. Pressure favors fewer.
3. Select option A.
Answer: (A) | equilibrium |
Heating drives the endothermic decomposition right, toward CaO and CO2. | null | che_0873 | worked_example | en | How does heating affect CaCO3(s) <-> CaO(s) + CO2(g)? | [
"Decomposition needs heat.",
"Adding heat shifts right.",
"More CO2 forms."
] | chemistry | Answer: Shifts right.
Steps:
1. Decomposition needs heat.
2. Adding heat shifts right.
3. More CO2 forms. | equilibrium |
Zahřátí posune rozklad doprava, k CaO a CO2. | null | che_0874 | worked_example | cs | Kam posune zahřátí rozklad CaCO3? | [
"Rozklad potřebuje teplo.",
"Přidání tepla.",
"Posun doprava."
] | chemistry | Answer: Doprava.
Steps:
1. Rozklad potřebuje teplo.
2. Přidání tepla.
3. Posun doprava. | equilibrium |
Breaking CaCO3 apart requires heat input. | {
"A": "Exothermic",
"B": "Endothermic",
"C": "Athermal",
"D": "Nuclear"
} | che_0875 | multiple_choice | en | Is CaCO3 decomposition endo- or exothermic? | [
"Needs furnace heat.",
"Heat absorbed.",
"Select option B."
] | chemistry | Answer: Endothermic.
Steps:
1. Needs furnace heat.
2. Heat absorbed.
3. Select option B.
Answer: (B) | equilibrium |
A shared ion pushes dissolution back, lowering solubility. | null | che_0876 | concept | en | How does a common ion affect solubility? | [
"Extra ion added.",
"Equilibrium shifts left.",
"Less solid dissolves."
] | chemistry | Answer: Lowers solubility.
Steps:
1. Extra ion added.
2. Equilibrium shifts left.
3. Less solid dissolves. | equilibrium |
Added chloride suppresses AgCl dissolving, so solubility drops. | null | che_0877 | worked_example | en | What happens to AgCl solubility in NaCl solution? | [
"Cl- is the common ion.",
"Dissolution shifts back.",
"Solubility falls."
] | chemistry | Answer: Solubility drops.
Steps:
1. Cl- is the common ion.
2. Dissolution shifts back.
3. Solubility falls. | equilibrium |
Common ions push dissolution equilibrium back toward the solid. | {
"A": "Decreases it",
"B": "Increases it",
"C": "No effect",
"D": "Dissolves all"
} | che_0878 | multiple_choice | en | What does a common ion do to solubility? | [
"Shared ion added.",
"Shift goes left.",
"Select option A."
] | chemistry | Answer: Decreases it.
Steps:
1. Shared ion added.
2. Shift goes left.
3. Select option A.
Answer: (A) | equilibrium |
Selective precipitation separates ions by their different Ksp values. | null | che_0879 | concept | en | What is selective precipitation? | [
"Add reagent slowly.",
"Lowest Ksp falls first.",
"Ions separate in order."
] | chemistry | Answer: Separation by Ksp order.
Steps:
1. Add reagent slowly.
2. Lowest Ksp falls first.
3. Ions separate in order. | equilibrium |
Selektivní srážení odděluje ionty podle rozdílných součinů rozpustnosti. | null | che_0880 | concept | cs | Co je selektivní srážení? | [
"Postupné přidávání činidla.",
"Nejdřív nejmenší Ksp.",
"Ionty se oddělí."
] | chemistry | Answer: Oddělení podle Ksp.
Steps:
1. Postupné přidávání činidla.
2. Nejdřív nejmenší Ksp.
3. Ionty se oddělí. | equilibrium |
The least soluble salt hits its Ksp first and falls out. | {
"A": "Higher Ksp",
"B": "Higher Q",
"C": "Lower Ksp",
"D": "Larger ion"
} | che_0881 | multiple_choice | en | Which salt precipitates first? | [
"Compare Ksp values.",
"Lowest saturates first.",
"Select option C."
] | chemistry | Answer: Lower Ksp.
Steps:
1. Compare Ksp values.
2. Lowest saturates first.
3. Select option C.
Answer: (C) | equilibrium |
The solubility is 2.14 x 10^-4 M, from (3.9 x 10^-11 / 4)^(1/3). | null | che_0882 | worked_example | en | CaF2 has Ksp = 3.9 x 10^-11. Compute its molar solubility. | [
"Write: Ksp = 4s^3.",
"s^3 = 3.9 x 10^-11 / 4 = 9.75 x 10^-12.",
"s = 2.14 x 10^-4 M."
] | chemistry | Answer: 2.14 x 10^-4 M.
Steps:
1. Write: Ksp = 4s^3.
2. s^3 = 3.9 x 10^-11 / 4 = 9.75 x 10^-12.
3. s = 2.14 x 10^-4 M. | equilibrium |
Reduction tables list E° values to compare oxidizing strength. | null | che_0883 | concept | en | How is a standard reduction table read? | [
"Higher E° on top.",
"Top species oxidize.",
"Bottom species reduce."
] | chemistry | Answer: Top oxidizes, bottom reduces.
Steps:
1. Higher E° on top.
2. Top species oxidize.
3. Bottom species reduce. | electrochemistry |
The galvanic cathode is positive as it draws electrons in. | {
"A": "Negative",
"B": "Positive",
"C": "Neutral",
"D": "Variable"
} | che_0884 | multiple_choice | en | What is the cathode's sign in a galvanic cell? | [
"Cathode accepts electrons.",
"It stays positive.",
"Select option B."
] | chemistry | Answer: Positive.
Steps:
1. Cathode accepts electrons.
2. It stays positive.
3. Select option B.
Answer: (B) | electrochemistry |
The cell potential is 2.00 V, from 0.34 - (-1.66). | null | che_0885 | worked_example | en | Compute E cell for Al/Cu with E(Al) = -1.66 V and E(Cu) = 0.34 V. | [
"Cathode is Cu at 0.34 V.",
"Anode is Al at -1.66 V.",
"E = 0.34 - (-1.66) = 2.00 V."
] | chemistry | Answer: 2.00 V.
Steps:
1. Cathode is Cu at 0.34 V.
2. Anode is Al at -1.66 V.
3. E = 0.34 - (-1.66) = 2.00 V. | electrochemistry |
Standardní vodíková elektroda je referenční poločlánek s potenciálem 0,00 V. | null | che_0886 | concept | cs | Co je standardní vodíková elektroda? | [
"Vodík na platině.",
"1 M H+ a 1 atm.",
"Definována 0,00 V."
] | chemistry | Answer: Referenční elektroda 0,00 V.
Steps:
1. Vodík na platině.
2. 1 M H+ a 1 atm.
3. Definována 0,00 V. | electrochemistry |
The hydrogen reference is defined as exactly 0.00 V. | {
"A": "0.00 V",
"B": "1.00 V",
"C": "-1.00 V",
"D": "0.34 V"
} | che_0887 | multiple_choice | en | What is the SHE potential? | [
"Reference defined zero.",
"All others relative.",
"Select option A."
] | chemistry | Answer: 0.00 V.
Steps:
1. Reference defined zero.
2. All others relative.
3. Select option A.
Answer: (A) | electrochemistry |
A cell can do electrical work equal to -nFE. | null | che_0888 | concept | en | How does cell potential relate to work? | [
"Charge times voltage.",
"Work = -nFE.",
"Spontaneous cells do work."
] | chemistry | Answer: Work equals -nFE.
Steps:
1. Charge times voltage.
2. Work = -nFE.
3. Spontaneous cells do work. | electrochemistry |
ΔG is -386 kJ, from -2 x 96485 x 2.00 J. | null | che_0889 | worked_example | en | Compute ΔG for n = 2 and E = 2.00 V (F = 96485 C/mol). | [
"Write: ΔG = -2 x 96485 x 2.00.",
"Multiply: -385,940 J.",
"ΔG = -386 kJ."
] | chemistry | Answer: -386 kJ.
Steps:
1. Write: ΔG = -2 x 96485 x 2.00.
2. Multiply: -385,940 J.
3. ΔG = -386 kJ. | electrochemistry |
The potential is 0.69 V, from 0.78 - 0.0296 x log(1000). | null | che_0890 | worked_example | en | E° = 0.78 V, n = 2, Q = 1000 at 25 C. Compute E. | [
"Write: E = 0.78 - 0.0296 x 3.",
"Subtract: 0.78 - 0.0888.",
"E = 0.69 V."
] | chemistry | Answer: 0.69 V.
Steps:
1. Write: E = 0.78 - 0.0296 x 3.
2. Subtract: 0.78 - 0.0888.
3. E = 0.69 V. | electrochemistry |
The base-10 log of 1000 is 3. | {
"A": "2",
"B": "4",
"C": "3",
"D": "10"
} | che_0891 | multiple_choice | en | What is log(1000)? | [
"1000 = 10^3.",
"Log gives exponent.",
"Select option C."
] | chemistry | Answer: 3.
Steps:
1. 1000 = 10^3.
2. Log gives exponent.
3. Select option C.
Answer: (C) | electrochemistry |
Brine electrolysis gives hydrogen, chlorine, and sodium hydroxide. | null | che_0892 | concept | en | What are the products of aqueous NaCl electrolysis? | [
"Cathode: water to H2.",
"Anode: Cl- to Cl2.",
"NaOH builds up."
] | chemistry | Answer: H2, Cl2, NaOH.
Steps:
1. Cathode: water to H2.
2. Anode: Cl- to Cl2.
3. NaOH builds up. | electrochemistry |
Vzniká vodík, chlor a hydroxid sodný. | null | che_0893 | worked_example | cs | Jaké produkty dá elektrolýza roztoku NaCl? | [
"Katoda: H2 z vody.",
"Anoda: Cl2.",
"Zbývá NaOH."
] | chemistry | Answer: H2, Cl2, NaOH.
Steps:
1. Katoda: H2 z vody.
2. Anoda: Cl2.
3. Zbývá NaOH. | electrochemistry |
The NaOH mass is 40.00 g, from 1.00 mol times 40.00 g/mol. | null | che_0894 | worked_example | en | What mass of NaOH forms from 1.00 mol of electrons in brine electrolysis? | [
"1 e- gives 1 NaOH: 1.00 mol.",
"Multiply: 1.00 x 40.00.",
"Mass = 40.00 g."
] | chemistry | Answer: 40.00 g.
Steps:
1. 1 e- gives 1 NaOH: 1.00 mol.
2. Multiply: 1.00 x 40.00.
3. Mass = 40.00 g. | electrochemistry |
Water reduces to hydrogen gas at the cathode. | {
"A": "Na",
"B": "H2",
"C": "Cl2",
"D": "O2"
} | che_0895 | multiple_choice | en | What forms at the cathode in aqueous NaCl electrolysis? | [
"Water reduces easier than Na+.",
"Product is H2.",
"Select option B."
] | chemistry | Answer: H2.
Steps:
1. Water reduces easier than Na+.
2. Product is H2.
3. Select option B.
Answer: (B) | electrochemistry |
The alkaline cell pairs zinc with manganese dioxide in KOH paste. | null | che_0896 | concept | en | How does an alkaline battery work? | [
"Zn anode oxidizes.",
"MnO2 cathode reduces.",
"KOH paste conducts."
] | chemistry | Answer: Zn/MnO2 in KOH.
Steps:
1. Zn anode oxidizes.
2. MnO2 cathode reduces.
3. KOH paste conducts. | electrochemistry |
The time is 2.68 h, from 96,485 C divided by 10.0 A. | null | che_0897 | worked_example | en | How long at 10.0 A to pass 1.00 mol of electrons? | [
"Q = 1.00 x 96485 = 96,485 C.",
"t = 96,485/10.0 = 9,649 s.",
"t = 2.68 h."
] | chemistry | Answer: 2.68 h.
Steps:
1. Q = 1.00 x 96485 = 96,485 C.
2. t = 96,485/10.0 = 9,649 s.
3. t = 2.68 h. | electrochemistry |
Suchý článek je galvanický článek s pastovým elektrolytem. | null | che_0898 | concept | cs | Co je suchý článek? | [
"Zinková nádobka.",
"Uhlíková tyčinka.",
"Pastový elektrolyt."
] | chemistry | Answer: Článek s pastou.
Steps:
1. Zinková nádobka.
2. Uhlíková tyčinka.
3. Pastový elektrolyt. | electrochemistry |
A standard dry cell delivers about 1.5 V. | {
"A": "12 V",
"B": "6 V",
"C": "3.7 V",
"D": "1.5 V"
} | che_0899 | multiple_choice | en | What voltage does a dry cell give? | [
"Zn/C chemistry.",
"About 1.5 V.",
"Select option D."
] | chemistry | Answer: 1.5 V.
Steps:
1. Zn/C chemistry.
2. About 1.5 V.
3. Select option D.
Answer: (D) | electrochemistry |
The oxygen mass is 16.00 g, from 0.500 mol times 32.00 g/mol. | null | che_0900 | worked_example | en | What mass of O2 comes from 2.00 mol of electrons? | [
"4 e- give 1 O2: 0.500 mol.",
"Multiply: 0.500 x 32.00.",
"Mass = 16.00 g."
] | chemistry | Answer: 16.00 g.
Steps:
1. 4 e- give 1 O2: 0.500 mol.
2. Multiply: 0.500 x 32.00.
3. Mass = 16.00 g. | electrochemistry |
Yields fall short from incomplete reaction, side products, and handling losses. | null | che_0901 | concept | en | Why is actual yield below theoretical? | [
"Reactions may not finish.",
"Side products consume feed.",
"Transfers lose material."
] | chemistry | Answer: Incomplete, side, lost.
Steps:
1. Reactions may not finish.
2. Side products consume feed.
3. Transfers lose material. | stoichiometry |
Potassium nitrate totals 101.10 g/mol from K, N, and 3 O. | {
"A": "85.10 g/mol",
"B": "69.10 g/mol",
"C": "39.10 g/mol",
"D": "101.10 g/mol"
} | che_0902 | multiple_choice | en | What is the molar mass of KNO3? | [
"39.10 + 14.01 + 48.00.",
"Total 101.10.",
"Select option D."
] | chemistry | Answer: 101.10 g/mol.
Steps:
1. 39.10 + 14.01 + 48.00.
2. Total 101.10.
3. Select option D.
Answer: (D) | stoichiometry |
The mass is 48.05 g, from 0.750 mol times 64.06 g/mol. | null | che_0903 | worked_example | en | What mass is 0.750 mol of SO2 (M = 64.06 g/mol)? | [
"Write: m = 0.750 x 64.06.",
"Multiply.",
"m = 48.05 g."
] | chemistry | Answer: 48.05 g.
Steps:
1. Write: m = 0.750 x 64.06.
2. Multiply.
3. m = 48.05 g. | stoichiometry |
Procentuální výtěžek je podíl skutečného a teoretického výtěžku krát sto. | null | che_0904 | concept | cs | Co je procentuální výtěžek? | [
"Skutečný děleno teoretickým.",
"Krát 100 procent.",
"Vždy pod 100 %."
] | chemistry | Answer: Podíl výtěžků krát 100.
Steps:
1. Skutečný děleno teoretickým.
2. Krát 100 procent.
3. Vždy pod 100 %. | stoichiometry |
There are 0.448 mol, from 25.0 g divided by 55.845 g/mol. | null | che_0905 | worked_example | en | How many moles are in 25.0 g of iron (M = 55.845 g/mol)? | [
"Write: n = 25.0/55.845.",
"Divide.",
"n = 0.448 mol."
] | chemistry | Answer: 0.448 mol.
Steps:
1. Write: n = 25.0/55.845.
2. Divide.
3. n = 0.448 mol. | stoichiometry |
The balanced equation is 2Cu + O2 -> 2CuO. | {
"A": "1, 1, 1",
"B": "2, 1, 2",
"C": "1, 2, 2",
"D": "2, 2, 2"
} | che_0906 | multiple_choice | en | What coefficients balance Cu + O2 -> CuO? | [
"Cu: 2 each side.",
"O: 2 each side.",
"Select option B."
] | chemistry | Answer: 2, 1, 2.
Steps:
1. Cu: 2 each side.
2. O: 2 each side.
3. Select option B.
Answer: (B) | stoichiometry |
Find the limiting reactant first, then subtract its need from the excess start. | null | che_0907 | concept | en | How is leftover excess reactant found? | [
"Identify limiting reactant.",
"Compute excess consumed.",
"Subtract from initial."
] | chemistry | Answer: Initial minus consumed.
Steps:
1. Identify limiting reactant.
2. Compute excess consumed.
3. Subtract from initial. | stoichiometry |
The water mass is 35.74 g, from 1.984 mol times 18.015 g/mol. | null | che_0908 | worked_example | en | What mass of H2O forms from 4.00 g H2 with excess O2? | [
"Moles H2 = 4.00/2.016 = 1.984.",
"Moles water = 1.984.",
"Mass = 1.984 x 18.015 = 35.74 g."
] | chemistry | Answer: 35.74 g.
Steps:
1. Moles H2 = 4.00/2.016 = 1.984.
2. Moles water = 1.984.
3. Mass = 1.984 x 18.015 = 35.74 g. | stoichiometry |
Objem je 89,7 l, protože 4 krát 22,414 se rovná 89,656. | null | che_0909 | worked_example | cs | Jaký objem mají 4 moly plynu za STP? | [
"Napiš vzorec: V = 4 x 22,414.",
"Vynásob.",
"V = 89,7 l."
] | chemistry | Answer: 89,7 l.
Steps:
1. Napiš vzorec: V = 4 x 22,414.
2. Vynásob.
3. V = 89,7 l. | stoichiometry |
The molarity is 0.50 M, from 2.00 mol divided by 4.00 L. | {
"A": "8.0 M",
"B": "2.0 M",
"C": "0.50 M",
"D": "0.25 M"
} | che_0910 | multiple_choice | en | What is the molarity of 2.00 mol in 4.00 L? | [
"Write: M = 2.00/4.00.",
"Divide: 0.50.",
"Select option C."
] | chemistry | Answer: 0.50 M.
Steps:
1. Write: M = 2.00/4.00.
2. Divide: 0.50.
3. Select option C.
Answer: (C) | stoichiometry |
Dimensional analysis chains conversion factors so units cancel to the target. | null | che_0911 | concept | en | What is dimensional analysis? | [
"Start with given units.",
"Multiply by ratios.",
"Cancel to target units."
] | chemistry | Answer: Unit-cancelling conversions.
Steps:
1. Start with given units.
2. Multiply by ratios.
3. Cancel to target units. | stoichiometry |
The pH scale runs 0 to 14, with 7 neutral at 25 C. | null | che_0912 | concept | en | What is the pH scale? | [
"0 to 14 range.",
"Below 7 acidic.",
"Above 7 basic."
] | chemistry | Answer: 0-14 acidity scale.
Steps:
1. 0 to 14 range.
2. Below 7 acidic.
3. Above 7 basic. | acids_bases |
Two pH units mean 10 squared, or 100 times the acidity. | {
"A": "2 times",
"B": "100 times",
"C": "1000 times",
"D": "Equal"
} | che_0913 | multiple_choice | en | How much more acidic is pH 3 than pH 5? | [
"Difference is 2 units.",
"Each unit is 10x.",
"Select option B."
] | chemistry | Answer: 100 times.
Steps:
1. Difference is 2 units.
2. Each unit is 10x.
3. Select option B.
Answer: (B) | acids_bases |
The pH is 1.22, from -log(0.060). | null | che_0914 | worked_example | en | Compute the pH of 0.060 M HCl. | [
"[H+] = 0.060 M.",
"pH = -log(0.060).",
"pH = 1.22."
] | chemistry | Answer: 1.22.
Steps:
1. [H+] = 0.060 M.
2. pH = -log(0.060).
3. pH = 1.22. | acids_bases |
Stupnice pH měří kyselost od 0 do 14, neutrální je 7. | null | che_0915 | concept | cs | Co je stupnice pH? | [
"Rozsah 0 až 14.",
"Pod 7 kyselé.",
"Nad 7 zásadité."
] | chemistry | Answer: Škála kyselosti 0-14.
Steps:
1. Rozsah 0 až 14.
2. Pod 7 kyselé.
3. Nad 7 zásadité. | acids_bases |
The concentration is 3.16 x 10^-9 M, from 10 raised to -8.50. | null | che_0916 | worked_example | en | A solution has pH 8.50. Compute [H+]. | [
"Write: [H+] = 10^(-8.50).",
"Evaluate.",
"[H+] = 3.16 x 10^-9 M."
] | chemistry | Answer: 3.16 x 10^-9 M.
Steps:
1. Write: [H+] = 10^(-8.50).
2. Evaluate.
3. [H+] = 3.16 x 10^-9 M. | acids_bases |
Adding a proton to HPO4(2-) gives H2PO4-. | {
"A": "H2PO4(-)",
"B": "PO4(3-)",
"C": "H3PO4",
"D": "H2O"
} | che_0917 | multiple_choice | en | What is the conjugate acid of HPO4(2-)? | [
"Add one H+.",
"Result is H2PO4-.",
"Select option A."
] | chemistry | Answer: H2PO4-.
Steps:
1. Add one H+.
2. Result is H2PO4-.
3. Select option A.
Answer: (A) | acids_bases |
Antacids are weak bases that neutralize excess stomach HCl. | null | che_0918 | concept | en | How do antacids work? | [
"Stomach holds HCl.",
"Base neutralizes it.",
"pH rises, pain eases."
] | chemistry | Answer: Neutralize stomach acid.
Steps:
1. Stomach holds HCl.
2. Base neutralizes it.
3. pH rises, pain eases. | acids_bases |
There are 10.0 mmol, from 5.00 mL times 2.00 M. | null | che_0919 | worked_example | en | How many millimoles are in 5.00 mL of 2.00 M acid? | [
"Write: mmol = 5.00 x 2.00.",
"Multiply.",
"mmol = 10.0."
] | chemistry | Answer: 10.0 mmol.
Steps:
1. Write: mmol = 5.00 x 2.00.
2. Multiply.
3. mmol = 10.0. | acids_bases |
pH je 1,30, protože -log(0,05) = 1,30. | null | che_0920 | worked_example | cs | Vypočítej pH roztoku 0,05 M HNO3. | [
"HNO3 disociuje plně.",
"pH = -log(0,05).",
"pH = 1,30."
] | chemistry | Answer: 1,30.
Steps:
1. HNO3 disociuje plně.
2. pH = -log(0,05).
3. pH = 1,30. | acids_bases |
Soap solutions release OH- from fatty acid salts, so they are basic. | {
"A": "Acidic",
"B": "Neutral",
"C": "Basic",
"D": "Cannot tell"
} | che_0921 | multiple_choice | en | Is soap solution acidic or basic? | [
"Soap is weak-acid salt.",
"Anion hydrolyzes.",
"Select option C."
] | chemistry | Answer: Basic.
Steps:
1. Soap is weak-acid salt.
2. Anion hydrolyzes.
3. Select option C.
Answer: (C) | acids_bases |
Dissolved CO2 forms carbonic acid that lowers ocean pH. | null | che_0922 | concept | en | What is ocean acidification? | [
"CO2 dissolves in water.",
"Carbonic acid forms.",
"pH falls, shells weaken."
] | chemistry | Answer: CO2 lowering ocean pH.
Steps:
1. CO2 dissolves in water.
2. Carbonic acid forms.
3. pH falls, shells weaken. | acids_bases |
Group 2 metals always take oxidation state +2 in compounds. | null | che_0923 | concept | en | What oxidation state do Group 2 metals take? | [
"Group 2 loses two electrons.",
"State is +2 always.",
"Example: Mg in MgO."
] | chemistry | Answer: +2 always.
Steps:
1. Group 2 loses two electrons.
2. State is +2 always.
3. Example: Mg in MgO. | redox |
Magnesium is +2, balanced by two chlorides at -1. | {
"A": "+2",
"B": "+1",
"C": "0",
"D": "-2"
} | che_0924 | multiple_choice | en | What is the oxidation state of Mg in MgCl2? | [
"Cl is -1 each.",
"Solve: Mg - 2 = 0.",
"Mg = +2; select A."
] | chemistry | Answer: +2.
Steps:
1. Cl is -1 each.
2. Solve: Mg - 2 = 0.
3. Mg = +2; select A.
Answer: (A) | redox |
Arsenic is +5, since 3(+1) + As + 4(-2) = 0. | null | che_0925 | worked_example | en | Find the oxidation number of As in H3AsO4. | [
"Assign H +1 and O -2.",
"Equation: 3 + As - 8 = 0.",
"As = +5."
] | chemistry | Answer: +5.
Steps:
1. Assign H +1 and O -2.
2. Equation: 3 + As - 8 = 0.
3. As = +5. | redox |
Redukce je přijetí elektronů spojené se snížením oxidačního čísla. | null | che_0926 | concept | cs | Co je redukce? | [
"Příjem elektronů.",
"Pokles oxidačního čísla.",
"Příklad: Cu2+ na Cu."
] | chemistry | Answer: Příjem elektronů.
Steps:
1. Příjem elektronů.
2. Pokles oxidačního čísla.
3. Příklad: Cu2+ na Cu. | redox |
The balanced equation is 2Ba + O2 -> 2BaO. | null | che_0927 | worked_example | en | Balance: Ba + O2 -> BaO. | [
"Balance O with 2 BaO.",
"Balance Ba with 2 Ba.",
"Result: 2Ba + O2 -> 2BaO."
] | chemistry | Answer: 2Ba + O2 -> 2BaO.
Steps:
1. Balance O with 2 BaO.
2. Balance Ba with 2 Ba.
3. Result: 2Ba + O2 -> 2BaO. | redox |
Hydrogen goes from 0 to +1, so it is oxidized. | {
"A": "Cl2",
"B": "H2",
"C": "HCl",
"D": "None"
} | che_0928 | multiple_choice | en | In H2 + Cl2 -> 2HCl, 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 |
Photosynthesis reduces CO2 to sugar while water is oxidized to O2. | null | che_0929 | concept | en | How is photosynthesis a redox process? | [
"CO2 gains electrons.",
"Water loses them.",
"Light powers the transfer."
] | chemistry | Answer: CO2 reduced, water oxidized.
Steps:
1. CO2 gains electrons.
2. Water loses them.
3. Light powers the transfer. | redox |
Two electrons are gained as I2 becomes 2I-. | null | che_0930 | worked_example | en | How many electrons are gained in I2 -> 2I-? | [
"Each I goes 0 to -1.",
"Two I gain 1 each.",
"Total gained = 2."
] | chemistry | Answer: 2 electrons.
Steps:
1. Each I goes 0 to -1.
2. Two I gain 1 each.
3. Total gained = 2. | redox |
Olovo má oxidační číslo +4, protože Pb + 2(-2) = 0. | null | che_0931 | worked_example | cs | Urči oxidační číslo olova v PbO2. | [
"Přiřaď O -2.",
"Rovnice: Pb - 4 = 0.",
"Pb = +4."
] | chemistry | Answer: +4.
Steps:
1. Přiřaď O -2.
2. Rovnice: Pb - 4 = 0.
3. Pb = +4. | redox |
Potassium is +1 as in all its compounds. | {
"A": "-1",
"B": "0",
"C": "+1",
"D": "+2"
} | che_0932 | multiple_choice | en | What is the oxidation state of K in KO2? | [
"Alkali metals are +1.",
"Applies here too.",
"Select option C."
] | chemistry | Answer: +1.
Steps:
1. Alkali metals are +1.
2. Applies here too.
3. Select option C.
Answer: (C) | redox |
Respiration oxidizes glucose to CO2 while oxygen is reduced to water. | null | che_0933 | concept | en | How is respiration a redox process? | [
"Glucose loses electrons.",
"Oxygen gains them.",
"Energy is released."
] | chemistry | Answer: Glucose oxidized, O2 reduced.
Steps:
1. Glucose loses electrons.
2. Oxygen gains them.
3. Energy is released. | redox |
Melting, vaporizing, and dissolving salts are common endothermic examples. | null | che_0934 | concept | en | What are examples of endothermic processes? | [
"Ice melting absorbs heat.",
"Water boiling absorbs heat.",
"Cold packs use salt dissolving."
] | chemistry | Answer: Melting, boiling, cold packs.
Steps:
1. Ice melting absorbs heat.
2. Water boiling absorbs heat.
3. Cold packs use salt dissolving. | thermochemistry |
Melting absorbs heat from the surroundings. | {
"A": "Heat in",
"B": "Heat out",
"C": "No heat",
"D": "Work only"
} | che_0935 | multiple_choice | en | Does ice melting need heat in or out? | [
"Bonds break.",
"Breaking needs energy.",
"Select option A."
] | chemistry | Answer: Heat in.
Steps:
1. Bonds break.
2. Breaking needs energy.
3. Select option A.
Answer: (A) | thermochemistry |
The heat is 18.8 kJ, from 50 x 4.184 x 90 J. | null | che_0936 | worked_example | en | How much heat warms 50 g of water from 10 C to 100 C? | [
"ΔT = 100 - 10 = 90 C.",
"q = 50 x 4.184 x 90.",
"q = 18,828 J = 18.8 kJ."
] | chemistry | Answer: 18.8 kJ.
Steps:
1. ΔT = 100 - 10 = 90 C.
2. q = 50 x 4.184 x 90.
3. q = 18,828 J = 18.8 kJ. | thermochemistry |
Tepelná kapacita tělesa je teplo potřebné k ohřátí o 1 °C. | null | che_0937 | concept | cs | Co je tepelná kapacita tělesa? | [
"Teplo na 1 °C.",
"Závisí na hmotnosti.",
"Jednotka J/°C."
] | chemistry | Answer: Teplo na 1 °C.
Steps:
1. Teplo na 1 °C.
2. Závisí na hmotnosti.
3. Jednotka J/°C. | thermochemistry |
The rise is 1.20 C, from 500 J divided by (100 x 4.184). | null | che_0938 | worked_example | en | 100 g of water gains 500 J. Find ΔT (c = 4.184 J/gC). | [
"Write: ΔT = 500/(100 x 4.184).",
"Divide: 500/418.4.",
"ΔT = 1.20 C."
] | chemistry | Answer: 1.20 C.
Steps:
1. Write: ΔT = 500/(100 x 4.184).
2. Divide: 500/418.4.
3. ΔT = 1.20 C. | thermochemistry |
One calorie equals 4.184 joules exactly. | {
"A": "1 J",
"B": "4.184 J",
"C": "1000 J",
"D": "0.239 J"
} | che_0939 | multiple_choice | en | How many joules are in one calorie? | [
"Definition: 4.184 J.",
"Eliminate others.",
"Select option B."
] | chemistry | Answer: 4.184 J.
Steps:
1. Definition: 4.184 J.
2. Eliminate others.
3. Select option B.
Answer: (B) | thermochemistry |
A food Calorie is one kilocalorie, or 1000 chemistry calories. | null | che_0940 | concept | en | What is a food Calorie? | [
"Capital C matters.",
"Equals 1 kcal.",
"Equals 4184 J."
] | chemistry | Answer: One kilocalorie.
Steps:
1. Capital C matters.
2. Equals 1 kcal.
3. Equals 4184 J. | thermochemistry |
The ΔH is -300 kJ, from -500 minus -200 kJ. | null | che_0941 | worked_example | en | Products total -500 kJ and reactants -200 kJ. Find ΔH. | [
"Write: ΔH = products - reactants.",
"Substitute: -500 - (-200).",
"ΔH = -300 kJ."
] | chemistry | Answer: -300 kJ.
Steps:
1. Write: ΔH = products - reactants.
2. Substitute: -500 - (-200).
3. ΔH = -300 kJ. | thermochemistry |
Potřebné teplo je 8368 J, protože 80 x 4,184 x 25 se rovná 8368. | null | che_0942 | worked_example | cs | Kolik tepla ohřeje 80 g vody o 25 °C? | [
"Napiš vzorec: q = 80 x 4,184 x 25.",
"Vynásob.",
"q = 8368 J."
] | chemistry | Answer: 8368 J.
Steps:
1. Napiš vzorec: q = 80 x 4,184 x 25.
2. Vynásob.
3. q = 8368 J. | thermochemistry |
Rising universe entropy marks spontaneous processes. | {
"A": "Spontaneity",
"B": "Speed",
"C": "Color",
"D": "Mass"
} | che_0943 | multiple_choice | en | What does ΔS of the universe determine? | [
"Second law criterion.",
"Positive means spontaneous.",
"Select option A."
] | chemistry | Answer: Spontaneity.
Steps:
1. Second law criterion.
2. Positive means spontaneous.
3. Select option A.
Answer: (A) | thermochemistry |
The third law states a perfect crystal has zero entropy at 0 K. | null | che_0944 | concept | en | What is the third law of thermodynamics? | [
"Perfect order at 0 K.",
"One microstate only.",
"Entropy equals zero."
] | chemistry | Answer: Zero entropy at 0 K.
Steps:
1. Perfect order at 0 K.
2. One microstate only.
3. Entropy equals zero. | thermochemistry |
Isopropyl and tert-butyl are common branched alkyl names. | null | che_0945 | concept | en | What are isopropyl and tert-butyl? | [
"Isopropyl: 3 carbons branched.",
"Tert-butyl: 4 carbons branched.",
"Retained common names."
] | chemistry | Answer: Branched alkyl groups.
Steps:
1. Isopropyl: 3 carbons branched.
2. Tert-butyl: 4 carbons branched.
3. Retained common names. | organic_naming |
Isopropyl attaches through the middle carbon, making it 2-propyl. | {
"A": "1-propyl",
"B": "Ethyl",
"C": "2-propyl",
"D": "Butyl"
} | che_0946 | multiple_choice | en | What is isopropyl systematically? | [
"Three-carbon branch.",
"Middle attachment.",
"Select option C."
] | chemistry | Answer: 2-propyl.
Steps:
1. Three-carbon branch.
2. Middle attachment.
3. Select option C.
Answer: (C) | organic_naming |
The name is 2,3-dimethylpentane: methyls on carbons 2 and 3. | null | che_0947 | worked_example | en | Name CH3CH2CH(CH3)CH(CH3)2. | [
"Longest chain: 5 = pentane.",
"Methyls on carbons 2 and 3.",
"Name: 2,3-dimethylpentane."
] | chemistry | Answer: 2,3-dimethylpentane.
Steps:
1. Longest chain: 5 = pentane.
2. Methyls on carbons 2 and 3.
3. Name: 2,3-dimethylpentane. | organic_naming |
Alkyl je uhlovodíkový zbytek vzniklý odtržením vodíku z alkanu. | null | che_0948 | concept | cs | Co je alkyl? | [
"Alkan bez jednoho H.",
"Obecný vzorec CnH2n+1.",
"Příklad: methyl, ethyl."
] | chemistry | Answer: Zbytek alkanu.
Steps:
1. Alkan bez jednoho H.
2. Obecný vzorec CnH2n+1.
3. Příklad: methyl, ethyl. | organic_naming |
The name is pentane: five carbons in a straight chain. | null | che_0949 | worked_example | en | Name CH3CH2CH2CH2CH3. | [
"Count 5 carbons.",
"Single bonds only.",
"Name: pentane."
] | chemistry | Answer: Pentane.
Steps:
1. Count 5 carbons.
2. Single bonds only.
3. Name: pentane. | organic_naming |
Pentane has five carbon atoms. | {
"A": "4",
"B": "5",
"C": "6",
"D": "3"
} | che_0950 | multiple_choice | en | How many carbons are in pentane? | [
"Pent- means five.",
"Straight chain.",
"Select option B."
] | chemistry | Answer: 5.
Steps:
1. Pent- means five.
2. Straight chain.
3. Select option B.
Answer: (B) | organic_naming |
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