Test di esercitazione AP Chemistry
Le domande del test restano in inglese.
Questo è un assaggio gratuito da 15 domande nello stile dell'esame AP Chemistry. Tocca tutte e nove le unità, dalla struttura atomica e dai legami fino a gas, cinetica, termochimica, equilibrio, acidi e basi, termodinamica ed elettrochimica, e mescola domande a scelta multipla, a risposta multipla e a completamento. Rispondi a tutte e 15, poi guarda il punteggio e leggi la spiegazione di ognuna.
A compound is found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?
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1. A compound is found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?
- A. CHO
- B. CH2O
- C. C2H4O
- D. C2H2O
Mostra risposta e spiegazione
Risposta: B. CH2O
Pretend you have 100 g of the compound so the percentages become grams, then convert each to moles: about 3.33 mol C, 6.7 mol H and 3.33 mol O. Dividing by the smallest value gives a ratio of 1 : 2 : 1, so the empirical formula is CH2O. The empirical formula is only the simplest ratio. To get the molecular formula you would also need the molar mass.
2. The first ionization energy of aluminum is lower than that of magnesium. Which explanation is best?
- A. Aluminum has a larger nuclear charge, so it holds its outer electron less tightly
- B. Magnesium has more protons than aluminum, so its outer electron is held more tightly
- C. Aluminum's outermost electron is in a 3p subshell, which is higher in energy and partly shielded by the 3s electrons
- D. Aluminum's outermost electron is in a half-filled subshell, which is unusually easy to remove
Mostra risposta e spiegazione
Risposta: C. Aluminum's outermost electron is in a 3p subshell, which is higher in energy and partly shielded by the 3s electrons
Across a period the nuclear charge rises, so the general trend is a rising ionization energy, but this pair is a known exception. Magnesium loses a 3s electron. Aluminum loses a 3p electron, which sits slightly farther out and is shielded a little by the filled 3s subshell, so it comes off with less energy. A larger nuclear charge on its own would raise the ionization energy, not lower it, and aluminum's 3p subshell holds a single electron, so it is not a half-filled subshell.
3. Complete the sentence.
The central xenon atom in XeF2 has _____ lone pairs of electrons, which gives the molecule a _____ shape.
Opzioni per lo spazio 1: 1, 2, 3, 4
Opzioni per lo spazio 2: bent, linear, trigonal planar, T-shaped
Mostra risposta e spiegazione
Risposta: 3, linear
Xenon brings 8 valence electrons and uses 2 of them to bond with the two fluorine atoms. The other 6 electrons form 3 lone pairs, so there are 5 electron domains and the electron geometry is trigonal bipyramidal. The three lone pairs sit in the equatorial plane, where they repel each other least, and that pushes the two fluorine atoms to opposite poles. The molecular shape is linear.
4. 1-Propanol (CH3CH2CH2OH) and methyl ethyl ether (CH3OCH2CH3) both have a molar mass of about 60 g/mol, yet 1-propanol boils at a much higher temperature. Why?
- A. Molecules of 1-propanol hydrogen bond to one another, and the ether cannot, so more energy is needed to separate them
- B. 1-Propanol has much stronger London dispersion forces because its molar mass is larger
- C. The covalent bonds inside 1-propanol are stronger, so they take more energy to break
- D. The ether has no intermolecular forces at all, so it vaporizes at almost any temperature
Mostra risposta e spiegazione
Risposta: A. Molecules of 1-propanol hydrogen bond to one another, and the ether cannot, so more energy is needed to separate them
Boiling separates molecules from each other and does not break the covalent bonds inside them. The two molar masses match, so dispersion forces are close to equal. The difference is that 1-propanol has an O-H bond, so one molecule's hydrogen can hydrogen bond to another molecule's oxygen. The ether has oxygen but no hydrogen attached to it, so it can accept hydrogen bonds from other molecules but cannot form them with its own kind. It still has dipole-dipole and dispersion forces.
5. A fixed amount of an ideal gas in a rigid container is heated from 300 K to 600 K. Which statement is correct?
- A. The average kinetic energy and the typical molecular speed both double
- B. The average kinetic energy is unchanged because the volume is unchanged
- C. The typical speed doubles and the average kinetic energy quadruples
- D. The average kinetic energy doubles and the typical speed increases by a factor of about 1.4
Mostra risposta e spiegazione
Risposta: D. The average kinetic energy doubles and the typical speed increases by a factor of about 1.4
Average kinetic energy is directly proportional to the Kelvin temperature, so doubling T doubles it. Kinetic energy depends on the square of the speed, so the typical (root-mean-square) speed grows by the square root of 2, about 1.4. The pressure doubles too, since the amount of gas and the volume are fixed. Using Celsius instead of Kelvin is the usual mistake on this kind of question.
6. A student uses a spectrophotometer to measure the absorbance of a colored solution. According to the Beer-Lambert law, which changes would increase the measured absorbance?
Seleziona tutte le risposte corrette.
- A. Using a solution with a higher concentration of the absorbing species
- B. Using a cuvette with a shorter path length
- C. Using a cuvette with a longer path length
- D. Setting the instrument to a wavelength the species absorbs more strongly
- E. Diluting the solution with water
Mostra risposta e spiegazione
Risposta: A. Using a solution with a higher concentration of the absorbing species, C. Using a cuvette with a longer path length, D. Setting the instrument to a wavelength the species absorbs more strongly
The Beer-Lambert law is A = abc, where a is the molar absorptivity: absorbance rises with the absorptivity at the chosen wavelength, the path length and the concentration. That is why you measure near the wavelength of maximum absorbance, and why a straight-line calibration of absorbance against concentration works. Diluting the solution or shortening the path length lowers the absorbance.
7. A reaction vessel contains 0.60 mol of Al(s) and 0.60 mol of Cl2(g), which react as 2 Al + 3 Cl2 -> 2 AlCl3. Which reactant is limiting, and how many moles of AlCl3 can form?
- A. Al is limiting; 0.60 mol of AlCl3
- B. Cl2 is limiting; 0.40 mol of AlCl3
- C. Cl2 is limiting; 0.60 mol of AlCl3
- D. Al is limiting; 0.40 mol of AlCl3
Mostra risposta e spiegazione
Risposta: B. Cl2 is limiting; 0.40 mol of AlCl3
Use the mole ratio from the balanced equation. The 0.60 mol of Al could make 0.60 mol of AlCl3, but the 0.60 mol of Cl2 can make only 0.60 x (2/3) = 0.40 mol. The reactant that gives the smaller amount runs out first, so Cl2 is limiting and 0.40 mol of AlCl3 forms. Check it the other way as well: 0.60 mol of Al would need 0.90 mol of Cl2, and there is only 0.60 mol.
8. Complete the sentence.
In a reaction between A and B, doubling [A] while holding [B] constant doubles the initial rate, and doubling [B] while holding [A] constant quadruples the initial rate. The reaction is _____ order in A and _____ order in B.
Opzioni per lo spazio 1: zero, first, second, third
Opzioni per lo spazio 2: zero, first, second, third
Mostra risposta e spiegazione
Risposta: first, second
Compare how the rate changes when one concentration changes and the other stays fixed. A factor of 2 in [A] gave a factor of 2 in rate, so the order in A is 1 (2^1 = 2). A factor of 2 in [B] gave a factor of 4, so the order in B is 2 (2^2 = 4). The rate law is rate = k[A][B]^2, and the overall order is 3.
9. Which statements about a catalyst are correct?
Seleziona tutte le risposte corrette.
- A. It increases the value of the equilibrium constant
- B. It provides an alternative pathway with a lower activation energy
- C. It leaves the enthalpy change of the overall reaction unchanged
- D. It is permanently used up, so it has to be replaced as the reaction proceeds
- E. It can be consumed in one elementary step and regenerated in a later step
Mostra risposta e spiegazione
Risposta: B. It provides an alternative pathway with a lower activation energy, C. It leaves the enthalpy change of the overall reaction unchanged, E. It can be consumed in one elementary step and regenerated in a later step
A catalyst speeds up a reaction by opening a pathway with a lower activation energy, often through a different mechanism, and it takes part in that mechanism: it is used in one step and made again in another. It does not appear in the net equation. Because it changes only the route, not the starting and ending energies, the enthalpy change is the same and the equilibrium constant is the same. It reaches equilibrium faster and does not shift where equilibrium lies.
10. Given N2(g) + O2(g) -> 2 NO(g), delta H = +180 kJ, and 2 NO(g) + O2(g) -> 2 NO2(g), delta H = -114 kJ, what is delta H for N2(g) + 2 O2(g) -> 2 NO2(g)?
- A. -294 kJ
- B. -66 kJ
- C. +66 kJ
- D. +294 kJ
Mostra risposta e spiegazione
Risposta: C. +66 kJ
Add the two steps. The 2 NO on the product side of the first equation cancels the 2 NO on the reactant side of the second, and one O2 from each side combines to leave N2 + 2 O2 -> 2 NO2. By Hess's law the enthalpy changes add: +180 + (-114) = +66 kJ. A positive value means the overall reaction is endothermic.
11. For A(g) + B(g) <=> C(g), K = 5.0 at a certain temperature. A mixture has [A] = 0.20 M, [B] = 0.20 M and [C] = 0.40 M. What happens next?
- A. Q = 2.0, so the reaction shifts toward the products
- B. Q = 10, so the reaction shifts toward the products
- C. Q = 5.0, so the mixture is already at equilibrium
- D. Q = 10, so the reaction shifts toward the reactants
Mostra risposta e spiegazione
Risposta: D. Q = 10, so the reaction shifts toward the reactants
The reaction quotient has the same form as K but uses the current concentrations: Q = [C] / ([A][B]) = 0.40 / (0.20 x 0.20) = 10. Since Q is larger than K = 5.0, the mixture has too much product to be at equilibrium, and the net reaction runs in reverse until Q falls to K. If Q were smaller than K, the shift would go toward the products instead.
12. Complete the sentence.
A buffer contains equal concentrations of acetic acid and acetate ion. The pKa of acetic acid is 4.74, so the pH of the buffer is _____. When a small amount of strong base is added, the pH _____.
Opzioni per lo spazio 1: 3.74, 4.74, 5.74, 7.00
Opzioni per lo spazio 2: rises slightly, rises sharply, falls slightly, does not change at all
Mostra risposta e spiegazione
Risposta: 4.74, rises slightly
The Henderson-Hasselbalch equation is pH = pKa + log([A-]/[HA]). With equal concentrations the ratio is 1 and log(1) = 0, so the pH equals the pKa, 4.74. Added hydroxide reacts with acetic acid and turns a little of it into acetate. That nudges the ratio up, so the pH rises slightly, but both buffer components are still present in large amounts, which is why the change is small and not sharp.
13. A weak monoprotic acid is titrated with a strong base. At the equivalence point, the solution is:
- A. basic, because the conjugate base of the weak acid reacts with water
- B. neutral, because the moles of acid and base are equal
- C. acidic, because some weak acid is always left over
- D. at a pH equal to the pKa of the acid
Mostra risposta e spiegazione
Risposta: A. basic, because the conjugate base of the weak acid reacts with water
At the equivalence point all of the weak acid has been converted to its conjugate base, and that base reacts with water to make a little hydroxide, so the pH is above 7. Equal moles of acid and base does not mean a pH of 7 unless both are strong. The pH equals the pKa at the half-equivalence point, where half of the acid has been converted.
14. Which processes would you expect to have a positive entropy change for the system?
Seleziona tutte le risposte corrette.
- A. I2(s) -> I2(g)
- B. H2O(g) -> H2O(l)
- C. 2 NO2(g) -> N2O4(g)
- D. CaCO3(s) -> CaO(s) + CO2(g)
- E. H2O(s) -> H2O(l) at the melting point
Mostra risposta e spiegazione
Risposta: A. I2(s) -> I2(g), D. CaCO3(s) -> CaO(s) + CO2(g), E. H2O(s) -> H2O(l) at the melting point
Entropy rises when matter becomes more dispersed or has more ways to arrange itself. Sublimation, melting and a solid breaking down into a solid plus a gas all spread the particles out, so each is positive. Condensing a gas and joining two gas molecules into one both reduce the number of free particles, so those are negative.
15. A galvanic cell has a positive standard cell potential. What does that tell you about the cell reaction?
- A. delta G is positive and K is greater than 1
- B. delta G is positive and K is less than 1
- C. delta G is negative and K is greater than 1
- D. delta G is negative and K is less than 1
Mostra risposta e spiegazione
Risposta: C. delta G is negative and K is greater than 1
The standard free energy change is delta G = -nFE, so a positive E gives a negative delta G. The same quantity is related to the equilibrium constant by delta G = -RT ln K, so a negative delta G means ln K is positive and K is greater than 1. All three signs point the same way: the reaction is thermodynamically favorable as written, and a spontaneous reaction is what a galvanic cell uses to do work.
Com'è fatto l'esame AP Chemistry e che peso ha ogni parte
L'esame AP Chemistry ha un punteggio da 1 a 5 e due sezioni dello stesso peso. La Sezione I ha 60 domande a scelta multipla in 90 minuti, alcune isolate e altre legate a una tabella, a un grafico o a un esperimento. La Sezione II ha sette domande a risposta aperta in 105 minuti: tre domande lunghe da 10 punti ciascuna e quattro brevi da 4 punti ciascuna. La calcolatrice è ammessa e viene fornito un foglio di riferimento, quindi l'esame non verifica se hai memorizzato una formula. Verifica se sai quale relazione usare e perché. La parte a scelta multipla si svolge ora nell'app Bluebook e le risposte aperte si scrivono a mano, quindi prima del giorno dell'esame controlla le regole aggiornate sulla pagina del College Board.
Il corso è organizzato in nove unità e il College Board pubblica quanto della sezione a scelta multipla spetta a ciascuna. Struttura atomica, struttura dei composti, reazioni chimiche, cinetica, termochimica, equilibrio e termodinamica con elettrochimica pesano ognuna circa il 7-9 per cento. Due unità pesano di più: proprietà di sostanze e miscele, che comprende forze intermolecolari, gas, soluzioni e spettroscopia, con circa il 18-22 per cento, e acidi e basi con circa l'11-15 per cento. Negli ultimi anni la descrizione del corso è stata riorganizzata, quindi i nomi delle unità e gli elenchi di argomenti in un vecchio libro di ripasso possono non coincidere. Usa come riferimento l'attuale Course and Exam Description.
Il 5 è il punteggio massimo, ma quanto vale lo decide ogni college. Alcuni riconoscono crediti con un 3, molti chiedono un 4 o un 5 e alcuni non danno crediti per AP Chemistry. Informati sulla politica delle università che ti interessano prima di decidere quanto spingere.
Come usare questo assaggio
Le 15 domande sono distribuite su tutte e nove le unità, con più domande sull'unità a cui l'esame dà più peso. Servono a mostrarti quale unità aprire per prima, non a prevedere un punteggio. Ognuna chiede il motivo, non solo la risposta: perché l'alluminio cede un elettrone più facilmente del magnesio, perché due molecole con la stessa massa bollono a temperature diverse, cosa cambia un catalizzatore e cosa non può toccare. È il tipo di ragionamento che la sezione a risposta aperta premia.
Svolgilo in una sola seduta con una calcolatrice e una tavola periodica, ma senza appunti, poi leggi la spiegazione di ogni domanda, comprese quelle a cui hai risposto bene. Tratta un errore come un argomento. Se hai sbagliato la domanda sul tampone, la risposta non è un pH da memorizzare, ma l'abitudine di chiederti se l'acido e la sua base coniugata sono presenti in quantità uguali. Quando ricontrolli, scrivi le unità di misura accanto a ogni numero, perché perdere un'unità è il modo più economico per lasciare un punto sulla sezione a risposta aperta.
Le domande sono originali. Sono scritte sul programma pubblicato del corso e non usano nessuna domanda d'esame già rilasciata. Questo assaggio copre lo stile di ragionamento a scelta multipla. Non esercita le lunghe domande a risposta aperta in più parti, quindi per quel lato dell'esame affiancagli le domande aperte degli anni passati pubblicate dal College Board.
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Scritto dal team di StudyPDF. Ultimo aggiornamento 2026-08-19.