answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
Glucose has molar mass 180.16 g/mol, from 6 C, 12 H, and 6 O atoms. | null | che_0151 | concept | en | How is molar mass found from a formula? | [
"Multiply each atomic mass by its count.",
"Sum all contributions.",
"Example: glucose totals 180.16 g/mol."
] | chemistry | Answer: Sum of atomic masses times counts.
Steps:
1. Multiply each atomic mass by its count.
2. Sum all contributions.
3. Example: glucose totals 180.16 g/mol. | stoichiometry |
There are 0.50 mol, from 90 g divided by 180.16 g/mol. | {
"A": "0.25 mol",
"B": "0.50 mol",
"C": "1.00 mol",
"D": "2.00 mol"
} | che_0152 | multiple_choice | en | How many moles are in 90 g of glucose (M = 180.16 g/mol)? | [
"Write: n = 90 / 180.16.",
"Compute: 0.50 mol.",
"Select option B."
] | chemistry | Answer: 0.50 mol.
Steps:
1. Write: n = 90 / 180.16.
2. Compute: 0.50 mol.
3. Select option B.
Answer: (B) | stoichiometry |
One mole of methane yields 44.01 g of CO2, a one-to-one mole ratio. | null | che_0153 | worked_example | en | What mass of CO2 forms from burning 1.00 mol of CH4? | [
"Balanced: 1 CH4 gives 1 CO2.",
"Moles CO2 = 1.00 mol.",
"Mass = 1.00 x 44.01 = 44.01 g."
] | chemistry | Answer: 44.01 g.
Steps:
1. Balanced: 1 CH4 gives 1 CO2.
2. Moles CO2 = 1.00 mol.
3. Mass = 1.00 x 44.01 = 44.01 g. | stoichiometry |
Stechiometrie počítá látková množství reaktantů a produktů z vyčíslených rovnic. | null | che_0154 | concept | cs | Co je stechiometrie? | [
"Vyčísli rovnici.",
"Převeď na moly.",
"Použij poměry koeficientů."
] | chemistry | Answer: Výpočty z chemických rovnic.
Steps:
1. Vyčísli rovnici.
2. Převeď na moly.
3. Použij poměry koeficientů. | stoichiometry |
Hydrogen limits because 2 mol H2 can only pair with 0.67 mol N2, leaving N2 in excess. | {
"A": "H2 is limiting",
"B": "N2 is limiting",
"C": "Both limit equally",
"D": "Cannot tell"
} | che_0155 | multiple_choice | en | For N2 + 3H2 -> 2NH3 with 1 mol N2 and 2 mol H2, which limits? | [
"Need 3 H2 per N2.",
"Have only 2 H2 for 1 N2.",
"Select option A."
] | chemistry | Answer: H2 is limiting.
Steps:
1. Need 3 H2 per N2.
2. Have only 2 H2 for 1 N2.
3. Select option A.
Answer: (A) | stoichiometry |
The theoretical yield is 36.03 g of water, from 2.00 mol times 18.015 g/mol. | null | che_0156 | worked_example | en | What mass of H2O forms from 2.00 mol H2 with excess O2? | [
"Ratio H2:H2O is 1:1.",
"Moles water = 2.00 mol.",
"Mass = 2.00 x 18.015 = 36.03 g."
] | chemistry | Answer: 36.03 g.
Steps:
1. Ratio H2:H2O is 1:1.
2. Moles water = 2.00 mol.
3. Mass = 2.00 x 18.015 = 36.03 g. | stoichiometry |
Percent composition gives each element's mass share; water is 11.2% H and 88.8% O. | null | che_0157 | concept | en | What is percent composition? | [
"Divide element mass by molar mass.",
"Multiply by 100.",
"Example: water 11.2% H, 88.8% O."
] | chemistry | Answer: Mass percent of each element.
Steps:
1. Divide element mass by molar mass.
2. Multiply by 100.
3. Example: water 11.2% H, 88.8% O. | stoichiometry |
Oxygen is 72.7% of CO2, from 32.00 divided by 44.01 times 100. | null | che_0158 | worked_example | en | Compute the mass percent of oxygen in CO2. | [
"O mass = 2 x 16.00 = 32.00.",
"Divide: 32.00 / 44.01 = 0.727.",
"Percent = 72.7%."
] | chemistry | Answer: 72.7%.
Steps:
1. O mass = 2 x 16.00 = 32.00.
2. Divide: 32.00 / 44.01 = 0.727.
3. Percent = 72.7%. | stoichiometry |
The ratio C:H:O is 1:2:1, giving the empirical formula CH2O. | {
"A": "CHO",
"B": "C2H4O",
"C": "CH2O",
"D": "CO2"
} | che_0159 | multiple_choice | en | A compound is 40.0% C, 6.7% H, 53.3% O. What is its empirical formula? | [
"Moles: C 3.33, H 6.65, O 3.33.",
"Divide by 3.33: 1, 2, 1.",
"Select option C."
] | chemistry | Answer: CH2O.
Steps:
1. Moles: C 3.33, H 6.65, O 3.33.
2. Divide by 3.33: 1, 2, 1.
3. Select option C.
Answer: (C) | stoichiometry |
Hmotnost je 20 g, protože 0,5 krát 40 se rovná 20. | null | che_0160 | worked_example | cs | Jaká je hmotnost 0,5 molu NaOH (M = 40 g/mol)? | [
"Napiš vzorec: m = n x M.",
"Dosaď: 0,5 x 40.",
"m = 20 g."
] | chemistry | Answer: 20 g.
Steps:
1. Napiš vzorec: m = n x M.
2. Dosaď: 0,5 x 40.
3. m = 20 g. | stoichiometry |
A hydrate like CuSO4.5H2O holds a fixed number of water molecules per formula unit. | null | che_0161 | concept | en | What is a hydrate? | [
"Define trapped water in crystals.",
"Dot shows fixed ratio.",
"Heating drives water off."
] | chemistry | Answer: Salt with fixed crystal water.
Steps:
1. Define trapped water in crystals.
2. Dot shows fixed ratio.
3. Heating drives water off. | stoichiometry |
The solution is 1.00 M, from 58.44 g (1.00 mol) in 1.00 L. | {
"A": "0.10 M",
"B": "0.50 M",
"C": "2.00 M",
"D": "1.00 M"
} | che_0162 | multiple_choice | en | 58.44 g of NaCl dissolves in 1.00 L. What is the molarity? | [
"Moles = 58.44/58.44 = 1.00.",
"Divide by 1.00 L.",
"Select option D."
] | chemistry | Answer: 1.00 M.
Steps:
1. Moles = 58.44/58.44 = 1.00.
2. Divide by 1.00 L.
3. Select option D.
Answer: (D) | stoichiometry |
There are 0.0500 mol, from 0.250 L times 0.200 M. | null | che_0163 | worked_example | en | How many moles are in 250 mL of 0.200 M solution? | [
"Convert: 250 mL = 0.250 L.",
"Multiply: 0.250 x 0.200.",
"n = 0.0500 mol."
] | chemistry | Answer: 0.0500 mol.
Steps:
1. Convert: 250 mL = 0.250 L.
2. Multiply: 0.250 x 0.200.
3. n = 0.0500 mol. | stoichiometry |
Dilution adds solvent while moles stay fixed, so M1V1 equals M2V2. | null | che_0164 | concept | en | What principle governs dilution? | [
"Moles before equal moles after.",
"Write M1V1 = M2V2.",
"Volume rises, M falls."
] | chemistry | Answer: Moles stay fixed; M1V1 = M2V2.
Steps:
1. Moles before equal moles after.
2. Write M1V1 = M2V2.
3. Volume rises, M falls. | stoichiometry |
Molární koncentrace je počet molů látky v jednom litru roztoku (mol/l). | null | che_0165 | concept | cs | Co je molární koncentrace? | [
"Moly děleno litry.",
"Jednotka mol/l (M).",
"Příklad: 1 mol v 1 l = 1 M."
] | chemistry | Answer: Moly na litr roztoku.
Steps:
1. Moly děleno litry.
2. Jednotka mol/l (M).
3. Příklad: 1 mol v 1 l = 1 M. | stoichiometry |
Nitrogen density is 1.25 g/L, from 28.02 g divided by 22.414 L. | {
"A": "0.63 g/L",
"B": "1.25 g/L",
"C": "2.50 g/L",
"D": "14.0 g/L"
} | che_0166 | multiple_choice | en | What is the density of N2 gas at STP? | [
"Mass of 1 mol = 28.02 g.",
"Volume = 22.414 L.",
"Density 1.25; select B."
] | chemistry | Answer: 1.25 g/L.
Steps:
1. Mass of 1 mol = 28.02 g.
2. Volume = 22.414 L.
3. Density 1.25; select B.
Answer: (B) | stoichiometry |
Oxygen density is 1.43 g/L, from 32.00 g divided by 22.414 L. | null | che_0167 | worked_example | en | Compute the density of O2 gas at STP. | [
"Mass of 1 mol = 32.00 g.",
"Divide: 32.00 / 22.414.",
"Density = 1.43 g/L."
] | chemistry | Answer: 1.43 g/L.
Steps:
1. Mass of 1 mol = 32.00 g.
2. Divide: 32.00 / 22.414.
3. Density = 1.43 g/L. | stoichiometry |
Objem je 11,2 l, protože 0,5 krát 22,414 se rovná 11,207. | null | che_0168 | worked_example | cs | Jaký objem má 0,5 molu plynu za STP (22,414 l/mol)? | [
"Napiš vzorec: V = n x 22,414.",
"Dosaď: 0,5 x 22,414.",
"V = 11,2 l."
] | chemistry | Answer: 11,2 l.
Steps:
1. Napiš vzorec: V = n x 22,414.
2. Dosaď: 0,5 x 22,414.
3. V = 11,2 l. | stoichiometry |
The excess reactant is left over after the limiting one is consumed. | {
"A": "Limits the yield",
"B": "Is fully consumed",
"C": "Remains after reaction",
"D": "Never participates"
} | che_0169 | multiple_choice | en | What is true of the excess reactant? | [
"Limiting reactant sets yield.",
"Excess is more than needed.",
"Select option C."
] | chemistry | Answer: Remains after reaction.
Steps:
1. Limiting reactant sets yield.
2. Excess is more than needed.
3. Select option C.
Answer: (C) | stoichiometry |
Strong acids dissociate fully while weak acids only partly ionize in water. | null | che_0170 | concept | en | How do strong and weak acids differ? | [
"Strong: ~100% dissociation.",
"Weak: equilibrium with Ka.",
"Example: HCl versus acetic acid."
] | chemistry | Answer: Full versus partial dissociation.
Steps:
1. Strong: ~100% dissociation.
2. Weak: equilibrium with Ka.
3. Example: HCl versus acetic acid. | acids_bases |
Sodium hydroxide is a strong base that dissociates fully. | {
"A": "NaOH",
"B": "NH3",
"C": "CH3NH2",
"D": "C5H5N"
} | che_0171 | multiple_choice | en | Which of these is a strong base? | [
"Group 1 hydroxides are strong.",
"Eliminate ammonia and amines.",
"Select option A."
] | chemistry | Answer: NaOH.
Steps:
1. Group 1 hydroxides are strong.
2. Eliminate ammonia and amines.
3. Select option A.
Answer: (A) | acids_bases |
The pOH is 11.00, since pH 3.00 gives 14 - 3 = 11. | null | che_0172 | worked_example | en | Compute the pOH of 0.0010 M HCl. | [
"pH = -log(0.0010) = 3.00.",
"pOH = 14 - pH.",
"pOH = 11.00."
] | chemistry | Answer: 11.00.
Steps:
1. pH = -log(0.0010) = 3.00.
2. pOH = 14 - pH.
3. pOH = 11.00. | acids_bases |
Silná kyselina ve vodě disociuje prakticky úplně. | null | che_0173 | concept | cs | Co je silná kyselina? | [
"Úplná disociace.",
"Příklady: HCl, HNO3, H2SO4.",
"Kontrast se slabými kyselinami."
] | chemistry | Answer: Kyselina s úplnou disociací.
Steps:
1. Úplná disociace.
2. Příklady: HCl, HNO3, H2SO4.
3. Kontrast se slabými kyselinami. | acids_bases |
Adding a proton to NH3 gives its conjugate acid NH4+. | {
"A": "NH2(-)",
"B": "NH4(+)",
"C": "NO3(-)",
"D": "N2"
} | che_0174 | multiple_choice | en | What is the conjugate acid of ammonia? | [
"Add one H+ to NH3.",
"Result is NH4+.",
"Select option B."
] | chemistry | Answer: NH4+.
Steps:
1. Add one H+ to NH3.
2. Result is NH4+.
3. Select option B.
Answer: (B) | acids_bases |
Amphiprotic species like water and HCO3- can both donate and accept protons. | null | che_0175 | concept | en | What does amphiprotic mean? | [
"Can act as acid or base.",
"Examples: H2O, HCO3-.",
"Depends on partner."
] | chemistry | Answer: Both donates and accepts protons.
Steps:
1. Can act as acid or base.
2. Examples: H2O, HCO3-.
3. Depends on partner. | acids_bases |
Hydroxide concentration is 1.0 x 10^-5 M, from pOH = 14 - 9 = 5. | null | che_0176 | worked_example | en | A solution has pH 9.00. Compute [OH-]. | [
"pOH = 14 - 9.00 = 5.00.",
"Write: [OH-] = 10^(-pOH).",
"[OH-] = 1.0 x 10^-5 M."
] | chemistry | Answer: 1.0 x 10^-5 M.
Steps:
1. pOH = 14 - 9.00 = 5.00.
2. Write: [OH-] = 10^(-pOH).
3. [OH-] = 1.0 x 10^-5 M. | acids_bases |
A larger Ka means a stronger acid that donates protons more readily. | null | che_0177 | concept | en | How does Ka relate to acid strength? | [
"Ka measures dissociation.",
"Bigger Ka means stronger acid.",
"pKa inverts the scale."
] | chemistry | Answer: Larger Ka means stronger acid.
Steps:
1. Ka measures dissociation.
2. Bigger Ka means stronger acid.
3. pKa inverts the scale. | acids_bases |
The acid with the largest Ka dissociates most and is strongest. | {
"A": "Smallest Ka",
"B": "Ka near 1",
"C": "Largest Ka",
"D": "Ka is irrelevant"
} | che_0178 | multiple_choice | en | Which acid is strongest? | [
"Compare Ka values.",
"Largest Ka wins.",
"Select option C."
] | chemistry | Answer: Largest Ka.
Steps:
1. Compare Ka values.
2. Largest Ka wins.
3. Select option C.
Answer: (C) | acids_bases |
pH je 1, protože -log(0,1) = 1. | null | che_0179 | worked_example | cs | Vypočítej pH roztoku 0,1 M HCl. | [
"HCl disociuje plně.",
"pH = -log(0,1).",
"pH = 1."
] | chemistry | Answer: 1.
Steps:
1. HCl disociuje plně.
2. pH = -log(0,1).
3. pH = 1. | acids_bases |
The common ion effect suppresses weak electrolyte dissociation when a shared ion is added. | null | che_0180 | concept | en | What is the common ion effect? | [
"Add ion shared with equilibrium.",
"Equilibrium shifts back.",
"Dissociation drops."
] | chemistry | Answer: Shared ion suppresses dissociation.
Steps:
1. Add ion shared with equilibrium.
2. Equilibrium shifts back.
3. Dissociation drops. | acids_bases |
Acetate suppresses acetic acid dissociation, so pH rises. | {
"A": "pH falls sharply",
"B": "pH rises",
"C": "Nothing changes",
"D": "Acid strengthens"
} | che_0181 | multiple_choice | en | What happens adding sodium acetate to acetic acid? | [
"Common acetate ion added.",
"Dissociation suppressed.",
"Select option B."
] | chemistry | Answer: pH rises.
Steps:
1. Common acetate ion added.
2. Dissociation suppressed.
3. Select option B.
Answer: (B) | acids_bases |
The buffer pH is 4.76, equal to pKa when the ratio is 1:1. | null | che_0182 | worked_example | en | An acetate buffer has equal acid and base (pKa = 4.76). Find pH. | [
"Write: pH = pKa + log(base/acid).",
"Ratio 1 gives log 1 = 0.",
"pH = 4.76."
] | chemistry | Answer: 4.76.
Steps:
1. Write: pH = pKa + log(base/acid).
2. Ratio 1 gives log 1 = 0.
3. pH = 4.76. | acids_bases |
The Henderson-Hasselbalch equation gives buffer pH from pKa and the base-to-acid ratio. | null | che_0183 | concept | en | What is the Henderson-Hasselbalch equation? | [
"pH = pKa + log([A-]/[HA]).",
"Use for buffers.",
"Equal parts give pH = pKa."
] | chemistry | Answer: Buffer pH from pKa and ratio.
Steps:
1. pH = pKa + log([A-]/[HA]).
2. Use for buffers.
3. Equal parts give pH = pKa. | acids_bases |
Pufr je roztok slabé kyseliny a její soli, který odolává změnám pH. | null | che_0184 | concept | cs | Co je pufr (tlumivý roztok)? | [
"Slabá kyselina + sůl.",
"Váže přidané H+ i OH-.",
"Příklad: octanový pufr."
] | chemistry | Answer: Roztok stabilizující pH.
Steps:
1. Slabá kyselina + sůl.
2. Váže přidané H+ i OH-.
3. Příklad: octanový pufr. | acids_bases |
Strong acid with strong base gives neutral salt water, so equivalence pH is 7. | {
"A": "pH 7",
"B": "pH 4",
"C": "pH 10",
"D": "pH 1"
} | che_0185 | multiple_choice | en | What is the equivalence pH of HCl titrated with NaOH? | [
"Salt NaCl is neutral.",
"Only water remains.",
"Select option A."
] | chemistry | Answer: pH 7.
Steps:
1. Salt NaCl is neutral.
2. Only water remains.
3. Select option A.
Answer: (A) | acids_bases |
There are 10 mmol, from 50 mL times 0.200 M. | null | che_0186 | worked_example | en | How many millimoles are in 50.0 mL of 0.200 M acid? | [
"Write: mmol = mL x M.",
"Substitute: 50.0 x 0.200.",
"mmol = 10.0."
] | chemistry | Answer: 10.0 mmol.
Steps:
1. Write: mmol = mL x M.
2. Substitute: 50.0 x 0.200.
3. mmol = 10.0. | acids_bases |
[OH-] je 1 x 10^-4 M, protože 10 na -4 odpovídá pOH 4. | null | che_0187 | worked_example | cs | Roztok má pOH 4. Vypočítej [OH-]. | [
"Napiš vzorec: [OH-] = 10^(-pOH).",
"Dosaď: 10^(-4).",
"[OH-] = 1 x 10^-4 M."
] | chemistry | Answer: 1 x 10^-4 M.
Steps:
1. Napiš vzorec: [OH-] = 10^(-pOH).
2. Dosaď: 10^(-4).
3. [OH-] = 1 x 10^-4 M. | acids_bases |
The anion hydrolyzes to make OH-, so the solution is basic. | {
"A": "Acidic",
"B": "Neutral",
"C": "Cannot tell",
"D": "Basic"
} | che_0188 | multiple_choice | en | A salt of a weak acid and strong base gives what solution? | [
"Anion of weak acid hydrolyzes.",
"OH- forms.",
"Select option D."
] | chemistry | Answer: Basic.
Steps:
1. Anion of weak acid hydrolyzes.
2. OH- forms.
3. Select option D.
Answer: (D) | acids_bases |
Oxidation numbers track electron bookkeeping: elements are 0, oxygen usually -2, hydrogen usually +1. | null | che_0189 | concept | en | What are the basic oxidation number rules? | [
"Elements: 0.",
"O usually -2, H usually +1.",
"Sum matches total charge."
] | chemistry | Answer: Bookkeeping rules summing to charge.
Steps:
1. Elements: 0.
2. O usually -2, H usually +1.
3. Sum matches total charge. | redox |
Iron is +3, since 2Fe + 3(-2) = 0 gives Fe = +3. | {
"A": "+1",
"B": "+2",
"C": "+3",
"D": "+6"
} | che_0190 | multiple_choice | en | What is the oxidation state of iron in Fe2O3? | [
"Assign O -2.",
"Solve: 2Fe - 6 = 0.",
"Fe = +3; select C."
] | chemistry | Answer: +3.
Steps:
1. Assign O -2.
2. Solve: 2Fe - 6 = 0.
3. Fe = +3; select C.
Answer: (C) | redox |
Nitrogen is +5, since N + 3(-2) = -1 gives N = +5. | null | che_0191 | worked_example | en | Find the oxidation number of N in NO3-. | [
"Assign O -2.",
"Equation: N - 6 = -1.",
"N = +5."
] | chemistry | Answer: +5.
Steps:
1. Assign O -2.
2. Equation: N - 6 = -1.
3. N = +5. | redox |
Oxidační číslo je pomyslný náboj atomu přiřazený podle elektronegativity. | null | che_0192 | concept | cs | Co je oxidační číslo? | [
"Pomyslný náboj atomu.",
"Součet = náboj částice.",
"Slouží k bilanci redoxu."
] | chemistry | Answer: Pomyslný náboj pro bilanci.
Steps:
1. Pomyslný náboj atomu.
2. Součet = náboj částice.
3. Slouží k bilanci redoxu. | redox |
The oxidized species loses electrons and its oxidation number rises. | {
"A": "Loses electrons",
"B": "Gains electrons",
"C": "Loses protons",
"D": "Stays unchanged"
} | che_0193 | multiple_choice | en | What happens to the species being oxidized? | [
"Oxidation means loss.",
"Electrons are lost.",
"Select option A."
] | chemistry | Answer: Loses electrons.
Steps:
1. Oxidation means loss.
2. Electrons are lost.
3. Select option A.
Answer: (A) | redox |
The balanced equation is Mg + 2HCl -> MgCl2 + H2. | null | che_0194 | worked_example | en | Balance: Mg + HCl -> MgCl2 + H2. | [
"Balance Cl with 2 HCl.",
"Check H: 2 each side.",
"Result: Mg + 2HCl -> MgCl2 + H2."
] | chemistry | Answer: Mg + 2HCl -> MgCl2 + H2.
Steps:
1. Balance Cl with 2 HCl.
2. Check H: 2 each side.
3. Result: Mg + 2HCl -> MgCl2 + H2. | redox |
The half-reaction method splits oxidation and reduction, balances each, then recombines them. | null | che_0195 | concept | en | What is the half-reaction method? | [
"Split redox into two halves.",
"Balance atoms and charge each.",
"Add halves, cancel electrons."
] | chemistry | Answer: Balance halves, then combine.
Steps:
1. Split redox into two halves.
2. Balance atoms and charge each.
3. Add halves, cancel electrons. | redox |
Three electrons are lost as Al goes from 0 to +3. | null | che_0196 | worked_example | en | How many electrons are lost in Al -> Al3+? | [
"Change: 0 to +3.",
"Loss = 3 units.",
"Electrons lost = 3."
] | chemistry | Answer: 3 electrons.
Steps:
1. Change: 0 to +3.
2. Loss = 3 units.
3. Electrons lost = 3. | redox |
Fluorine is the strongest elemental oxidizing agent due to extreme electronegativity. | {
"A": "Na",
"B": "Fe",
"C": "H2",
"D": "F2"
} | che_0197 | multiple_choice | en | Which is the strongest oxidizing agent? | [
"Oxidizers gain electrons eagerly.",
"Fluorine tops the scale.",
"Select option D."
] | chemistry | Answer: F2.
Steps:
1. Oxidizers gain electrons eagerly.
2. Fluorine tops the scale.
3. Select option D.
Answer: (D) | redox |
Síra má oxidační číslo +4, protože S + 2(-2) = 0 dává S = +4. | null | che_0198 | worked_example | cs | Urči oxidační číslo síry v SO2. | [
"Přiřaď O -2.",
"Rovnice: S - 4 = 0.",
"S = +4."
] | chemistry | Answer: +4.
Steps:
1. Přiřaď O -2.
2. Rovnice: S - 4 = 0.
3. S = +4. | redox |
Combustion oxidizes fuel while oxygen is reduced to water and carbon dioxide. | null | che_0199 | concept | en | Why is combustion a redox reaction? | [
"Fuel carbons oxidize.",
"Oxygen reduces to oxides.",
"Electrons transfer with heat release."
] | chemistry | Answer: Fuel oxidizes, oxygen reduces.
Steps:
1. Fuel carbons oxidize.
2. Oxygen reduces to oxides.
3. Electrons transfer with heat release. | redox |
Methane combustion changes oxidation states, unlike the non-redox choices. | {
"A": "HCl + NaOH -> NaCl + H2O",
"B": "CH4 + 2O2 -> CO2 + 2H2O",
"C": "NaCl dissolving",
"D": "Ice melting"
} | che_0200 | multiple_choice | en | Which reaction is a redox reaction? | [
"Need oxidation number change.",
"Only combustion qualifies.",
"Select option B."
] | chemistry | Answer: CH4 + 2O2 -> CO2 + 2H2O.
Steps:
1. Need oxidation number change.
2. Only combustion qualifies.
3. Select option B.
Answer: (B) | redox |
The balanced equation is 2Na + Cl2 -> 2NaCl. | null | che_0201 | worked_example | en | Balance: Na + Cl2 -> NaCl. | [
"Balance Cl with 2 NaCl.",
"Balance Na with 2 Na.",
"Result: 2Na + Cl2 -> 2NaCl."
] | chemistry | Answer: 2Na + Cl2 -> 2NaCl.
Steps:
1. Balance Cl with 2 NaCl.
2. Balance Na with 2 Na.
3. Result: 2Na + Cl2 -> 2NaCl. | redox |
Oxidační činidlo elektrony přijímá a redukuje se; redukční činidlo je odevzdává a oxiduje se. | null | che_0202 | concept | cs | Co je oxidační a redukční činidlo? | [
"Oxidovadlo: příjem elektronů.",
"Redukovadlo: výdej elektronů.",
"Příklad: Cu2+ a Zn."
] | chemistry | Answer: Příjemce versus dárce elektronů.
Steps:
1. Oxidovadlo: příjem elektronů.
2. Redukovadlo: výdej elektronů.
3. Příklad: Cu2+ a Zn. | redox |
The reducing agent is itself oxidized as it donates electrons. | {
"A": "Is reduced",
"B": "Stays unchanged",
"C": "Is oxidized",
"D": "Disappears"
} | che_0203 | multiple_choice | en | What happens to the reducing agent? | [
"Reducer donates electrons.",
"Losing electrons is oxidation.",
"Select option C."
] | chemistry | Answer: Is oxidized.
Steps:
1. Reducer donates electrons.
2. Losing electrons is oxidation.
3. Select option C.
Answer: (C) | redox |
Galvanizing coats iron with zinc, which oxidizes first and shields the steel. | null | che_0204 | concept | en | How does galvanizing prevent rust? | [
"Zinc coats the iron.",
"Zinc oxidizes preferentially.",
"Iron stays protected."
] | chemistry | Answer: Zinc sacrifices itself first.
Steps:
1. Zinc coats the iron.
2. Zinc oxidizes preferentially.
3. Iron stays protected. | redox |
Vyčíslená rovnice je 2Ca + O2 -> 2CaO. | null | che_0205 | worked_example | cs | Vyčísli: Ca + O2 -> CaO. | [
"Kyslík: 2 CaO a 1 O2.",
"Vápník: 2 Ca.",
"Výsledek: 2Ca + O2 -> 2CaO."
] | chemistry | Answer: 2Ca + O2 -> 2CaO.
Steps:
1. Kyslík: 2 CaO a 1 O2.
2. Vápník: 2 Ca.
3. Výsledek: 2Ca + O2 -> 2CaO. | redox |
Hydrogen in metal hydrides is -1 because the metal is less electronegative. | {
"A": "+1",
"B": "-1",
"C": "0",
"D": "+2"
} | che_0206 | multiple_choice | en | What is the oxidation state of H in NaH? | [
"Na is +1.",
"Sum must be 0.",
"H = -1; select B."
] | chemistry | Answer: -1.
Steps:
1. Na is +1.
2. Sum must be 0.
3. H = -1; select B.
Answer: (B) | redox |
Five electrons are gained as manganese drops from +7 to +2. | null | che_0207 | worked_example | en | How many electrons are gained in MnO4- -> Mn2+? | [
"Mn goes +7 to +2.",
"Change = 5 units down.",
"Electrons gained = 5."
] | chemistry | Answer: 5 electrons.
Steps:
1. Mn goes +7 to +2.
2. Change = 5 units down.
3. Electrons gained = 5. | redox |
The first law states energy is conserved: internal energy changes by heat plus work. | null | che_0208 | concept | en | What is the first law of thermodynamics? | [
"Energy cannot be created.",
"ΔU = q + w.",
"Heat and work interconvert."
] | chemistry | Answer: Energy is conserved.
Steps:
1. Energy cannot be created.
2. ΔU = q + w.
3. Heat and work interconvert. | thermochemistry |
Heat entering the system is positive by the chemistry sign convention. | {
"A": "Negative",
"B": "Positive",
"C": "Zero",
"D": "Undefined"
} | che_0209 | multiple_choice | en | What sign does absorbed heat (q) have? | [
"System gains energy.",
"Gain is positive.",
"Select option B."
] | chemistry | Answer: Positive.
Steps:
1. System gains energy.
2. Gain is positive.
3. Select option B.
Answer: (B) | thermochemistry |
The internal energy change is +300 J, from +500 plus -200 J. | null | che_0210 | worked_example | en | A system absorbs 500 J of heat and does 200 J of work. Find ΔU. | [
"Write: ΔU = q + w.",
"Substitute: 500 + (-200).",
"ΔU = +300 J."
] | chemistry | Answer: +300 J.
Steps:
1. Write: ΔU = q + w.
2. Substitute: 500 + (-200).
3. ΔU = +300 J. | thermochemistry |
Měrná tepelná kapacita je teplo potřebné k ohřátí 1 g látky o 1 °C. | null | che_0211 | concept | cs | Co je měrná tepelná kapacita? | [
"Teplo na gram a stupeň.",
"Jednotka J/g°C.",
"Voda má 4,184 J/g°C."
] | chemistry | Answer: Teplo na 1 g a 1 °C.
Steps:
1. Teplo na gram a stupeň.
2. Jednotka J/g°C.
3. Voda má 4,184 J/g°C. | thermochemistry |
Specific heat is heat capacity per gram of substance. | {
"A": "Per mole only",
"B": "Per liter",
"C": "Per gram",
"D": "Per reaction"
} | che_0212 | multiple_choice | en | How does specific heat relate to heat capacity? | [
"Heat capacity is total.",
"Divide by mass.",
"Select option C."
] | chemistry | Answer: Per gram.
Steps:
1. Heat capacity is total.
2. Divide by mass.
3. Select option C.
Answer: (C) | thermochemistry |
The aluminum releases 3588 J, from 50 x 0.897 x (-80). | null | che_0213 | worked_example | en | 50 g of Al (c = 0.897 J/gC) cools from 100 C to 20 C. Find q. | [
"ΔT = 20 - 100 = -80 C.",
"q = 50 x 0.897 x (-80).",
"q = -3588 J released."
] | chemistry | Answer: -3588 J.
Steps:
1. ΔT = 20 - 100 = -80 C.
2. q = 50 x 0.897 x (-80).
3. q = -3588 J released. | thermochemistry |
Fusion melts solid to liquid and vaporization boils liquid to gas, both absorbing heat. | null | che_0214 | concept | en | What are heats of fusion and vaporization? | [
"Fusion: solid to liquid.",
"Vaporization: liquid to gas.",
"Both need heat input."
] | chemistry | Answer: Heats of melting and boiling.
Steps:
1. Fusion: solid to liquid.
2. Vaporization: liquid to gas.
3. Both need heat input. | thermochemistry |
Melting needs 12.02 kJ, from 2.00 mol times 6.01 kJ/mol. | null | che_0215 | worked_example | en | How much heat melts 36.0 g of ice (ΔHfus = 6.01 kJ/mol)? | [
"Moles = 36.0/18.015 = 2.00.",
"Multiply: 2.00 x 6.01.",
"Heat = 12.02 kJ."
] | chemistry | Answer: 12.02 kJ.
Steps:
1. Moles = 36.0/18.015 = 2.00.
2. Multiply: 2.00 x 6.01.
3. Heat = 12.02 kJ. | thermochemistry |
Vaporization absorbs heat to break liquid bonds, so it is endothermic. | {
"A": "Endothermic",
"B": "Exothermic",
"C": "Athermal",
"D": "Spontaneous only"
} | che_0216 | multiple_choice | en | Is vaporization endothermic or exothermic? | [
"Bonds break going to gas.",
"Breaking needs energy.",
"Select option A."
] | chemistry | Answer: Endothermic.
Steps:
1. Bonds break going to gas.
2. Breaking needs energy.
3. Select option A.
Answer: (A) | thermochemistry |
K roztátí je třeba 6,01 kJ, protože 18 g je 1 mol a ΔH je 6,01 kJ/mol. | null | che_0217 | worked_example | cs | Kolik tepla roztaví 18 g ledu (ΔH = 6,01 kJ/mol)? | [
"Moly = 18/18 = 1.",
"Vynásob: 1 x 6,01.",
"Teplo = 6,01 kJ."
] | chemistry | Answer: 6,01 kJ.
Steps:
1. Moly = 18/18 = 1.
2. Vynásob: 1 x 6,01.
3. Teplo = 6,01 kJ. | thermochemistry |
Standard enthalpy of formation is ΔH for making one mole from elements in reference states. | null | che_0218 | concept | en | What is standard enthalpy of formation? | [
"Form 1 mol of compound.",
"Start from elements.",
"Standard states apply."
] | chemistry | Answer: ΔH to form 1 mol from elements.
Steps:
1. Form 1 mol of compound.
2. Start from elements.
3. Standard states apply. | thermochemistry |
Elements in reference states have ΔHf of zero by definition. | {
"A": "100 kJ/mol",
"B": "0 kJ/mol",
"C": "-100 kJ/mol",
"D": "Undefined"
} | che_0219 | multiple_choice | en | What is ΔHf of O2(g)? | [
"O2 is the reference state.",
"Reference elements equal zero.",
"Select option B."
] | chemistry | Answer: 0 kJ/mol.
Steps:
1. O2 is the reference state.
2. Reference elements equal zero.
3. Select option B.
Answer: (B) | thermochemistry |
The combustion ΔH is -890.3 kJ, from -965.1 minus -74.8 kJ. | null | che_0220 | worked_example | en | Use ΔHf: CH4 -74.8, CO2 -393.5, H2O -285.8 kJ/mol. Find ΔH of CH4 + 2O2 -> CO2 + 2H2O. | [
"Products: -393.5 + 2(-285.8) = -965.1.",
"Reactants: -74.8 + 0.",
"ΔH = -965.1 - (-74.8) = -890.3 kJ."
] | chemistry | Answer: -890.3 kJ.
Steps:
1. Products: -393.5 + 2(-285.8) = -965.1.
2. Reactants: -74.8 + 0.
3. ΔH = -965.1 - (-74.8) = -890.3 kJ. | thermochemistry |
Spalné teplo je teplo uvolněné úplným spálením 1 molu látky. | null | che_0221 | concept | cs | Co je spalné teplo? | [
"Úplné spálení v kyslíku.",
"Na 1 mol látky.",
"Vždy exotermické."
] | chemistry | Answer: Teplo úplného spálení.
Steps:
1. Úplné spálení v kyslíku.
2. Na 1 mol látky.
3. Vždy exotermické. | thermochemistry |
Entropy has units of energy per kelvin, joules per kelvin. | {
"A": "kJ/mol",
"B": "J/gC",
"C": "atm L",
"D": "J/K"
} | che_0222 | multiple_choice | en | What are the units of entropy? | [
"S = q/T.",
"Joules over kelvin.",
"Select option D."
] | chemistry | Answer: J/K.
Steps:
1. S = q/T.
2. Joules over kelvin.
3. Select option D.
Answer: (D) | thermochemistry |
The entropy change is 20 J/K, from 6000 J divided by 300 K. | null | che_0223 | worked_example | en | 6000 J of heat transfers reversibly at 300 K. Compute ΔS. | [
"Write: ΔS = q/T.",
"Substitute: 6000/300.",
"ΔS = 20 J/K."
] | chemistry | Answer: 20 J/K.
Steps:
1. Write: ΔS = q/T.
2. Substitute: 6000/300.
3. ΔS = 20 J/K. | thermochemistry |
Entropie je míra neuspořádanosti soustavy. | null | che_0224 | concept | cs | Co je entropie? | [
"Míra neuspořádání.",
"Plyn má vyšší než pevná látka.",
"Roste u samovolných dějů vesmíru."
] | chemistry | Answer: Míra neuspořádanosti.
Steps:
1. Míra neuspořádání.
2. Plyn má vyšší než pevná látka.
3. Roste u samovolných dějů vesmíru. | thermochemistry |
Zero ΔG means the system sits at equilibrium with no net drive either way. | {
"A": "Spontaneous",
"B": "Nonspontaneous",
"C": "At equilibrium",
"D": "Impossible"
} | che_0225 | multiple_choice | en | What does ΔG = 0 mean? | [
"No free energy drive.",
"Forward equals reverse.",
"Select option C."
] | chemistry | Answer: At equilibrium.
Steps:
1. No free energy drive.
2. Forward equals reverse.
3. Select option C.
Answer: (C) | thermochemistry |
The crossover temperature is 500 K, from 50 kJ divided by 0.100 kJ/K. | null | che_0226 | worked_example | en | ΔH = +50 kJ and ΔS = +0.100 kJ/K. Above what T is the reaction spontaneous? | [
"Set ΔG = 0: T = ΔH/ΔS.",
"Substitute: 50 / 0.100.",
"T = 500 K."
] | chemistry | Answer: 500 K.
Steps:
1. Set ΔG = 0: T = ΔH/ΔS.
2. Substitute: 50 / 0.100.
3. T = 500 K. | thermochemistry |
Name branched alkanes by longest chain, lowest numbers, then alphabetical substituents. | null | che_0227 | concept | en | What are the steps for naming branched alkanes? | [
"Find longest chain.",
"Number for lowest locants.",
"List substituents alphabetically."
] | chemistry | Answer: Chain, numbers, substituents.
Steps:
1. Find longest chain.
2. Number for lowest locants.
3. List substituents alphabetically. | organic_naming |
2-methylbutane has five carbons: four in the chain plus one methyl branch. | {
"A": "4",
"B": "5",
"C": "6",
"D": "3"
} | che_0228 | multiple_choice | en | How many carbons are in 2-methylbutane? | [
"Butane chain = 4.",
"Methyl adds 1.",
"Total 5; select B."
] | chemistry | Answer: 5.
Steps:
1. Butane chain = 4.
2. Methyl adds 1.
3. Total 5; select B.
Answer: (B) | organic_naming |
The name is 3-methylpentane: five carbons with a methyl on carbon 3. | null | che_0229 | worked_example | en | Name CH3CH2CH(CH3)CH2CH3. | [
"Longest chain: 5 = pentane.",
"Methyl on carbon 3.",
"Name: 3-methylpentane."
] | chemistry | Answer: 3-methylpentane.
Steps:
1. Longest chain: 5 = pentane.
2. Methyl on carbon 3.
3. Name: 3-methylpentane. | organic_naming |
Řetězec se čísluje tak, aby substituenty měly co nejnižší čísla. | null | che_0230 | concept | cs | Jak se čísluje uhlíkatý řetězec? | [
"Začni od bližšího konce.",
"Substituenty co nejníže.",
"Při shodě rozhodne abeceda."
] | chemistry | Answer: Od konce s nižšími čísly.
Steps:
1. Začni od bližšího konce.
2. Substituenty co nejníže.
3. Při shodě rozhodne abeceda. | organic_naming |
Ethyne is commonly called acetylene, used in welding torches. | {
"A": "Ethylene",
"B": "Ethane",
"C": "Benzene",
"D": "Acetylene"
} | che_0231 | multiple_choice | en | What is the common name of ethyne? | [
"Recall C2H2 trivial name.",
"Acetylene is standard.",
"Select option D."
] | chemistry | Answer: Acetylene.
Steps:
1. Recall C2H2 trivial name.
2. Acetylene is standard.
3. Select option D.
Answer: (D) | organic_naming |
Cycloalkanes add the cyclo- prefix to the alkane name for the ring size. | null | che_0232 | concept | en | How are cycloalkanes named? | [
"Count ring carbons.",
"Add cyclo- prefix.",
"Example: cyclohexane."
] | chemistry | Answer: Cyclo- plus alkane name.
Steps:
1. Count ring carbons.
2. Add cyclo- prefix.
3. Example: cyclohexane. | organic_naming |
The name is methylbenzene, commonly called toluene. | null | che_0233 | worked_example | en | Name C6H5CH3. | [
"Benzene ring present.",
"Methyl substituent attached.",
"Name: methylbenzene (toluene)."
] | chemistry | Answer: Methylbenzene (toluene).
Steps:
1. Benzene ring present.
2. Methyl substituent attached.
3. Name: methylbenzene (toluene). | organic_naming |
Aldehydes carry a terminal -CHO group and use the -al suffix. | null | che_0234 | concept | en | How are aldehydes named? | [
"Find terminal -CHO.",
"Use -al suffix.",
"Example: ethanal."
] | chemistry | Answer: Terminal -CHO with -al.
Steps:
1. Find terminal -CHO.
2. Use -al suffix.
3. Example: ethanal. | organic_naming |
HCHO is methanal, commonly known as formaldehyde. | {
"A": "Methanal",
"B": "Ethanal",
"C": "Methanol",
"D": "Methane"
} | che_0235 | multiple_choice | en | What is the IUPAC name of formaldehyde (HCHO)? | [
"One carbon aldehyde.",
"Name is methanal.",
"Select option A."
] | chemistry | Answer: Methanal.
Steps:
1. One carbon aldehyde.
2. Name is methanal.
3. Select option A.
Answer: (A) | organic_naming |
CH3CH2CHO je propanal. | null | che_0236 | worked_example | cs | Pojmenuj sloučeninu CH3CH2CHO. | [
"Koncová skupina -CHO.",
"Tři uhlíky.",
"Název: propanal."
] | chemistry | Answer: Propanal.
Steps:
1. Koncová skupina -CHO.
2. Tři uhlíky.
3. Název: propanal. | organic_naming |
Amines are named as alkylamine, with N- prefixes for substituents on nitrogen. | null | che_0237 | concept | en | How are simple amines named? | [
"Name alkyl group.",
"Add -amine suffix.",
"Example: ethylamine."
] | chemistry | Answer: Alkyl plus -amine.
Steps:
1. Name alkyl group.
2. Add -amine suffix.
3. Example: ethylamine. | organic_naming |
CH3NH2 is methylamine: one carbon attached to nitrogen. | {
"A": "Ethylamine",
"B": "Methylamine",
"C": "Aniline",
"D": "Ammonia"
} | che_0238 | multiple_choice | en | What is the name of CH3NH2? | [
"One-carbon alkyl group.",
"Amine suffix applies.",
"Select option B."
] | chemistry | Answer: Methylamine.
Steps:
1. One-carbon alkyl group.
2. Amine suffix applies.
3. Select option B.
Answer: (B) | organic_naming |
The name is methoxymethane, commonly dimethyl ether. | null | che_0239 | worked_example | en | Name CH3OCH3. | [
"Oxygen links two carbons.",
"Two methyl groups present.",
"Name: dimethyl ether."
] | chemistry | Answer: Dimethyl ether.
Steps:
1. Oxygen links two carbons.
2. Two methyl groups present.
3. Name: dimethyl ether. | organic_naming |
Ethers have an oxygen bridging two carbon groups, R-O-R. | null | che_0240 | concept | en | What is the ether functional group? | [
"Oxygen between carbons.",
"General form R-O-R.",
"Example: diethyl ether."
] | chemistry | Answer: R-O-R linkage.
Steps:
1. Oxygen between carbons.
2. General form R-O-R.
3. Example: diethyl ether. | organic_naming |
Funkční skupina je atom či skupina určující typické reakce molekuly. | null | che_0241 | concept | cs | Co je funkční skupina? | [
"Reaktivní část molekuly.",
"Určuje třídu sloučeniny.",
"Příklady: -OH, -COOH, -NH2."
] | chemistry | Answer: Reaktivní část molekuly.
Steps:
1. Reaktivní část molekuly.
2. Určuje třídu sloučeniny.
3. Příklady: -OH, -COOH, -NH2. | organic_naming |
The -C triple N group defines a nitrile. | {
"A": "Amine",
"B": "Amide",
"C": "Nitrile",
"D": "Nitro"
} | che_0242 | multiple_choice | en | Which group is -C≡N? | [
"Triple bond C to N.",
"That is a nitrile.",
"Select option C."
] | chemistry | Answer: Nitrile.
Steps:
1. Triple bond C to N.
2. That is a nitrile.
3. Select option C.
Answer: (C) | organic_naming |
The name is propanoic acid: three carbons ending in -COOH. | null | che_0243 | worked_example | en | Name CH3CH2COOH. | [
"Count 3 carbons.",
"-COOH gives -oic acid.",
"Name: propanoic acid."
] | chemistry | Answer: Propanoic acid.
Steps:
1. Count 3 carbons.
2. -COOH gives -oic acid.
3. Name: propanoic acid. | organic_naming |
C2H5OH je ethanol. | null | che_0244 | worked_example | cs | Pojmenuj sloučeninu C2H5OH. | [
"Dva uhlíky.",
"Skupina -OH = -ol.",
"Název: ethanol."
] | chemistry | Answer: Ethanol.
Steps:
1. Dva uhlíky.
2. Skupina -OH = -ol.
3. Název: ethanol. | organic_naming |
Common names like acetone persist alongside systematic names like propanone. | null | che_0245 | concept | en | How do common and IUPAC names relate? | [
"IUPAC is systematic.",
"Common names are historical.",
"Both may be accepted."
] | chemistry | Answer: Systematic versus historical names.
Steps:
1. IUPAC is systematic.
2. Common names are historical.
3. Both may be accepted. | organic_naming |
Reaction coordinate diagrams plot energy against progress, showing barriers and intermediates. | null | che_0246 | concept | en | What is a reaction coordinate diagram? | [
"X-axis is progress.",
"Y-axis is energy.",
"Peaks are transition states."
] | chemistry | Answer: Energy versus progress plot.
Steps:
1. X-axis is progress.
2. Y-axis is energy.
3. Peaks are transition states. | organic_mechanisms |
The peak of the energy barrier is the transition state. | {
"A": "Intermediate",
"B": "Transition state",
"C": "Product",
"D": "Reactant"
} | che_0247 | multiple_choice | en | What is the highest point on an energy diagram? | [
"Peak means max energy.",
"That is the transition state.",
"Select option B."
] | chemistry | Answer: Transition state.
Steps:
1. Peak means max energy.
2. That is the transition state.
3. Select option B.
Answer: (B) | organic_mechanisms |
The SN2 rate quadruples to 4x because both concentrations double. | null | che_0248 | worked_example | en | SN2 rate = k[RX][Nu]. What happens if both concentrations double? | [
"Write: rate = k x 2[RX] x 2[Nu].",
"Factor: 4 x k[RX][Nu].",
"Rate becomes 4x."
] | chemistry | Answer: Rate quadruples.
Steps:
1. Write: rate = k x 2[RX] x 2[Nu].
2. Factor: 4 x k[RX][Nu].
3. Rate becomes 4x. | organic_mechanisms |
SN2 je jednokroková substituce se současným atakem nukleofilu a odchodem skupiny. | null | che_0249 | concept | cs | Co je substituce SN2? | [
"Jeden krok, zadní atak.",
"Rychlost závisí na obou.",
"Inverze konfigurace."
] | chemistry | Answer: Jednokroková substituce.
Steps:
1. Jeden krok, zadní atak.
2. Rychlost závisí na obou.
3. Inverze konfigurace. | organic_mechanisms |
Good leaving groups are weak bases like iodide that depart with the electron pair. | null | che_0250 | concept | en | What makes a good leaving group? | [
"Must take electron pair.",
"Weak bases leave best.",
"Order: I- > Br- > Cl- > F-."
] | chemistry | Answer: Weak base departing with pair.
Steps:
1. Must take electron pair.
2. Weak bases leave best.
3. Order: I- > Br- > Cl- > F-. | organic_mechanisms |
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