answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
Iodide leaves fastest as the weakest base among the halides. | {
"A": "F(-)",
"B": "Cl(-)",
"C": "Br(-)",
"D": "I(-)"
} | che_0251 | multiple_choice | en | Which is the best leaving group? | [
"Weakest base leaves best.",
"I- is weakest base.",
"Select option D."
] | chemistry | Answer: I-.
Steps:
1. Weakest base leaves best.
2. I- is weakest base.
3. Select option D.
Answer: (D) | organic_mechanisms |
The major product is 2-butene, the more substituted Zaitsev alkene. | null | che_0252 | worked_example | en | What is the major E2 product of 2-bromobutane with strong base? | [
"Remove H from adjacent carbons.",
"Zaitsev favors more substituted alkene.",
"Product: 2-butene."
] | chemistry | Answer: 2-butene.
Steps:
1. Remove H from adjacent carbons.
2. Zaitsev favors more substituted alkene.
3. Product: 2-butene. | organic_mechanisms |
Zaitsev's rule favors the most substituted alkene in elimination. | null | che_0253 | concept | en | What is Zaitsev's rule? | [
"Elimination removes HX.",
"Most substituted alkene wins.",
"More stable product forms."
] | chemistry | Answer: Most substituted alkene favored.
Steps:
1. Elimination removes HX.
2. Most substituted alkene wins.
3. More stable product forms. | organic_mechanisms |
Eliminace odstraňuje atomy ze sousedních uhlíků za vzniku dvojné vazby. | null | che_0254 | concept | cs | Co je eliminace? | [
"Odchod HX ze sousedů.",
"Vzniká alken.",
"Opak adice."
] | chemistry | Answer: Odstranění za vzniku dvojné vazby.
Steps:
1. Odchod HX ze sousedů.
2. Vzniká alken.
3. Opak adice. | organic_mechanisms |
Markovnikov's rule puts H on the less substituted carbon of the double bond. | {
"A": "H to less substituted C",
"B": "H to more substituted C",
"C": "Random addition",
"D": "No addition occurs"
} | che_0255 | multiple_choice | en | What does Markovnikov's rule state? | [
"H adds first as electrophile.",
"Stable carbocation forms.",
"Select option A."
] | chemistry | Answer: H to less substituted C.
Steps:
1. H adds first as electrophile.
2. Stable carbocation forms.
3. Select option A.
Answer: (A) | organic_mechanisms |
Acid-catalyzed hydration of ethene gives ethanol. | null | che_0256 | worked_example | en | What is the product of ethene + H2O with acid catalyst? | [
"H+ adds to double bond.",
"Water captures carbocation.",
"Product: ethanol."
] | chemistry | Answer: Ethanol.
Steps:
1. H+ adds to double bond.
2. Water captures carbocation.
3. Product: ethanol. | organic_mechanisms |
Alkene hydration adds water across the double bond following Markovnikov's rule. | null | che_0257 | concept | en | What is hydration of alkenes? | [
"Water adds to C=C.",
"Acid catalyzes it.",
"Alcohol forms."
] | chemistry | Answer: Water addition to alkenes.
Steps:
1. Water adds to C=C.
2. Acid catalyzes it.
3. Alcohol forms. | organic_mechanisms |
The balanced equation is C2H5OH + 3O2 -> 2CO2 + 3H2O. | null | che_0258 | worked_example | en | Balance ethanol combustion: C2H5OH + O2 -> CO2 + H2O. | [
"Balance C with 2 CO2.",
"Balance H with 3 H2O.",
"Balance O with 3 O2."
] | chemistry | Answer: C2H5OH + 3O2 -> 2CO2 + 3H2O.
Steps:
1. Balance C with 2 CO2.
2. Balance H with 3 H2O.
3. Balance O with 3 O2. | organic_mechanisms |
Tertiary radicals are most stable due to hyperconjugation from three alkyl groups. | {
"A": "Methyl",
"B": "Primary",
"C": "Tertiary",
"D": "Vinyl"
} | che_0259 | multiple_choice | en | Which carbon radical is most stable? | [
"Alkyl groups stabilize radicals.",
"Tertiary has three.",
"Select option C."
] | chemistry | Answer: Tertiary.
Steps:
1. Alkyl groups stabilize radicals.
2. Tertiary has three.
3. Select option C.
Answer: (C) | organic_mechanisms |
Hydrogenací propenu vzniká propan. | null | che_0260 | worked_example | cs | Jaký produkt vznikne hydrogenací propenu (H2, Ni)? | [
"Vodík se aduje na dvojnou vazbu.",
"Každý uhlík získá vodík.",
"Produkt: propan."
] | chemistry | Answer: Propan.
Steps:
1. Vodík se aduje na dvojnou vazbu.
2. Každý uhlík získá vodík.
3. Produkt: propan. | organic_mechanisms |
Addition polymerization links alkene monomers into long chains through their double bonds. | null | che_0261 | concept | en | What is addition polymerization? | [
"Monomers have C=C bonds.",
"Bonds open and link.",
"Chain grows long."
] | chemistry | Answer: Linking alkenes into chains.
Steps:
1. Monomers have C=C bonds.
2. Bonds open and link.
3. Chain grows long. | organic_mechanisms |
The repeat unit is -CH2-CH2-, from opened ethene monomers. | null | che_0262 | worked_example | en | What is the repeat unit of polyethene? | [
"Monomer is CH2=CH2.",
"Double bond opens.",
"Repeat: -CH2-CH2-."
] | chemistry | Answer: -CH2-CH2-.
Steps:
1. Monomer is CH2=CH2.
2. Double bond opens.
3. Repeat: -CH2-CH2-. | organic_mechanisms |
The nitronium ion NO2+ is the electrophile in aromatic nitration. | {
"A": "OH(-)",
"B": "NO2(+)",
"C": "Cl(-)",
"D": "NH3"
} | che_0263 | multiple_choice | en | Which species acts as an electrophile? | [
"Need electron-poor species.",
"NO2+ is cationic.",
"Select option B."
] | chemistry | Answer: NO2+.
Steps:
1. Need electron-poor species.
2. NO2+ is cationic.
3. Select option B.
Answer: (B) | organic_mechanisms |
Katalyzátor zrychluje reakci snížením aktivační energie a sám se nespotřebuje. | null | che_0264 | concept | cs | Co je katalyzátor? | [
"Alternativní rychlejší cesta.",
"Nižší aktivační energie.",
"Na konci nezměněn."
] | chemistry | Answer: Látka zrychlující reakci.
Steps:
1. Alternativní rychlejší cesta.
2. Nižší aktivační energie.
3. Na konci nezměněn. | organic_mechanisms |
K changes with temperature: heating favors the endothermic direction. | null | che_0265 | concept | en | How does temperature affect K? | [
"K depends on T only.",
"Heat favors endothermic side.",
"K rises or falls accordingly."
] | chemistry | Answer: K shifts with temperature.
Steps:
1. K depends on T only.
2. Heat favors endothermic side.
3. K rises or falls accordingly. | equilibrium |
Only temperature changes K; concentration and pressure only shift position. | {
"A": "Temperature only",
"B": "Pressure only",
"C": "Concentration only",
"D": "Catalysts"
} | che_0266 | multiple_choice | en | What changes the value of K? | [
"K is temperature-defined.",
"Other stresses shift, not change K.",
"Select option A."
] | chemistry | Answer: Temperature only.
Steps:
1. K is temperature-defined.
2. Other stresses shift, not change K.
3. Select option A.
Answer: (A) | equilibrium |
The reverse K is 0.25, the reciprocal of 4. | null | che_0267 | worked_example | en | Forward K = 4 for A <-> B. What is K for the reverse? | [
"Reverse inverts the ratio.",
"K_rev = 1/K_fwd.",
"K_rev = 1/4 = 0.25."
] | chemistry | Answer: 0.25.
Steps:
1. Reverse inverts the ratio.
2. K_rev = 1/K_fwd.
3. K_rev = 1/4 = 0.25. | equilibrium |
Dynamická rovnováha znamená stejné rychlosti obou směrů při stálých koncentracích. | null | che_0268 | concept | cs | Co je dynamická rovnováha? | [
"Oba směry běží dál.",
"Rychlosti se rovnají.",
"Koncentrace se nemění."
] | chemistry | Answer: Stejné rychlosti obou směrů.
Steps:
1. Oba směry běží dál.
2. Rychlosti se rovnají.
3. Koncentrace se nemění. | equilibrium |
Doubling all coefficients squares the equilibrium constant. | {
"A": "K halves",
"B": "K doubles",
"C": "K is squared",
"D": "K unchanged"
} | che_0269 | multiple_choice | en | What happens to K when the equation is doubled? | [
"Exponents double in K.",
"That squares the value.",
"Select option C."
] | chemistry | Answer: K is squared.
Steps:
1. Exponents double in K.
2. That squares the value.
3. Select option C.
Answer: (C) | equilibrium |
The overall K is 6, from 2 times 3. | null | che_0270 | worked_example | en | Step 1 has K1 = 2 and step 2 has K2 = 3. Find overall K. | [
"Adding steps multiplies K.",
"K = K1 x K2.",
"K = 2 x 3 = 6."
] | chemistry | Answer: 6.
Steps:
1. Adding steps multiplies K.
2. K = K1 x K2.
3. K = 2 x 3 = 6. | equilibrium |
The reaction quotient Q has the K form but uses current, not equilibrium, concentrations. | null | che_0271 | concept | en | What is the reaction quotient Q? | [
"Same form as K.",
"Uses current concentrations.",
"Compare to K for direction."
] | chemistry | Answer: Current-state version of K.
Steps:
1. Same form as K.
2. Uses current concentrations.
3. Compare to K for direction. | equilibrium |
Excess products push the system back toward reactants. | {
"A": "Shifts right",
"B": "Shifts left",
"C": "No shift",
"D": "K changes"
} | che_0272 | multiple_choice | en | If Q > K, which way does the system shift? | [
"Too many products.",
"System consumes them.",
"Select option B."
] | chemistry | Answer: Shifts left.
Steps:
1. Too many products.
2. System consumes them.
3. Select option B.
Answer: (B) | equilibrium |
Q is 0.50, from 1.0 divided by (2.0 x 1.0). | null | che_0273 | worked_example | en | For A + B <-> C with [A] = 2.0, [B] = 1.0, [C] = 1.0 M, compute Q. | [
"Write: Q = [C]/([A][B]).",
"Substitute: 1.0/(2.0 x 1.0).",
"Q = 0.50."
] | chemistry | Answer: 0.50.
Steps:
1. Write: Q = [C]/([A][B]).
2. Substitute: 1.0/(2.0 x 1.0).
3. Q = 0.50. | equilibrium |
Odebrání produktu posune rovnováhu doprava, k produktům. | null | che_0274 | worked_example | cs | Kam se posune rovnováha po odebrání produktu? | [
"Produktů je málo.",
"Soustava je doplní.",
"Posun doprava."
] | chemistry | Answer: Doprava.
Steps:
1. Produktů je málo.
2. Soustava je doplní.
3. Posun doprava. | equilibrium |
The system makes more product to replace what was removed. | {
"A": "Shifts right",
"B": "Shifts left",
"C": "No shift",
"D": "Stops reacting"
} | che_0275 | multiple_choice | en | What happens when a product is removed at equilibrium? | [
"Products are scarce.",
"System replaces them.",
"Select option A."
] | chemistry | Answer: Shifts right.
Steps:
1. Products are scarce.
2. System replaces them.
3. Select option A.
Answer: (A) | equilibrium |
Shrinking volume raises pressure, shifting gas equilibrium toward fewer moles. | null | che_0276 | concept | en | How does decreasing volume affect gas equilibrium? | [
"Smaller volume raises pressure.",
"System favors fewer gas moles.",
"No effect if moles are equal."
] | chemistry | Answer: Shifts toward fewer gas moles.
Steps:
1. Smaller volume raises pressure.
2. System favors fewer gas moles.
3. No effect if moles are equal. | equilibrium |
Higher pressure favors ammonia because 4 gas moles become 2. | null | che_0277 | worked_example | en | For N2 + 3H2 <-> 2NH3, which side does high pressure favor? | [
"Count gas moles: 4 left, 2 right.",
"Pressure favors fewer moles.",
"High pressure favors NH3."
] | chemistry | Answer: Products (NH3).
Steps:
1. Count gas moles: 4 left, 2 right.
2. Pressure favors fewer moles.
3. High pressure favors NH3. | equilibrium |
Inert gas at fixed volume changes no partial pressures, so there is no shift. | {
"A": "Shifts right",
"B": "Shifts left",
"C": "K doubles",
"D": "No shift"
} | che_0278 | multiple_choice | en | What does adding inert gas at constant volume do? | [
"Partial pressures stay same.",
"Q stays equal to K.",
"Select option D."
] | chemistry | Answer: No shift.
Steps:
1. Partial pressures stay same.
2. Q stays equal to K.
3. Select option D.
Answer: (D) | equilibrium |
Ksp is the equilibrium constant for a solid dissolving into its ions. | null | che_0279 | concept | en | What is the solubility product Ksp? | [
"Solid salt dissolves to ions.",
"Ksp multiplies ion concentrations.",
"Larger Ksp means more soluble."
] | chemistry | Answer: Equilibrium constant of dissolving.
Steps:
1. Solid salt dissolves to ions.
2. Ksp multiplies ion concentrations.
3. Larger Ksp means more soluble. | equilibrium |
Součin rozpustnosti je rovnovážná konstanta rozpouštění málo rozpustné soli. | null | che_0280 | concept | cs | Co je součin rozpustnosti? | [
"Sůl se rozpouští na ionty.",
"Ksp = součin koncentrací.",
"Větší Ksp = rozpustnější."
] | chemistry | Answer: Konstanta rozpouštění soli.
Steps:
1. Sůl se rozpouští na ionty.
2. Ksp = součin koncentrací.
3. Větší Ksp = rozpustnější. | equilibrium |
Ksp of silver chloride equals [Ag+] times [Cl-]. | {
"A": "[AgCl]",
"B": "[Ag+][Cl-]",
"C": "[Ag+]/[Cl-]",
"D": "[Ag+] + [Cl-]"
} | che_0281 | multiple_choice | en | What is the Ksp expression for AgCl? | [
"AgCl gives Ag+ and Cl-.",
"Multiply ion concentrations.",
"Select option B."
] | chemistry | Answer: [Ag+][Cl-].
Steps:
1. AgCl gives Ag+ and Cl-.
2. Multiply ion concentrations.
3. Select option B.
Answer: (B) | equilibrium |
The solubility is 1.34 x 10^-5 M, the square root of 1.8 x 10^-10. | null | che_0282 | worked_example | en | AgCl has Ksp = 1.8 x 10^-10. Compute its molar solubility. | [
"Write: s^2 = Ksp.",
"Substitute: s = sqrt(1.8 x 10^-10).",
"s = 1.34 x 10^-5 M."
] | chemistry | Answer: 1.34 x 10^-5 M.
Steps:
1. Write: s^2 = Ksp.
2. Substitute: s = sqrt(1.8 x 10^-10).
3. s = 1.34 x 10^-5 M. | equilibrium |
Cell notation lists anode on the left and cathode on the right, split by a double bar. | null | che_0283 | concept | en | How is galvanic cell notation written? | [
"Anode half on the left.",
"Cathode half on the right.",
"Double bar is the salt bridge."
] | chemistry | Answer: Anode || cathode.
Steps:
1. Anode half on the left.
2. Cathode half on the right.
3. Double bar is the salt bridge. | electrochemistry |
By convention the anode (oxidation) half-cell is written on the left. | {
"A": "Anode",
"B": "Cathode",
"C": "Salt bridge",
"D": "Voltmeter"
} | che_0284 | multiple_choice | en | Which half-cell goes on the left in cell notation? | [
"Recall anode-left convention.",
"Eliminate cathode, bridge, meter.",
"Select option A."
] | chemistry | Answer: Anode.
Steps:
1. Recall anode-left convention.
2. Eliminate cathode, bridge, meter.
3. Select option A.
Answer: (A) | electrochemistry |
The cell potential is 0.46 V, from 0.80 - 0.34. | null | che_0285 | worked_example | en | Compute E cell for Ag/Cu with E(Ag) = 0.80 V and E(Cu) = 0.34 V. | [
"Cathode is Ag at 0.80 V.",
"Anode is Cu at 0.34 V.",
"E = 0.80 - 0.34 = 0.46 V."
] | chemistry | Answer: 0.46 V.
Steps:
1. Cathode is Ag at 0.80 V.
2. Anode is Cu at 0.34 V.
3. E = 0.80 - 0.34 = 0.46 V. | electrochemistry |
Elektrodový potenciál udává sklon poločlánku redukovat se vůči vodíkové elektrodě. | null | che_0286 | concept | cs | Co je elektrodový potenciál? | [
"Měří se proti vodíku.",
"Kladnější = snazší redukce.",
"Rozdíl dává napětí článku."
] | chemistry | Answer: Sklon k redukci proti vodíku.
Steps:
1. Měří se proti vodíku.
2. Kladnější = snazší redukce.
3. Rozdíl dává napětí článku. | electrochemistry |
A more positive reduction potential means the species reduces more easily. | {
"A": "Harder to reduce",
"B": "Easier to reduce",
"C": "Never reacts",
"D": "Always an anode"
} | che_0287 | multiple_choice | en | What does a more positive E° mean? | [
"E° ranks reduction ease.",
"Higher means easier gain of electrons.",
"Select option B."
] | chemistry | Answer: Easier to reduce.
Steps:
1. E° ranks reduction ease.
2. Higher means easier gain of electrons.
3. Select option B.
Answer: (B) | electrochemistry |
Positive E° gives negative ΔG, so the cell reaction is spontaneous. | null | che_0288 | worked_example | en | A cell has E° = +0.46 V. Is it spontaneous? | [
"ΔG = -nFE°.",
"Positive E° makes ΔG negative.",
"Yes, spontaneous."
] | chemistry | Answer: Yes, spontaneous.
Steps:
1. ΔG = -nFE°.
2. Positive E° makes ΔG negative.
3. Yes, spontaneous. | electrochemistry |
A concentration cell uses the same electrodes with different ion concentrations to generate voltage. | null | che_0289 | concept | en | What is a concentration cell? | [
"Same metal both sides.",
"Different concentrations.",
"Voltage evens them out."
] | chemistry | Answer: Same electrodes, different concentrations.
Steps:
1. Same metal both sides.
2. Different concentrations.
3. Voltage evens them out. | electrochemistry |
The cell voltage is 0.0592 V, from (0.0592/2) x log(100). | null | che_0290 | worked_example | en | A concentration cell (n = 2) has 1.0 M cathode and 0.010 M anode. Compute E at 25 C. | [
"E° = 0 for same electrodes.",
"E = (0.0592/2) x log(1.0/0.010).",
"E = 0.0296 x 2 = 0.0592 V."
] | chemistry | Answer: 0.0592 V.
Steps:
1. E° = 0 for same electrodes.
2. E = (0.0592/2) x log(1.0/0.010).
3. E = 0.0296 x 2 = 0.0592 V. | electrochemistry |
Series connection adds the individual cell voltages together. | {
"A": "Voltages cancel",
"B": "Voltage stays same",
"C": "Voltages add",
"D": "Current reverses"
} | che_0291 | multiple_choice | en | What happens wiring batteries in series? | [
"Positive meets negative.",
"Potentials sum up.",
"Select option C."
] | chemistry | Answer: Voltages add.
Steps:
1. Positive meets negative.
2. Potentials sum up.
3. Select option C.
Answer: (C) | electrochemistry |
The lead-acid battery pairs lead and lead dioxide plates in sulfuric acid, rechargeable by reversing the reaction. | null | che_0292 | concept | en | How does a lead-acid battery work? | [
"Pb oxidizes, PbO2 reduces.",
"Sulfuric acid is consumed.",
"Charging reverses the chemistry."
] | chemistry | Answer: Pb/PbO2 in sulfuric acid.
Steps:
1. Pb oxidizes, PbO2 reduces.
2. Sulfuric acid is consumed.
3. Charging reverses the chemistry. | electrochemistry |
Náboj je 600 C, protože 1 A krát 600 s se rovná 600 C. | null | che_0293 | worked_example | cs | Jaký náboj proteče při 1 A za 10 minut? | [
"Převeď čas: 10 min = 600 s.",
"Q = I x t = 1 x 600.",
"Q = 600 C."
] | chemistry | Answer: 600 C.
Steps:
1. Převeď čas: 10 min = 600 s.
2. Q = I x t = 1 x 600.
3. Q = 600 C. | electrochemistry |
There are 0.200 mol of electrons, from 19,300 C divided by 96,485 C/mol. | null | che_0294 | worked_example | en | How many moles of electrons are in 19,300 C? | [
"Write: n = Q / F.",
"Substitute: 19,300 / 96,485.",
"n = 0.200 mol."
] | chemistry | Answer: 0.200 mol.
Steps:
1. Write: n = Q / F.
2. Substitute: 19,300 / 96,485.
3. n = 0.200 mol. | electrochemistry |
The plated object is the cathode, where metal ions are reduced. | {
"A": "Anode",
"B": "Cathode",
"C": "Salt bridge",
"D": "Electrolyte"
} | che_0295 | multiple_choice | en | In electroplating, what is the object to be plated? | [
"Metal ions must reduce onto it.",
"Reduction happens at cathode.",
"Select option B."
] | chemistry | Answer: Cathode.
Steps:
1. Metal ions must reduce onto it.
2. Reduction happens at cathode.
3. Select option B.
Answer: (B) | electrochemistry |
A sacrificial anode of reactive metal corrodes in place of the protected structure. | null | che_0296 | concept | en | What is a sacrificial anode? | [
"Attach reactive metal.",
"It oxidizes preferentially.",
"Structure stays intact."
] | chemistry | Answer: Metal corroding instead of structure.
Steps:
1. Attach reactive metal.
2. It oxidizes preferentially.
3. Structure stays intact. | electrochemistry |
The plated silver mass is 2.01 g, from 1800 C giving 0.01866 mol Ag. | null | che_0297 | worked_example | en | What mass of Ag plates at 1.00 A for 30.0 min (M = 107.87 g/mol)? | [
"Q = 1.00 x 1800 = 1800 C.",
"Mol e- = 1800/96485 = 0.01866 mol Ag.",
"Mass = 0.01866 x 107.87 = 2.01 g."
] | chemistry | Answer: 2.01 g.
Steps:
1. Q = 1.00 x 1800 = 1800 C.
2. Mol e- = 1800/96485 = 0.01866 mol Ag.
3. Mass = 0.01866 x 107.87 = 2.01 g. | electrochemistry |
Koroze je samovolná oxidace kovů; brání jí nátěry, pozinkování a katodická ochrana. | null | che_0298 | concept | cs | Co je koroze a jak jí bránit? | [
"Oxidace kovu vzduchem a vodou.",
"Izoluj povrch nátěrem.",
"Obětovaná anoda chrání katodicky."
] | chemistry | Answer: Oxidace kovů; ochrana izolací a anodou.
Steps:
1. Oxidace kovu vzduchem a vodou.
2. Izoluj povrch nátěrem.
3. Obětovaná anoda chrání katodicky. | electrochemistry |
Iron rusts only when both oxygen and water are present. | {
"A": "Oxygen alone",
"B": "Water alone",
"C": "Neither",
"D": "Both oxygen and water"
} | che_0299 | multiple_choice | en | What does iron need to rust? | [
"Dry air gives no rust.",
"Air-free water gives no rust.",
"Select option D."
] | chemistry | Answer: Both oxygen and water.
Steps:
1. Dry air gives no rust.
2. Air-free water gives no rust.
3. Select option D.
Answer: (D) | electrochemistry |
The cell potential is 1.10 V, from 212,000 J divided by (2 x 96,485 C). | null | che_0300 | worked_example | en | A cell has ΔG = -212 kJ and n = 2. Compute E° (F = 96485 C/mol). | [
"Write: E° = -ΔG/(nF).",
"Substitute: 212,000/(2 x 96,485).",
"E° = 1.10 V."
] | chemistry | Answer: 1.10 V.
Steps:
1. Write: E° = -ΔG/(nF).
2. Substitute: 212,000/(2 x 96,485).
3. E° = 1.10 V. | electrochemistry |
The law of definite proportions states a compound always has fixed element mass ratios. | null | che_0301 | concept | en | What is the law of definite proportions? | [
"Compounds have fixed formulas.",
"Mass ratios stay constant.",
"Example: water always 11.2% H."
] | chemistry | Answer: Fixed mass ratios in compounds.
Steps:
1. Compounds have fixed formulas.
2. Mass ratios stay constant.
3. Example: water always 11.2% H. | stoichiometry |
Two moles contain 1.204 x 10^24 molecules, double Avogadro's number. | {
"A": "6.022 x 10^23",
"B": "3.011 x 10^23",
"C": "1.204 x 10^24",
"D": "2.408 x 10^24"
} | che_0302 | multiple_choice | en | How many molecules are in 2.00 mol of O2? | [
"Multiply: 2.00 x 6.022 x 10^23.",
"Result 1.204 x 10^24.",
"Select option C."
] | chemistry | Answer: 1.204 x 10^24.
Steps:
1. Multiply: 2.00 x 6.022 x 10^23.
2. Result 1.204 x 10^24.
3. Select option C.
Answer: (C) | stoichiometry |
There are 1.506 x 10^23 particles, from 0.250 times Avogadro's number. | null | che_0303 | worked_example | en | How many particles are in 0.250 mol of salt? | [
"Write: N = 0.250 x 6.022 x 10^23.",
"Multiply: 1.506 x 10^23.",
"N = 1.506 x 10^23."
] | chemistry | Answer: 1.506 x 10^23.
Steps:
1. Write: N = 0.250 x 6.022 x 10^23.
2. Multiply: 1.506 x 10^23.
3. N = 1.506 x 10^23. | stoichiometry |
Avogadrova konstanta udává počet částic v jednom molu: 6,022 x 10^23. | null | che_0304 | concept | cs | Co je Avogadrova konstanta? | [
"Počet částic na mol.",
"Hodnota 6,022 x 10^23.",
"Spojí mikro a makro svět."
] | chemistry | Answer: 6,022 x 10^23 částic/mol.
Steps:
1. Počet částic na mol.
2. Hodnota 6,022 x 10^23.
3. Spojí mikro a makro svět. | stoichiometry |
The mass is 72.92 g, from 3.00 mol times 24.305 g/mol. | null | che_0305 | worked_example | en | What mass is 3.00 mol of magnesium (M = 24.305 g/mol)? | [
"Write: m = 3.00 x 24.305.",
"Multiply.",
"m = 72.92 g."
] | chemistry | Answer: 72.92 g.
Steps:
1. Write: m = 3.00 x 24.305.
2. Multiply.
3. m = 72.92 g. | stoichiometry |
4 g of H2 is about 2.0 mol, more than any other choice at 1.0 mol. | {
"A": "18 g H2O",
"B": "44 g CO2",
"C": "4 g H2",
"D": "32 g O2"
} | che_0306 | multiple_choice | en | Which sample contains the most moles? | [
"A: 18/18 = 1.0 mol.",
"B: 44/44 = 1.0; D: 32/32 = 1.0.",
"C: 4/2 = 2.0; select C."
] | chemistry | Answer: 4 g H2.
Steps:
1. A: 18/18 = 1.0 mol.
2. B: 44/44 = 1.0; D: 32/32 = 1.0.
3. C: 4/2 = 2.0; select C.
Answer: (C) | stoichiometry |
Theoretical yield is the stoichiometric maximum while actual yield is what is really obtained. | null | che_0307 | concept | en | How do theoretical and actual yield differ? | [
"Theoretical from balanced equation.",
"Actual measured in lab.",
"Ratio gives percent yield."
] | chemistry | Answer: Calculated maximum versus measured amount.
Steps:
1. Theoretical from balanced equation.
2. Actual measured in lab.
3. Ratio gives percent yield. | stoichiometry |
Percent yield is 80%, from 64 g divided by 80 g times 100. | null | che_0308 | worked_example | en | Theoretical yield is 80 g and actual is 64 g. Compute percent yield. | [
"Divide: 64/80 = 0.80.",
"Multiply by 100.",
"Yield = 80%."
] | chemistry | Answer: 80%.
Steps:
1. Divide: 64/80 = 0.80.
2. Multiply by 100.
3. Yield = 80%. | stoichiometry |
Výtěžek je 85 %, protože 85 děleno 100 krát 100 se rovná 85. | null | che_0309 | worked_example | cs | Teoretický výtěžek je 100 g a skutečný 85 g. Vypočítej procentuální výtěžek. | [
"Vyděl: 85/100 = 0,85.",
"Vynásob stem.",
"Výtěžek = 85 %."
] | chemistry | Answer: 85 %.
Steps:
1. Vyděl: 85/100 = 0,85.
2. Vynásob stem.
3. Výtěžek = 85 %. | stoichiometry |
The balanced equation is 4Al + 3O2 -> 2Al2O3. | {
"A": "2, 3, 2",
"B": "4, 3, 2",
"C": "2, 1, 2",
"D": "4, 1, 2"
} | che_0310 | multiple_choice | en | What coefficients balance Al + O2 -> Al2O3? | [
"Al: 4 left, 2x2 right.",
"O: 3x2 left, 2x3 right.",
"Select option B."
] | chemistry | Answer: 4, 3, 2.
Steps:
1. Al: 4 left, 2x2 right.
2. O: 3x2 left, 2x3 right.
3. Select option B.
Answer: (B) | stoichiometry |
Coefficients give the mole ratios for converting between reactants and products. | null | che_0311 | concept | en | What is a mole ratio? | [
"Read coefficients.",
"Form ratio of species.",
"Convert moles with it."
] | chemistry | Answer: Coefficient ratio for conversion.
Steps:
1. Read coefficients.
2. Form ratio of species.
3. Convert moles with it. | stoichiometry |
Four moles of NH3 form, from 6 mol H2 times the 2/3 ratio. | null | che_0312 | worked_example | en | How many moles of NH3 form from 6.0 mol H2 (N2 + 3H2 -> 2NH3)? | [
"Ratio NH3:H2 = 2:3.",
"Multiply: 6.0 x 2/3.",
"NH3 = 4.0 mol."
] | chemistry | Answer: 4.0 mol.
Steps:
1. Ratio NH3:H2 = 2:3.
2. Multiply: 6.0 x 2/3.
3. NH3 = 4.0 mol. | stoichiometry |
The new molarity is 0.40 M, from 2.0 x 100 / 500. | {
"A": "0.40 M",
"B": "1.00 M",
"C": "2.50 M",
"D": "10.0 M"
} | che_0313 | multiple_choice | en | 100 mL of 2.0 M solution dilutes to 500 mL. What is the new M? | [
"Use M1V1 = M2V2.",
"M2 = 2.0 x 100/500.",
"Select option A."
] | chemistry | Answer: 0.40 M.
Steps:
1. Use M1V1 = M2V2.
2. M2 = 2.0 x 100/500.
3. Select option A.
Answer: (A) | stoichiometry |
Ředění snižuje koncentraci přidáním rozpouštědla při stálém množství látky. | null | che_0314 | concept | cs | Co je ředění roztoků? | [
"Moly zůstávají stejné.",
"Objem roste.",
"Platí M1V1 = M2V2."
] | chemistry | Answer: Snížení koncentrace rozpouštědlem.
Steps:
1. Moly zůstávají stejné.
2. Objem roste.
3. Platí M1V1 = M2V2. | stoichiometry |
The needed mass is 14.61 g, from 0.250 mol times 58.44 g/mol. | null | che_0315 | worked_example | en | What mass of NaCl makes 500 mL of 0.500 M solution? | [
"Moles = 0.500 x 0.500 = 0.250.",
"Multiply: 0.250 x 58.44.",
"Mass = 14.61 g."
] | chemistry | Answer: 14.61 g.
Steps:
1. Moles = 0.500 x 0.500 = 0.250.
2. Multiply: 0.250 x 58.44.
3. Mass = 14.61 g. | stoichiometry |
One mole of ideal gas occupies 22.414 L at STP. | {
"A": "1.00 L",
"B": "11.2 L",
"C": "24.5 L",
"D": "22.414 L"
} | che_0316 | multiple_choice | en | What is the molar volume of gas at STP? | [
"Recall STP value.",
"22.414 L per mole.",
"Select option D."
] | chemistry | Answer: 22.414 L.
Steps:
1. Recall STP value.
2. 22.414 L per mole.
3. Select option D.
Answer: (D) | stoichiometry |
The ideal gas law PV = nRT links pressure, volume, moles, and temperature. | null | che_0317 | concept | en | What is the ideal gas law? | [
"Write PV = nRT.",
"R = 8.314 J/mol K.",
"Assume point particles."
] | chemistry | Answer: PV = nRT.
Steps:
1. Write PV = nRT.
2. R = 8.314 J/mol K.
3. Assume point particles. | stoichiometry |
Je to 2,0 molu, protože 44,8 děleno 22,414 se rovná 2,0. | null | che_0318 | worked_example | cs | Kolik molů je ve 44,8 l plynu za STP? | [
"Vyděl molárním objemem.",
"44,8 / 22,414.",
"n = 2,0 mol."
] | chemistry | Answer: 2,0 mol.
Steps:
1. Vyděl molárním objemem.
2. 44,8 / 22,414.
3. n = 2,0 mol. | stoichiometry |
The multiplier is 6, from 180 divided by 30. | {
"A": "3",
"B": "6",
"C": "2",
"D": "12"
} | che_0319 | multiple_choice | en | Empirical mass is 30 and molecular mass is 180. What is the multiplier? | [
"Divide: 180/30.",
"Result = 6.",
"Select option B."
] | chemistry | Answer: 6.
Steps:
1. Divide: 180/30.
2. Result = 6.
3. Select option B.
Answer: (B) | stoichiometry |
Water autoionizes into H+ and OH- with Kw = 1.0 x 10^-14 at 25 C. | null | che_0320 | concept | en | What is the autoionization of water? | [
"H2O gives H+ and OH-.",
"Kw = [H+][OH-].",
"Kw = 1.0 x 10^-14 at 25 C."
] | chemistry | Answer: Water splits with Kw = 1e-14.
Steps:
1. H2O gives H+ and OH-.
2. Kw = [H+][OH-].
3. Kw = 1.0 x 10^-14 at 25 C. | acids_bases |
Pure water has [H+] of 1.0 x 10^-7 M, giving pH 7. | {
"A": "1.0 x 10^-7 M",
"B": "1.0 x 10^-14 M",
"C": "1.0 M",
"D": "0 M"
} | che_0321 | multiple_choice | en | What is [H+] in pure water at 25 C? | [
"Square root of Kw.",
"sqrt(1e-14) = 1e-7.",
"Select option A."
] | chemistry | Answer: 1.0 x 10^-7 M.
Steps:
1. Square root of Kw.
2. sqrt(1e-14) = 1e-7.
3. Select option A.
Answer: (A) | acids_bases |
The pH is 2.30, from -log(0.0050). | null | che_0322 | worked_example | en | Compute the pH of 0.0050 M HCl. | [
"[H+] = 0.0050 M.",
"pH = -log(0.0050).",
"pH = 2.30."
] | chemistry | Answer: 2.30.
Steps:
1. [H+] = 0.0050 M.
2. pH = -log(0.0050).
3. pH = 2.30. | acids_bases |
Autoprotolýza vody je její částečné štěpení na H+ a OH- s Kw = 1 x 10^-14. | null | che_0323 | concept | cs | Co je autoprotolýza vody? | [
"Voda se štěpí na ionty.",
"Kw = [H+][OH-].",
"Při 25 °C je 1 x 10^-14."
] | chemistry | Answer: Štěpení vody na ionty.
Steps:
1. Voda se štěpí na ionty.
2. Kw = [H+][OH-].
3. Při 25 °C je 1 x 10^-14. | acids_bases |
Removing a proton from HCO3- gives carbonate, CO3(2-). | {
"A": "H2CO3",
"B": "HCO3(-)",
"C": "CO3(2-)",
"D": "OH(-)"
} | che_0324 | multiple_choice | en | What is the conjugate base of HCO3-? | [
"Remove one H+.",
"Charge drops by one.",
"Select option C."
] | chemistry | Answer: CO3(2-).
Steps:
1. Remove one H+.
2. Charge drops by one.
3. Select option C.
Answer: (C) | acids_bases |
The pOH is 1.70, from -log(0.020). | null | che_0325 | worked_example | en | Compute the pOH of 0.020 M NaOH. | [
"[OH-] = 0.020 M.",
"pOH = -log(0.020).",
"pOH = 1.70."
] | chemistry | Answer: 1.70.
Steps:
1. [OH-] = 0.020 M.
2. pOH = -log(0.020).
3. pOH = 1.70. | acids_bases |
Ammonia accepts a proton from water, forming NH4+ and OH- with Kb. | null | che_0326 | concept | en | How does ammonia act as a weak base? | [
"NH3 takes H+ from water.",
"NH4+ and OH- form.",
"Equilibrium has small Kb."
] | chemistry | Answer: Takes H+ to make NH4+ and OH-.
Steps:
1. NH3 takes H+ from water.
2. NH4+ and OH- form.
3. Equilibrium has small Kb. | acids_bases |
The pH is 11.13, from pOH 2.87 via [OH-] = 1.34 x 10^-3 M. | null | che_0327 | worked_example | en | Compute the pH of 0.10 M NH3 (Kb = 1.8 x 10^-5). | [
"[OH-] = sqrt(1.8 x 10^-5 x 0.10) = 1.34 x 10^-3.",
"pOH = 2.87.",
"pH = 14 - 2.87 = 11.13."
] | chemistry | Answer: 11.13.
Steps:
1. [OH-] = sqrt(1.8 x 10^-5 x 0.10) = 1.34 x 10^-3.
2. pOH = 2.87.
3. pH = 14 - 2.87 = 11.13. | acids_bases |
Ammonium hydrolyzes to release H+, making the solution acidic. | {
"A": "Basic",
"B": "Acidic",
"C": "Neutral",
"D": "Cannot tell"
} | che_0328 | multiple_choice | en | A 0.10 M NH4Cl solution is what? | [
"NH4+ is acid of weak base.",
"It donates H+.",
"Select option B."
] | chemistry | Answer: Acidic.
Steps:
1. NH4+ is acid of weak base.
2. It donates H+.
3. Select option B.
Answer: (B) | acids_bases |
pH je 2, protože -log(0,01) = 2. | null | che_0329 | worked_example | cs | Vypočítej pH roztoku 0,01 M HNO3. | [
"HNO3 disociuje plně.",
"pH = -log(0,01).",
"pH = 2."
] | chemistry | Answer: 2.
Steps:
1. HNO3 disociuje plně.
2. pH = -log(0,01).
3. pH = 2. | acids_bases |
Polyprotic acids lose protons in steps, each with its own smaller Ka. | null | che_0330 | concept | en | How do polyprotic acids dissociate? | [
"First proton easiest.",
"Each step has Ka.",
"Ka1 > Ka2 > Ka3."
] | chemistry | Answer: Stepwise with falling Ka.
Steps:
1. First proton easiest.
2. Each step has Ka.
3. Ka1 > Ka2 > Ka3. | acids_bases |
The first proton of sulfuric acid dissociates fully and strongly. | {
"A": "Strong",
"B": "Weak",
"C": "Not acidic",
"D": "Amphiprotic"
} | che_0331 | multiple_choice | en | How does the first proton of H2SO4 behave? | [
"First step fully dissociates.",
"Only second is weak.",
"Select option A."
] | chemistry | Answer: Strong.
Steps:
1. First step fully dissociates.
2. Only second is weak.
3. Select option A.
Answer: (A) | acids_bases |
Exactly 20.0 mL of NaOH is needed, a one-to-one mole match. | null | che_0332 | worked_example | en | What volume of 0.100 M NaOH neutralizes 20.0 mL of 0.100 M HCl? | [
"Moles HCl = 0.0200 x 0.100 = 0.00200.",
"Same moles NaOH needed.",
"V = 0.00200/0.100 = 20.0 mL."
] | chemistry | Answer: 20.0 mL.
Steps:
1. Moles HCl = 0.0200 x 0.100 = 0.00200.
2. Same moles NaOH needed.
3. V = 0.00200/0.100 = 20.0 mL. | acids_bases |
Indicators change color across a narrow pH range near their pKa. | null | che_0333 | concept | en | How do acid-base indicators work? | [
"Weak acid with colored forms.",
"Color flips near pKa.",
"Range spans ~2 pH units."
] | chemistry | Answer: Color change near pKa.
Steps:
1. Weak acid with colored forms.
2. Color flips near pKa.
3. Range spans ~2 pH units. | acids_bases |
Titrace je stanovení koncentrace postupným přidáváním činidla o známé koncentraci. | null | che_0334 | concept | cs | Co je titrace? | [
"Byreta dávkuje titrační činidlo.",
"Indikátor ukáže konec.",
"Výpočet z ekvivalence."
] | chemistry | Answer: Stanovení koncentrace titrací.
Steps:
1. Byreta dávkuje titrační činidlo.
2. Indikátor ukáže konec.
3. Výpočet z ekvivalence. | acids_bases |
Methyl orange is red below pH 3 and yellow above pH 4.4. | {
"A": "Yellow in acid, red in base",
"B": "Blue in both",
"C": "Colorless in both",
"D": "Red in acid, yellow in base"
} | che_0335 | multiple_choice | en | How does methyl orange change color? | [
"Acid form is red.",
"Base form is yellow.",
"Select option D."
] | chemistry | Answer: Red in acid, yellow in base.
Steps:
1. Acid form is red.
2. Base form is yellow.
3. Select option D.
Answer: (D) | acids_bases |
The pH is 2.70, from -log(2.0 x 10^-3). | null | che_0336 | worked_example | en | A solution has [H+] = 2.0 x 10^-3 M. Compute pH. | [
"Write: pH = -log[H+].",
"Substitute: -log(2.0 x 10^-3).",
"pH = 2.70."
] | chemistry | Answer: 2.70.
Steps:
1. Write: pH = -log[H+].
2. Substitute: -log(2.0 x 10^-3).
3. pH = 2.70. | acids_bases |
pOH je 1, protože -log(0,1) = 1. | null | che_0337 | worked_example | cs | Vypočítej pOH roztoku 0,1 M KOH. | [
"KOH disociuje plně.",
"pOH = -log(0,1).",
"pOH = 1."
] | chemistry | Answer: 1.
Steps:
1. KOH disociuje plně.
2. pOH = -log(0,1).
3. pOH = 1. | acids_bases |
Buffers resist pH change by neutralizing small added acid or base. | {
"A": "Fix pH exactly forever",
"B": "Resist pH change",
"C": "Always give pH 7",
"D": "Destroy all ions"
} | che_0338 | multiple_choice | en | What does a buffer do? | [
"Conjugate pair absorbs H+/OH-.",
"pH shifts only slightly.",
"Select option B."
] | chemistry | Answer: Resist pH change.
Steps:
1. Conjugate pair absorbs H+/OH-.
2. pH shifts only slightly.
3. Select option B.
Answer: (B) | acids_bases |
Oxygen is -2 except in peroxides (-1) and with fluorine (positive). | null | che_0339 | concept | en | What are the exceptions to oxygen's oxidation state? | [
"Normal rule: O is -2.",
"Peroxides: O is -1.",
"With F: O is positive."
] | chemistry | Answer: Peroxides -1, with F positive.
Steps:
1. Normal rule: O is -2.
2. Peroxides: O is -1.
3. With F: O is positive. | redox |
Each oxygen in hydrogen peroxide is -1, since 2(+1) + 2O = 0. | {
"A": "-2",
"B": "-1",
"C": "0",
"D": "+1"
} | che_0340 | multiple_choice | en | What is oxygen's oxidation state in H2O2? | [
"H is +1 each.",
"Solve: 2 + 2O = 0.",
"O = -1; select B."
] | chemistry | Answer: -1.
Steps:
1. H is +1 each.
2. Solve: 2 + 2O = 0.
3. O = -1; select B.
Answer: (B) | redox |
Phosphorus is +5, since 3(+1) + P + 4(-2) = 0. | null | che_0341 | worked_example | en | Find the oxidation number of P in H3PO4. | [
"Assign H +1 and O -2.",
"Equation: 3 + P - 8 = 0.",
"P = +5."
] | chemistry | Answer: +5.
Steps:
1. Assign H +1 and O -2.
2. Equation: 3 + P - 8 = 0.
3. P = +5. | redox |
Disproporcionace je reakce, kde se jeden prvek současně oxiduje i redukuje. | null | che_0342 | concept | cs | Co je disproporcionace? | [
"Jeden prvek, dva osudy.",
"Část se oxiduje, část redukuje.",
"Příklad: chlor ve vodě."
] | chemistry | Answer: Současná oxidace i redukce.
Steps:
1. Jeden prvek, dva osudy.
2. Část se oxiduje, část redukuje.
3. Příklad: chlor ve vodě. | redox |
Copper ions gain electrons to form copper metal, so Cu2+ is reduced. | {
"A": "Zn",
"B": "SO4(2-)",
"C": "Cu2+",
"D": "H+"
} | che_0343 | multiple_choice | en | In Zn + CuSO4 -> ZnSO4 + Cu, what is reduced? | [
"Cu goes +2 to 0.",
"That is a gain of electrons.",
"Select option C."
] | chemistry | Answer: Cu2+.
Steps:
1. Cu goes +2 to 0.
2. That is a gain of electrons.
3. Select option C.
Answer: (C) | redox |
The balanced equation is 2K + Br2 -> 2KBr. | null | che_0344 | worked_example | en | Balance: K + Br2 -> KBr. | [
"Balance Br with 2 KBr.",
"Balance K with 2 K.",
"Result: 2K + Br2 -> 2KBr."
] | chemistry | Answer: 2K + Br2 -> 2KBr.
Steps:
1. Balance Br with 2 KBr.
2. Balance K with 2 K.
3. Result: 2K + Br2 -> 2KBr. | redox |
Half-reactions are scaled to equal electrons, then added with electrons cancelled. | null | che_0345 | concept | en | How are half-reactions combined? | [
"Balance each half.",
"Scale to equal electrons.",
"Add and cancel electrons."
] | chemistry | Answer: Scale, add, cancel electrons.
Steps:
1. Balance each half.
2. Scale to equal electrons.
3. Add and cancel electrons. | redox |
The net ionic equation is Zn + Cu2+ -> Zn2+ + Cu. | null | che_0346 | worked_example | en | Write the net ionic equation for Zn + CuSO4 -> ZnSO4 + Cu. | [
"Split strong electrolytes to ions.",
"Cancel spectator SO4(2-).",
"Net: Zn + Cu2+ -> Zn2+ + Cu."
] | chemistry | Answer: Zn + Cu2+ -> Zn2+ + Cu.
Steps:
1. Split strong electrolytes to ions.
2. Cancel spectator SO4(2-).
3. Net: Zn + Cu2+ -> Zn2+ + Cu. | redox |
Oxygen gains electrons to form MgO, so it is the oxidizing agent. | {
"A": "O2",
"B": "Mg",
"C": "MgO",
"D": "Heat"
} | che_0347 | multiple_choice | en | In 2Mg + O2 -> 2MgO, what is the oxidizing agent? | [
"Oxygen is reduced.",
"Reduced species is the oxidizer.",
"Select option A."
] | chemistry | Answer: O2.
Steps:
1. Oxygen is reduced.
2. Reduced species is the oxidizer.
3. Select option A.
Answer: (A) | redox |
Dusík má oxidační číslo -3, protože N + 3(+1) = 0. | null | che_0348 | worked_example | cs | Urči oxidační číslo dusíku v NH3. | [
"Přiřaď H +1.",
"Rovnice: N + 3 = 0.",
"N = -3."
] | chemistry | Answer: -3.
Steps:
1. Přiřaď H +1.
2. Rovnice: N + 3 = 0.
3. N = -3. | redox |
Batteries split redox into half-cells so electron flow does electrical work. | null | che_0349 | concept | en | How is redox used in batteries? | [
"Oxidation at one electrode.",
"Reduction at the other.",
"Electrons flow via wire."
] | chemistry | Answer: Split halves drive current.
Steps:
1. Oxidation at one electrode.
2. Reduction at the other.
3. Electrons flow via wire. | redox |
Rusting oxidizes iron metal to iron oxides. | {
"A": "Reduction of Fe",
"B": "Sublimation",
"C": "Neutralization",
"D": "Oxidation of Fe"
} | che_0350 | multiple_choice | en | What chemical change is rusting? | [
"Fe goes 0 to +3.",
"That is oxidation.",
"Select option D."
] | chemistry | Answer: Oxidation of Fe.
Steps:
1. Fe goes 0 to +3.
2. That is oxidation.
3. Select option D.
Answer: (D) | redox |
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