Number of molecules = 0.25 x 6.022 x 10^23 = 1.506 x 10^23 molecules. Each CO2 molecule has 3 atoms (1 C + 2 O), so number of atoms = 3 x 1.506 x 10^23 = 4.52 x 10^23 atoms.

N = n x N_A

Divide each percentage by atomic mass: C = 24.27/12 = 2.02, H = 4.07/1 = 4.07, Cl = 71.65/35.5 = 2.02. Divide by the smallest (2.02): C = 1, H = 2, Cl = 1. Empirical formula = CH2Cl (empirical mass = 12 + 2 + 35.5 = 49.5). n = molar mass / empirical mass = 98.96/49.5 = 2. Molecular formula = C2H4Cl2.

n = molar mass / empirical formula mass

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles of NaOH = 5/40 = 0.125 mol. Volume = 450 mL = 0.450 L. Molarity = 0.125 / 0.450 = 0.278 M (approximately 0.28 M).

M = n / V(L)

Moles of H2 = 3/2 = 1.5 mol; moles of O2 = 29/32 = 0.906 mol. From the equation, 2 mol H2 need 1 mol O2. For 1.5 mol H2, O2 required = 0.75 mol, but 0.906 mol O2 is available, so O2 is in excess and H2 is the limiting reagent. Water formed = moles of H2 (since 2H2 -> 2H2O) = 1.5 mol = 1.5 x 18 = 27 g.

mass = moles x molar mass

1) The electron revolves around the nucleus only in certain fixed circular orbits of definite energy called stationary states, without radiating energy. 2) Angular momentum of the electron is quantised: mvr = nh/2pi, where n = 1, 2, 3... 3) Energy is emitted or absorbed only when an electron jumps from one orbit to another, and the energy difference equals hv (delta E = E2 - E1 = h v).

mvr = nh/2pi ; delta E = h v

For a hydrogen atom, r_n = n^2 x r_1. For n = 3: r_3 = 3^2 x 0.529 = 9 x 0.529 = 4.761 Angstrom.

r_n = n^2 x r_1

de Broglie wavelength lambda = h / (m v) = (6.626 x 10^-34) / (9.1 x 10^-31 x 2.05 x 10^7). Denominator = 1.866 x 10^-23. lambda = 3.55 x 10^-11 m = 0.355 Angstrom.

lambda = h / (m v)

1) Principal quantum number (n): gives the main energy level/shell and size of the orbital (n = 1, 2, 3...). 2) Azimuthal (angular momentum) quantum number (l): gives the subshell and shape of the orbital (l = 0 to n-1, i.e. s, p, d, f). 3) Magnetic quantum number (m_l): gives the orientation of the orbital in space (m_l = -l to +l). 4) Spin quantum number (m_s): gives the direction of electron spin (+1/2 or -1/2).

l = 0 to (n-1); m_l = -l ... +l

Cr (Z = 24): [Ar] 3d5 4s1 (not 3d4 4s2). Cu (Z = 29): [Ar] 3d10 4s1 (not 3d9 4s2). One 4s electron shifts to 3d because exactly half-filled (3d5) and completely filled (3d10) subshells have extra stability due to symmetrical distribution of electrons and greater exchange energy.

Across a period (left to right), atomic radius decreases because nuclear charge increases while electrons are added to the same shell, so the increased effective nuclear charge pulls the electron cloud closer. Down a group, atomic radius increases because a new shell is added at each step and the number of inner shielding electrons increases, outweighing the rise in nuclear charge.

Ionization enthalpy is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Across a period it increases because nuclear charge increases and atomic size decreases, so the electron is held more tightly. Down a group it decreases because atomic size increases and shielding by inner electrons increases, so the outer electron is more easily removed.

M(g) -> M+(g) + e-

Elements are classified by the subshell into which the last (differentiating) electron enters: s-block (last electron in s), p-block (in p), d-block (in d) and f-block (in f). d-block elements are called transition elements because they lie between the s-block (metals) and p-block (non-metals) and show a gradual transition in properties; their atoms or common ions have partially filled d orbitals.

BF3: central B has 3 bond pairs and no lone pair -> trigonal planar, bond angle 120 deg. NH3: central N has 3 bond pairs and 1 lone pair -> pyramidal (trigonal pyramidal), bond angle about 107 deg. H2O: central O has 2 bond pairs and 2 lone pairs -> bent/angular, bond angle about 104.5 deg. Lone pairs repel more than bond pairs, reducing the angle from the ideal 109.5 deg in NH3 and H2O.

In ethyne each carbon is sp hybridised. The two sp hybrid orbitals on each carbon form sigma bonds: one C-C sigma bond and one C-H sigma bond, giving a linear molecule with a bond angle of 180 deg. The remaining two unhybridised p orbitals on each carbon overlap sideways to form two pi bonds, so there is a triple bond (1 sigma + 2 pi) between the carbon atoms.

H-C(triple bond)C-H

O2 has 16 electrons. Configuration: sigma(1s)2 sigma*(1s)2 sigma(2s)2 sigma*(2s)2 sigma(2pz)2 pi(2px)2 pi(2py)2 pi*(2px)1 pi*(2py)1. Bonding electrons Nb = 10, antibonding Na = 6. Bond order = (Nb - Na)/2 = (10 - 6)/2 = 2. The two unpaired electrons in the pi* antibonding orbitals make O2 paramagnetic.

Bond order = (Nb - Na)/2

Fajans' rules: covalent character in an ionic compound increases when (1) the cation is small, (2) the anion is large, and (3) the cation has a high charge (all three increase the polarising power/polarisability). Also, cations with a pseudo-noble gas (18-electron) configuration cause greater polarisation than those with a noble gas configuration. Greater polarisation of the anion by the cation increases covalent character.

Dipole moment (mu) is the product of the magnitude of charge and the distance between the centres of positive and negative charge; it is a vector quantity (unit: debye). CO2 is linear (O=C=O) and its two C=O bond dipoles are equal and opposite, so they cancel and the net dipole moment is zero. H2O is bent/angular, so its two O-H bond dipoles do not cancel and add up to give a net dipole moment (1.85 D). Hence CO2 is non-polar but H2O is polar.

mu = q x d

Convert temperatures to Kelvin: T1 = 27 + 273 = 300 K, T2 = 127 + 273 = 400 K. By Charles's law V1/T1 = V2/T2, so V2 = V1 x T2/T1 = 300 x 400/300 = 400 mL.

V1/T1 = V2/T2

Using PV = nRT, V = nRT/P = (2 x 0.0821 x 300) / 2 = 49.26/2 = 24.63 L.

V = nRT / P

1) A gas consists of a large number of tiny particles (molecules) whose actual volume is negligible compared with the volume of the container. 2) There are no forces of attraction or repulsion between the molecules. 3) The molecules are in constant, rapid, random motion and collide with one another and with the walls of the container. 4) The collisions are perfectly elastic (no loss of kinetic energy). 5) The average kinetic energy of the molecules is directly proportional to the absolute temperature.

KE(avg) proportional to T