AP Chemistry-øveprøve: enhed 1 til 9
Testens spørgsmål er på engelsk.
Det her er en gratis smagsprøve på 15 spørgsmål i stil med AP Chemistry-eksamen. Den berører alle ni enheder, fra atomets opbygning og bindinger over gasser, kinetik, termokemi, ligevægt og syrer og baser til termodynamik med elektrokemi, og den blander multiple choice, vælg alle der passer og udfyld det tomme felt. Svar på alle 15 spørgsmål, se så din score og læs forklaringen til hvert spørgsmål.
A compound is found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?
Alle quizzer i Gymnasium og optagelse
- Øveprøve · 15 spørgsmål
AP Biology studieguide
- Øveprøve · 15 spørgsmål
AP Psychology-øveprøve
- Øveprøve · 15 spørgsmål
AP World History studieguide

Lav din egen
Upload din bog, dine slides eller dine noter, eller fortæl bare Bo, hvad du vil lære. Du behøver ikke et dokument for at komme i gang.
Alle 15 spørgsmål i overblik
Vil du hellere læse, før du går i gang? Alle spørgsmål i smagsprøven står herunder. Svar og forklaringer er foldet sammen, indtil du åbner dem.
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
Vis svar og forklaring
Svar: 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
Vis svar og forklaring
Svar: 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.
Muligheder for felt 1: 1, 2, 3, 4
Muligheder for felt 2: bent, linear, trigonal planar, T-shaped
Vis svar og forklaring
Svar: 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
Vis svar og forklaring
Svar: 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
Vis svar og forklaring
Svar: 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?
Vælg alle, der passer.
- 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
Vis svar og forklaring
Svar: 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
Vis svar og forklaring
Svar: 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.
Muligheder for felt 1: zero, first, second, third
Muligheder for felt 2: zero, first, second, third
Vis svar og forklaring
Svar: 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?
Vælg alle, der passer.
- 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
Vis svar og forklaring
Svar: 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
Vis svar og forklaring
Svar: 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
Vis svar og forklaring
Svar: 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 _____.
Muligheder for felt 1: 3.74, 4.74, 5.74, 7.00
Muligheder for felt 2: rises slightly, rises sharply, falls slightly, does not change at all
Vis svar og forklaring
Svar: 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
Vis svar og forklaring
Svar: 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?
Vælg alle, der passer.
- 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
Vis svar og forklaring
Svar: 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
Vis svar og forklaring
Svar: 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.
Hvordan AP Chemistry-eksamen ser ud, og hvordan den vægtes
AP Chemistry-eksamen scores fra 1 til 5 og har to dele med lige stor vægt. Del I er 60 multiple choice-spørgsmål på 90 minutter, nogle enkeltstående og nogle samlet om en tabel, en graf eller et forsøg. Del II er syv fri besvarelser på 105 minutter: tre lange spørgsmål til 10 point hver og fire korte spørgsmål til 4 point hver. Lommeregner er tilladt, og et formelark udleveres, så eksamen tester ikke, om du har lært formler udenad. Den tester, om du ved, hvilken sammenhæng du skal bruge, og hvorfor. Multiple choice-delen tages nu i appen Bluebook, og de frie besvarelser skrives i hånden, så tjek College Boards side for de aktuelle regler før prøvedagen.
Kurset er opdelt i ni enheder, og College Board offentliggør, hvor meget af multiple choice-delen hver af dem får. Atomets opbygning, forbindelsers opbygning, kemiske reaktioner, kinetik, termokemi, ligevægt og termodynamik med elektrokemi fylder hver omkring 7 til 9 procent. To enheder er tungere: stoffers og blandingers egenskaber, som dækker intermolekylære kræfter, gasser, opløsninger og spektroskopi, med omkring 18 til 22 procent, og syrer og baser med omkring 11 til 15 procent. Kursusbeskrivelsen er blevet omorganiseret de seneste år, så enhedsnavne og emnelister i en ældre repetitionsbog passer måske ikke. Brug den aktuelle Course and Exam Description som reference.
En 5'er er den højeste score, men hvad en score er værd, afgør hvert college selv. Nogle giver merit for et 3-tal, mange kræver et 4 eller et 5, og nogle giver ingen for AP Chemistry. Slå politikken op på de skoler, du interesserer dig for, før du beslutter, hvor hårdt du vil presse.
Sådan bruger du smagsprøven
De 15 spørgsmål er spredt over alle ni enheder, med flest spørgsmål på den enhed, eksamen vægter højest. De er tænkt som en hjælp til at se, hvilken enhed du skal åbne næste gang, ikke som en forudsigelse af din score. Hvert spørgsmål beder om begrundelsen, ikke kun svaret: hvorfor aluminium afgiver en elektron lettere end magnesium, hvorfor to molekyler med samme masse koger ved forskellige temperaturer, hvad en katalysator ændrer, og hvad den ikke kan røre. Det er den slags ræsonnement, de frie besvarelser belønner.
Tag den i ét stræk med lommeregner og periodesystem, men uden noter, og læs så forklaringen til hvert spørgsmål, også dem du svarede rigtigt på. Behandl en fejl som et emne. Hvis du missede bufferspørgsmålet, er svaret ikke en pH-værdi, du skal lære udenad, men vanen med at spørge, om syren og dens konjugerede base er til stede i lige store mængder. Skriv enheder på hvert tal, når du tjekker dit arbejde, for en tabt enhed er den billigste måde at miste et point på i de frie besvarelser.
Spørgsmålene er originale. De er skrevet efter den offentliggjorte kursusbeskrivelse og bruger ikke noget frigivet eksamensspørgsmål. Denne smagsprøve dækker den ræsonnementsstil, multiple choice kræver. Den træner ikke de lange spørgsmål i flere dele i de frie besvarelser, så supplér med tidligere frie besvarelser fra College Board til den side af eksamen.
Sådan bygger StudyPDF AP Chemistry-øveprøver ud fra dine noter fra timerne
Femten spørgsmål kan vise dig, hvilke enheder der er usikre. De kan ikke gøre dig klar alene, og en generisk spørgsmålsbank rammer sjældent de laboratorieøvelser og opgavesæt, din lærer faktisk har givet.
StudyPDF tager i stedet udgangspunkt i dit materiale. Upload dine lærebogskapitler, din lærers slides, en laboratorierapport eller dit repetitionshæfte, tryk på Nyt materiale, og StudyPDF bygger øveprøver ud fra præcis de sider. Hvert spørgsmål bygger på dit materiale, og hver forklaring henviser til, hvor svaret kommer fra, så du kan gå direkte tilbage til siden, når et trin ikke giver mening.
Du behøver ikke uploade noget for at komme i gang. Nævn i stedet et emne, for eksempel buffere og titreringskurver eller reaktionsmekanismer, så sætter Bo det op med dets nøglebegreber, og du kan bygge quizzer og flashcards ud fra det. Dine resultater følger, hvilke begreber du bliver ved med at misse, så den næste øveprøve kan målrettes dem. Det er gratis at starte.
Skrevet af StudyPDF-teamet. Sidst opdateret 2026-08-19.