Placement TestFind your level and where to start

1 L = 1000 mL, so 2.5 L = 2.5 × 1000 = 2500 mL. Converting to a smaller unit makes the number bigger.

A chemical change makes a new substance. Rusting turns iron, oxygen, and water into rust, a new substance. Melting and evaporating only change the state, and breaking only changes the shape: those are physical changes.

An atom is neutral, so the number of electrons equals the number of protons: 11. Protons + neutrons = 23 is the mass number; neutrons carry no charge and don’t affect the electron count.

Elements in the same column (group) have the same number of outer electrons and similar chemical properties. O and S are both in group 16, with 6 outer electrons. N, F, and Ne share O’s period (row), not its group.

The positive and negative charges must add up to zero. Two Al give +6 and three O give −6, so the formula is $\ce{Al2O3}$. A shortcut: cross the charges over as subscripts, 2 on Al and 3 on O.

Balance C and H first: 1 $\ce{CH4}$ gives 1 $\ce{CO2}$ and 2 $\ce{H2O}$. The right side then has 2 + 2 = 4 O atoms, so 2 $\ce{O2}$ are needed: $\ce{CH4 + 2O2 -> CO2 + 2H2O}$. Check: 1 C, 4 H, and 4 O on each side.

Mass percent = mass of solute ÷ mass of solution × 100%, and the solution’s mass is solute + solvent = 20 + 80 = 100 g. 20 ÷ 100 × 100% = 20%. Dividing by 80 gives 25%, which mistakes the water for the whole solution.

A pH above 7 means the solution is basic. Limewater is a solution of calcium hydroxide, a base. Vinegar (acetic acid) and lemon juice (citric acid) are acidic, with pH below 7, and salt water is neutral, pH 7.

In multiplication and division, the answer keeps as many significant figures as the least precise value. 4.25 has 3 and 2.1 has only 2, so 4.25 × 2.1 = 8.925 rounds to 2: 8.9 cm².

Fill 16 electrons from lowest energy up: 2 in 1s, 2 in 2s, 6 in 2p, 2 in 3s, and the last 4 in 3p: 1s²2s²2p⁶3s²3p⁴. B has 18 electrons (Ar, or $\ce{S^2-}$); an s sublevel holds at most 2 electrons, so C is impossible; D skips 3s.

All four have polar bonds, so look at the shape. $\ce{CO2}$ (linear), $\ce{BF3}$ (trigonal planar), and $\ce{CH4}$ (tetrahedral) are symmetric, so their bond dipoles cancel. $\ce{H2O}$ is bent and unsymmetric, so it is polar (dipole moment 1.85 D).

$\ce{SO4^2-}$ is sulfate ($\ce{SO3^2-}$ is sulfite and $\ce{S^2-}$ is sulfide). Let the charge on Fe be x: 2x + 3(−2) = 0, so x = +3. A Roman numeral gives iron’s charge: iron(III) sulfate.

The molar mass of $\ce{H2O}$ is 18.02 g/mol. n = m ÷ M = 9.0 ÷ 18.02 = 0.50 mol. Multiplying by the molar mass (162) uses the relationship backward.

With temperature and amount fixed, pressure is inversely proportional to volume (Boyle’s law): P₁V₁ = P₂V₂, so P₂ = 1.0 × 3.0 ÷ 1.5 = 2.0 atm. Halving the volume doubles the pressure.

Molarity = moles ÷ liters of solution, and 250 mL = 0.250 L. M = 0.50 ÷ 0.250 = 2.0 M. Dividing by 250 or multiplying by the volume is wrong.

HCl is a strong acid and ionizes completely, so [H⁺] = 0.010 M = $10^{-2}$ M and pH = −log[H⁺] = 2.00. 12.00 is what you get by mixing up pH and pOH; an acid’s pH is always below 7.

The ionization energy jumps about 4.2 times between I₃ (2745) and I₄ (11577): the first 3 electrons are valence electrons, and the 4th must come from the inner 2p level. The period 3 element with 3 valence electrons is Al. Na’s jump comes after I₁ and Mg’s after I₂.

$\ce{CCl4}$ has four polar C–Cl bonds, but the molecule is tetrahedral, so the bond dipoles cancel completely and the dipole moment is 0. $\ce{SO2}$ and $\ce{H2O}$ are bent and $\ce{NH3}$ is trigonal pyramidal; their lone pairs make them unsymmetric and polar.

Boiling points (CRC): HCl −85.1 °C, HBr −66.4 °C, HI −35.6 °C, HF 19.5 °C. Among HCl, HBr, and HI, more electrons mean stronger dispersion forces and a higher boiling point, but HF molecules hydrogen-bond, which is far stronger, so HF boils highest.

A weak acid stays mostly as molecules, so it is written as HClO; KOH is a strong base, written as $\ce{K+}$ and $\ce{OH-}$, and $\ce{K+}$ is a spectator. A applies only to strong acids, C is a molecular equation, and D keeps the spectator ion.

Compare trials 1 and 2: [B] is constant and [A] doubles, and the rate goes up 4.0 times = 2², so the reaction is second order in A. Compare trials 1 and 3: [A] is constant and [B] doubles, with no change in rate, so it is zero order in B. Rate = k[A]². Orders come from experiment, not from the coefficients.

q = mcΔT = 50.0 × 4.18 × (30.0 − 20.0) = 2.09 × 10³ J = 2.09 kJ. ΔT is the change in temperature; using the final temperature, 30.0 °C, instead gives 6.27 kJ.

Q = [B]² ÷ [A] = 0.50² ÷ 1.0 = 0.25, which is less than K = 0.50. Q < K means there is too little product, so the reaction proceeds forward until Q = K.

By the Henderson–Hasselbalch equation, pH = pKa + log([F⁻] ÷ [HF]). The two concentrations are equal, so the ratio is 1 and log 1 = 0: pH = pKa = −log($6.4 \times 10^{-4}$) = 3.19. 1.96 would be the pH of 0.20 M HF alone, without the NaF.