Question 3

Which of the following correctly identifies which has the higher
 first-ionization energy, Cl or Ar, and supplies the best
 justification? (A) (B) (C) (D) Cl, because of its higher
 electronegativity Cl, because of its higher electron affinity Ar,
 because of its completely filled valence shell Ar, because of its
 higher effective nuclear char e

The first ionization energy varies in a predictable way across the
 periodic table. The ionization energy decreases from top to bottom in
 groups, and increases from left to right across a period. Thus, helium
 has the largest first ionization energy, while francium has one of the
 lowest. • From top to bottom in a group, orbitals corresponding to
 higher values of the principal quantum number (n) are being added,
 which are on average further away from the nucleus. Since the
 outermost electrons are further away, they are less strongly attracted
 by the nucleus, and are easier to remove, corresponding to a lower
 value for the first ionization energy. • From left to right across a
 period, more protons are being added to the nucleus, but the number of
 electrons in the inner, lower-energy nuclear char — the sum of the
 charges on the protons in the nucleus and the charges on the inner,
 core electrons. The valence electrons are therefore held more tightly,
 the atom decreases in size (see atomic radius), and it becomes
 increasingly difficult to remove them, corresponding to a higher value
 for the first ionization energy.

Question 4

  • Sealed rigid vessel = constant volume

    There are two cases on which equilibrium depends. These are:
Addition of an inert gas at constant volume: When an inert gas is
added to the system in equilibrium at constant volume, the total
pressure will increase. But the concentrations of the products and
reactants (i.e. ratio of their moles to the volume of the container)
Hence, when an inert gas is added to the system in equilibrium at
constant volume there will be no effect on the equilibrium. Addition
of an inert gas When an inert gas is added to the system in
equilibrium at constant pressure, then the total volume will increase.
Hence, the number of moles per unit volume of various reactants and
products will decrease. Hence, the equilibrium will shift towards the
direction in which there is increase in number of moles of gases.

Question 6

Real Gases and Limitations of the Kinetic Theory (a) I ligh pressure
 (b) I -ow temperature The ideal gas model breaks down at high
 pressures and low temperatures. • high pressure: volume of particles
 no longer negligible • low temperature: particles move slowly enough
 to interact

Question 12

Ionization Energies for Period 2 Main Grou Elements Element Lithium
 (Li) Beryllium (Be) Boron (B) Carbon (C) Ni trogen (N) Oxygen (O)
 Fluorine (F) Neon (Ne) Electron Configuration \[He\] 2sl \[He\] 2s
 \[He\] 2s 2P \[He\] 2s 2P \[He\] 2s 2P \[He\] 2s 2P \[He\] 2s 2P
 Ionization Energy 520 kJ/m01 899 kJ/m01 801 kJ/m01 1086 kJ/m01 1400
 kJ/m01 1314 kJ/m01 1680 kJ/m01 2081 kJ/m01 @ 000 2s @ 000 2s

Why the drop between groups 5 and 6 (N-O and P-S) ? Once again, you
 might expect the ionisation energy of the group 6 element to be higher
 than that of group 5 because of the extra proton. What is offsetting
 it this time? N 1s22s22px12py12pz1 1st I.E. 1400 kJ mol-I 0
 1s22s22px22py12pz1 1st I.E. 1310 kJ mol-I The screening is identical
 (from the 1s2 and, to some extent, from the 2s2 electrons), and the
 electron is being removed from an identical orbital. The difference is
 that in the oxygen case the electron being removed is one of the 2px2
 pair. The electrons in the same orbital means that the electron is
 easier to be. The drop in ionisation energy at sulphur is accounted
 for in the same way.

Question 27

10 Cuvette 1 I —1 cm Microplate (96- or 384-well) 1 10 I is variable
 —5 < mm < —10


Question 30

  • Common Hydgron bonds: H-O, H-N

    Hydrogen bond Water molecule Hydrogen bond ...H-N-H Ammonia molecule

    fluorine oxygen oxygen nitrogen nitrogen sulfur Donors (in an acid)
(in water, or alcohols) Donor strength medium medium weak weak
Acceptoe -o-p,-o-s-, H20, H-o\_c, C=N-C 7t -electrons Acceptor
strength very strong Strong strong medium—strong medium medium
medium—weak weak

Question 38

Because of the incomplete dissociation of the acid, the reaction is
 in equilibrium, with an acid dissociation constant, Ka, which is
 specific to that acid. point are the same. Therefore, at the
 half-equivalence point, the pH is equal to the pKa. Experiment 6
 Titration Il
 Expt6V-Titration2. pdf

25.00 0.100 M equivalence point 1/2 equivalence point 0.100 M NaOH

Question 40

Which of the following best explains why more energy is required for
 the process occurring at 110 K than for the process occurring at 90 K
 ? (A) Intermolecular attractions are completely overcome d va
 rization. (B) Intermolecular attractions in the solid phase are weaker
 than in the liquid phase. (C) Electron clouds of methane molecules are
 less polarizable at lower temperatures. (D) Vaporization involves a
 large increase in temperature.

Question 49

Trend \#2 — For molecules with a given functional group, boiling
 point increases with molecular weight. Look at the dramatic increases
 in boi ing points as you increase molecular weight in all of these
 series: In a given series, boiling point increases with molecular
 weight. Why? Boiling point -42 oc ALCOHOLS HO Boiling point CIDS
 Boiling point Boiling oint 97 ac 141 oc -24 oc HO 117 oc Ha 164 oc 35
 ac 36 •c HO 138 oc Ha 186 oc 89 ac 69 •c 158 oc Ha 202 oc Increasing
 surface area gives rise to increased Van Der Waals interactions Here's
 the question: How, exactly do intermolecular forces increase as
 molecular weight increases? Well, the key force that is acting here
 are Van der Waals dispersion forces, which are proportional to surface
 area. So as you increase the length of the chain, you also increase
 the surface area, which means that you increase the ability of
 individual molecules to attract each other.

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