Answers to Frequent Response Questions:

Chapter 1:

1.)

Chapter 2:

1.)

Chapter 3:

1.)

Chapter 4: Representing Aqueous Reactions

1.)

Guiding principles:

Each reaction is worth a total of three points.Reactants are +1 point, products are +2 points.Ignore balancing and states.Inappropriate ionization = maximum one point penalty per equation.

a) CN¯ + Ag+ ---> Ag(CN)2¯

Note: any complex ion of Ag+ with cyanide with consistent charge earns 3 points; AgCN given as product earns one product point.

b) Mn2+ + S2¯ ---> MnS

Note: If Mg is used instead of Mn, maximum possible score is two points.

c) P4O10 (or P2O5) + H2O ---> H3PO4

Note: Acidic species (H+ or oxyacid of phosphorous) earns one product point; P in +5 oxidation state in oxyanion earns one product point; anions of oxyacids of phosphorous require H+ for full credit for products.

d) (NH4)2CO3 ---> NH3 + H2O + CO2

Note: any one product earns one point; all three earn two points. NH4OH + CO2 earns one product point. NH3 + H2CO3 earns one product point.

e) CO2 + OH¯ ---> HCO3¯

Note: CO32¯ + H2O as products earns two product points. CO3

2¯ alone as product earns one product point. HCO3¯ + H2O earns one product point.

f) H+ + Cl¯ + KMnO4 ---> K+ + Mn2+ + Cl2 + H2O

Note: HCl and MnO4¯ acceptable as reactants. Any valid redox product earns one point. All four product earns two points. K+ and/or H2O only as products earns no credit. If both H+ and H2O omitted, then maximum of two points possible.

g) Na + H2O ---> H2 + Na+ + OH¯

Note: all three products earn two product points. Any valid redox product earns one product point.

h) Cr2O72¯ + Fe2+ + H+ ---> Cr3+ + Fe3+ + H2O

Note: All three products earn two product points. Any valid redox product earns one product point. H2O only earns no credit. If Cl¯ ---> Cl2 instead of Fe2+ ---> Fe3+, then maximum of two points possible.

2.)

3 points per reaction: 1 point for correct reactants, 2 points for correct products. Partial credit patterns were developed, taking into account the nature of each reaction. For some of the reactions, formulas other than those shown below received credit.

a) Zn2+ + PO43¯ ---> Zn3(PO4)2

b) Ag+ + Br¯ ---> AgBr

c) Cl2 + OH¯ ---> OCl¯ + Cl¯ + H2O

d) H+ + SO32¯ ---> H2O + SO2 (or H2SO3)

e) Sn2+ + H+ + MnO4¯ ---> Sn4+ + Mn2+ + H2O

f) Fe3+ + SCN¯ ---> Fe(SCN)2+ (or Fe(SCN)63¯)

g) BCl3 + NH3 ---> Cl3BNH3

h) CS2 + O2 ---> CO2 + SO2 (or SO3)

3.)

a) two points

Water boils at a lower temperature in Denver than in New York City because the atmospheric pressure is less at high altitudes.

At a lower temperature, the cooking process is slower, so the time to prepare a hard-boiled egg is longer.

b) two points

S + O2 ---> SO2 (as coal is burned)

SO2 + H2O ---> H2SO3 (in the atmosphere)

SO2 + 1/2 O2 + H2O ---> H2SO4 (in the atmosphere)

c) two points

Vaporization or evaporation of sweat from the skin.

These processes are endothermic and so cool the skin.

d) two points

Colligative properties, which depend on the number of particles present, are involved.

Solute (the antifreeze) causes the lowering of the vapor pressure of the solvent. When the vapor pressure of the solvent is lowered, the freezing point is lowered and the boiling point is raised.

Chapter 5: Gas’s

1.)

a) three points

n = PV ÷ RT = [(721) (0.090)] ÷ [(62.4) (298)]= 3.49 x 10¯3 mol H2

Each worth one point:

25°C to 298 K745 - 24 = 721 mm Hgcalculation of moles H2

Note: 62.4 is R in units of L torr per mol K

b) two points

[(23.8) (0.090)] ÷ [(62.4 ) (298)] = 1.15 x 10¯4 mol H2O

(1.15 x 10¯4) (6.02 x 1023) = 6.92 x 1019 molecules H2O

c) two points (one for formula, one for calculation)

The average kinetic energies are equal, so:

(1/2 mv2)H2O = (1/2 mv2)H2

vH2 / vH2O = square root [MMH2O / MMH2]

= square root (18 / 2) = 3

Note: credit also given for correct use of vrms = square root (3RT / M)

d) two points

H2O deviates more from ideal behavior.

Explanation:

1) The volume of the H2O molecule is larger than that of the H2 molecule

OR

2) The intermolecular forces among the H2O molecules are stronger than those among

H2 molecules.

Chapter 6: Thermochemistry

1.)

Chapter 7:

1.)

Chapter 8: Periodic Properties of Elements

1.)

a) two points

Ca2+ has fewer electrons, thus it is smaller than Ca.

The outermost electron in Ca is in a 4s orbital, whereas the outermost electron in Ca2+ is in a 3p orbital.

Note: The first point is earned for indicating the loss of electrons, the second point for indicating the outermost electrons are in different shells -- must account for the magnitude of the size difference between Ca and Ca2+.

b) two points

U for CaO is more negative than U for K2O, so it is more difficult to break up the CaO lattices (stronger bonds in CaO). Ca2+ is smaller than K+, so internuclear separations (between cations and O2¯) are less.

OR

Ca2+ is more highly charged than K+, thus cation-O2 bonds are stronger

Note: understanding what "lattice energy" is earns 1 point; size or charge explanation needed for the second point. Responses that use Lewis structures or otherwise indicate molecules rather than ionic lattice earn no points.

c) two points

i) Ca has ore protons and is smaller. The outermost electrons are more strongly held by the nuclear charge of Ca.

ii) The outermost electrons in Ca are in the 4s, which is a higher energy orbital (more shielded) than the 3p electrons in K.

Note: for (i), the idea of attraction between nucleus and electrons must be present; for (ii), a "noble-gas configuration" argument must be tied to an energy argument in order to earn credit.

d) two points

the highest energy (outermost) electrons in Al is in a 3p orbital, whereas that electron in Mg is in a 3s orbital.

The 3p electron in Al is of higher energy (is more shielded) than is the 3s electron in Mg.

Note: noting that different orbitals are involved earns the first point; a correct energy argument earns the second point.

Responses that attribute the greater stability of Ca over K (or K+ over Ca+, or Mg over Al) to the stability of a completely filled (vs. half or partially filled) orbital earn NO credit.

2.)

a) two points

Alakali metals have metallic bonds: cations in a sea of electrons.

As cations increase in size (Li to Cs), charge density decreases and attractive forces (and melting points) decreases.

b) two points

Halogen molecules are held in place by dispersion (van der Waals) forces: bonds due to temporary dipoles caused by polarization of electron clouds.

As molecules increase in size (F2 to I2), the larger electrons clouds are more readily polarized, and the attractive forces (and melting points) increase.

c) four points

Melting point order: LiF > NaCl > KBr > CsI

Compounds are ionic

Larger radii of ions as listed

Larger radii leads to smaller attraction and lower melting points.

Chapter 9:

1.)

Chapter 10: Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and MO Theory

1.)

a) 1 point for each correct diagram, 1 point for any two structure names, 1 more point for all three structure names.

b) three points. A maximum of 1 point was deducted if repulsion of atoms rather than of electron pairs was mentioned.

CF4 - 4 bonding pairs around the C at corners of regular tetrahedron to minimize repulsion (maximize bond angles).

XeF4 - 4 bonding pairs and 2 lone pairs give octahedral shape with lone pairs on opposite sides of Xe atoms.

ClF3 - 3 bonding pairs and 2 lone pairs give trigonal bipyramid with lone pairs in equatorial positions 120° apart.

Chapter 11: Liquids, Solids, and IMFs

1.)

a) two points

The addition of a solute lowers the freezing point of water.

A mole of NaCl contains (dissociates into) 2 moles of ions/particles, whereas a mole of CaCl2 contains (dissociates into) 3 moles of ions. Therefore, CaCl2 is more effective.

b) two points

Hydrogen bonding is the most important intermolecular attractive force between molecules of H2O and between molecules of NH3.

Water is a liquid because the hydrogen-bonding forces are stronger between the adjacent H2O molecules than between adjacent NH3 molecules.

c) two points

Graphite's structure consists of 2-dimensional sheets of covalently bonded carbon atoms.

The attractive forces between sheets (layers) are weak London (dispersion) forces, which allow the sheets to slide easily over one another. (Note: must indicate layers and sliding to earn point.)

Diamond consists of an extended 3-dimensional covalent network of carbon atoms. This makes diamond a very hard substance.

d) two points

Vinegar, a dilute solution of acetic acid, reacts with the white solid, which contains metal carbonates, in a neutralization reaction to form gaseous CO2.

2.)

Two points (to a maximum of eight) were awarded for each correct test with the expected results. Possibilities include the following:

1. Flame test: K+ lavender; (NH4)2CO3, no lavender

2. Add Cl2and CH2Cl2: KI, pink color in organic layer; (NH4)2CO3, no change

3. Add Pb2+: KI, yellow precipitate of PbI2; (NH4)2CO3, white precipitate of PbCO3

4. Add Ag+: KI, pale yellow precipitate of AgI; (NH4)2CO3, white precipitate of

Ag2CO3

5. Add I2: KI, brown color of I3¯; (NH4)2CO3, no change

6. Add good oxidizing agent: KI, brown color of I3¯; (NH4)2CO3, no change

7. Add strong base: KI, no change: (NH4)2CO3, odor of NH3 or color change of red

litmus

8. Add Ba2+ or Ca2+ or Mg2+: KI, no change; (NH4)2CO3, with precipitate of a

carbonate

9. Dissolve salts and use litmus: KI, neutral; (NH4)2CO3, basic solution

10. Add nonoxidizing acid: KI, no change; (NH4)2CO3, bubbles of CO2

11. Test melting points: KI, high; (NH4)2CO3, decomposes before melting

3.)

Physical properties: Metals NonmetalsMelting points Relatively high Relatively lowElectrical conductivity Good InsulatorsLuster High Little or nonePhysical state Most are solids Gases, liquids, or solids

Chemical properties: Metals Nonmetals

redox Reducing agents Oxidizingor reducing agents

attraction to electrons Electropositive Electronegative

acid/base character Oxides basicor amphoteric Oxides acidic

reactivity React with nonmetals React with both metalsand nonmetals

Electron configurations (1 point each statement): Metals: Valence electrons in s or d sublevels of their atoms. (A few heavy elements have atoms with one or two electrons in p sublevels.) Nonmetals: Valence electrons in the s and p sublevels of their atoms.

There are more metals than nonmetals because filling d orbitals in a given energy level involves the atoms of ten element and filling the f orbitals involves the atoms of 14 elements. In the same energy levels, the maximum number of elements with atoms receiving p electrons is six.

Chapter 12:

1.)

Chapter 13: Reaction Rates

1.)

a) three points (point for each order must include justification)

From exps. 1 and 2: doubling [H2] while keeping [NO] constant doubles the rate, therefore the reaction is first order in [H2].

From exps. 3 and 4; doubling [NO] while keeping [H2] constant quadruples the rate, therefore the reaction is second order in [NO].

Rate = k [H2] [NO]2

Note: full credit is earned for the rate expression as long as it is consistent with orders described by student.

b) two points (one for value and one for units)

k= Rate / ([H2] [NO]2)

From exp. 1: k = 1.8 x 10¯4 M/min / [(1.0 x 10¯3 M) (6.0 x 10¯3 M)2]

= 5.0 x 103 M¯2 min¯1

Note: the unit is often written as L2 mol¯2 min¯1

c) one point

Stoichiometry: NO : H2 is 1:1

When 0.0010 mole of H2 had reacted , it must have reacted with 0.0010 mole NO; thus [NO] remaining = 0.0060 - 0.0010 = 0.0050 M

d) three points

(i)

For I : Keq = [N2O2] / [NO]2

For II: Rate= k[H2] [N2O2]

[N2O2] = Keq [NO]2

Rate = k' [H2][NO]2

Note: there must be some clear algebraic manipulation showing that [N2O2] is proportional (NOT equal) to [NO]2.

Step II is the rate determining step.

(ii)

I: NO + NO ---> N2O2

II: N2O2 + H2 ---> H2O + N2OIII: N2O + H2 ---> N2 + H2OI + II + III: 2NO + 2H2 ---> N2 + 2 H2O

2.)

a) two points

Effective concentration are increased

So collision frequency is increased.

b) two points

Slight increase in collision frequency occurs.

More molecules have enough energy that many more collisions have the necessary activation energy. Raises reaction rate a great deal.

c) two points

Catalytic nickel lowers the activation eneryg needed for a reaction.

More often molecules have the needed energy for a reaction.

d) two points

Greater surface area with powdered Ni.

More catalytic sites means a greater rate.

3.)

a) three points; one point for correct form of law and two points for correct methodology without an error; one point for correct methodology with an error

Rate = k[Y]

b) two points

7.0 x 10¯4 mole / L sec = k (0.10 mole / L)

k = 7.0 x 10¯3 sec¯1

c) two points

2.3 log co / c = k t

2.3 log 0.60 / 0.40 = (7.0 x 10¯3) (t)

t = 58 s

d) two point

Mechanism 3 is correct.

The rate law shows that the slow reaction must involve one Y, consistent with mechanism 3.

Mechanisms 1 and 2 would involve both [X] and [Y] in the rate law, not consistent with the rate law.

4.)

Chapter 14: Chemical Equilibrium

1.)

a) two points

Volume decreases in beaker A; the concentration decreases in beaker B (either obs. earns one point provided the other is not wrong.)

The vapor pressure of pure H2O is greater than the vapor pressure of H2O in solution.

OR

The rate of evaporation of H2O molecules from pure H2O is greater than that from the sugar solution, while the condensation rates are the same.

b) two points

The water will begin to boil (or evaporate).

The external pressure on the water will become equal to the vapor pressure of the water, causing it to boil.

OR

The drop in external pressure causes the boiling point to drop to the temperature of the water.

c) two points

Solid copper is deposited on the zinc strip; the zinc strip goes into solution. No reaction occurs with silver.

Zinc is a better reducing agent or a more active metal than copper and will be oxidized. Silver is a less reactive metal than copper is.

d) two points

Two layers will form, one of which is colored. Iodine is nonpolar and will dissolve in TTE. Water is polar and will not dissolve in TTE.

Placement of I2 must be correctly indicated for second point.

Chapter 15: Acids and Bases

1.)

a) two points

PO43¯ + H+ ---> HPO4

2¯

HPO42¯ + H+ ---> H2PO4¯

Note: any proton transfer to any PxOy species earns one point.

b) two points

explicit 32 e¯

explicit 2¯ charge (somewhere)

not more than 1 double P-O bond

Note: HPO42¯ (formula only) or other PxOy species with correct diagram earns one

point.

c) three points

Graph goes from upper left to the lower right. (pH decreases)

In either direction:

Two protons transferredTwo "buffers"Two "equivalences"

Explain/correctly label at least one "buffer" or "equivalence" region

d) one point

H2PO4¯ + H+ <===> H3PO4

Note: other proton transfer earns one point if consistent with product in part (a)

2.)

a) two points

Mol. wt. = (1.3717 grams / (0.2211 mol/L) x (0.0523 L)

= 1.3717 grams / 7.789 x 10¯5 mole = 176.1 grams/mol

b) three points

pH = 4.23 therefore [H+] = 5.9 x 10¯5 M

[A¯] = [(0.01523 L) (0.2211 mol/L)] / (0.07000 L)

= 0.004422 mol / 0.07000 L = 0.06317 M

[HA] = [(0.01523 L) (0.2211 mol/L)] / (0.07000 L)

= 0.00367 mol / 0.07000 L = 0.04810 M

K = ([H+] [A¯]) / [HA] = [(5.9 x 10¯5) (0.06317)] / (0.04810) = 7.7 x 10¯5

c) one point

A¯ + H2O = HA + OH¯

K= ([HA] [OH¯]) / [A¯] = Kw / Ka

= (1.0 x 10¯14) / (7.7 x 10¯5) = 1.3 x 10¯10

d) three points

[A¯] at equiv. point = 7.789 x 10¯3 mol / 0.08523 L = 9.14 x 10¯2 M

[OH¯]2 = (1.3 x 10¯10) (9.14 x 10¯2) = 1.2 x 10¯11

[OH¯] = 3.4 x 10¯6 M

pOH = - log (3.4 x 10¯6) = 5.47

pH = 14 - 5.47 = 8.53

3.)

a) two points

pH = 8.60

[H+] = 1 x 10¯8.60 M = 2.5 x 10¯9 M

[OH¯] = (1 x 10¯14) / (2.5 x 10¯9) M

= 4.0 x 10¯6 M

b) three points

C6H5CO2¯ + HOH ---> C6H5CO2H + OH¯

K = ([C6H5CO2H][OH¯]) ÷ [C6H5CO2¯]

= ((4.0 x 10¯8) (4.0 x 10¯6)) ÷ (0.10 - 4.0 x 10¯6)

= 1.6 x 10¯10

c) one point

C6H5CO2H <===> H+ + C6H5CO2¯

Ka = ([H+][C6H5CO2¯]) ÷ [C6H5CO2H]

= ((2.5 x 10¯9) (0.10)) ÷ (4.0 x 10¯6)

= 6.3 x 10¯5

d) two points

C6H5CO2H <===> C6H5CO2¯ + H¯

pH = 2.88

[H+ ] = 1 x 10¯2.88M = 1.3 x 10¯3

Ka = ([H+ ][C6H5CO2¯]) ÷ [C6H5CO2H]

6.3 x 10¯8 = ((1.3 x 10¯3) (1.3 x 10¯3)) ÷ x

x = 2.8 x 10¯2 M

Total C6H5CO2H in solution =

(2.8 x 10¯2 + 1.3 x 10¯3) M =

2.9 x 10¯2 M

4.)

Points awarded are indicated at the end of each statement.

An indicator signals the endpoint of a titration by changing color. (2)

An indicator is a weak acid or a weak base where the acid form and the basic form of the indicator are different colors. (2)

An indicator usually changes color when the concentrations of the acid form and the basic form are about equal; that is, when the pH is near the pK for the indicator. (1)

When an indicator is selected, the pH at which the indicator changes color should bracket the pH of the titration solution at the equivalence point. (2)

For example, when a strong acid is titrated with a strong base, the pH at the equivalence point is 7, so an indicator that changes color at a pH of 7 should be used. (1)

5.)

Chapter 16: Aqueous Ionic Equilibrium

1.)

a) two points

Ksp = [Mg2+][F¯]2

= (1.21 x 10¯3) (2 x 1.21 x 10¯3)2

= 7.09 x 10¯9

b) two points

Ksp = [Mg2+] (2x + 0.100)2

since 2x is much less than 0.100

= 7.09 x 10¯9 = [Mg2+] (0.010)2

[Mg2+] = (7.09 x 10¯9) / (10¯2)

= 7.09 x 10¯7 M

Note: OK if 0.102 is used for [F¯], then Ksp = 6.76 x 10¯7

c) three points (first point earned if both concentrations are correct; correct substitution and calculation of the wrong concentration values earns the second point [calc. of Q], but not the first

[Mg2+]:100.0 x 3.00 x 10¯3 = 300.0 x [Mg2+][Mg2+] = 1.00 x 10¯3 M

[F¯]:200.0 x 2.00 x 10¯3 = 300.0 x [F¯][F¯] = 1.33 x 10¯3 M

Q = Ion Product = [Mg2+] [F¯]2

= (1.00 x 10¯3) (1.33 x 10¯3)2

= 1.77 x 10¯9

Since Q < Ksp, no precipitate will form.

Note: conclusion must be consistent with Q value.

d) two points

Solubility of MgF2 decreases with the increasing temperature, thus dissolution process is exothermic.

MgF2 ---> Mg2+ + 2F¯ + Q (or H)

Reason:

i) Increased temperature puts a stress on the system (LeChâtelier). The system will reduce the stress by shifting the equilibrium in the endothermic (left) direction.

OR

ii) A data supported argument such as comparing ion concentrations, calculating second Ksp and giving proper interpretations.

2.)

Chapter 17: Free Energy and Thermodynamics

1.)

Note: for parts (a), (b), and (c), just writing an equation is not sufficient for the 'explanation" point. To earn credit, the student must connect the equation to the issue to be explained.

a) two points

Statement that [delta]S° is negative

3 moles of gas ---> 2 moles of gas plus solid (3 moles ---> 2 moles earns no points),

OR

2 gases ---> one gas + solid

OR

use of [delta]G° = [delta]H° - T[delta]S° with [delta]G° = 0

b) two points

[delta]G° is less negative, goes to 0, goes positive, gets larger

Explanation using [delta]G° = [delta]H° - T[delta]S°

c) two points

Keq decreases (exponent goes more negative) as T increases

OR

Keq goes from > 1, to 1, to < 1, as T increases

Correct explanation using the equation

[delta]G° = - RT ln K (or ln(k1 / k2) = ([delta]H° / R) (1/T2 - 1/T1)

OR

higher T favors the reverse reaction (Le Châtelier) because the forward reaction is exothermic.

Note: if answer for (a) is that [delta]S° is positive, then statement that Keq will decrease or increase depending on the relative magnitudes of T and [delta]G° change earns two points.

d) two points

Since [delta]G° = 0 at this point, the equation is T= [delta]H° / [delta]S°

([delta]G° = [delta]H° - T [delta]S°S is NOT sufficient without [delta]G° = 0.)

Prediction is not exact since [delta]H° and/or [delta]S° vary with T.

2.)

a) two points

[delta]G° = - RT lnK

= - (8.31 J mol¯1 K¯1) (298 K) (ln 0.281)

= 3.14 x 103 J / mol

b) four points

[delta]H° = (193 J/g) (160. g/mol) = 3.08 x 104 J / mol

[delta]G° = [delta]H° - T [delta]S°

[delta]S° = ([delta]H° - [delta]G°) / T

= [(3.084 x 104) - (3.14 x 103)] / 298

= (2.770 x 104 J / mol) / 298 K = 92.9 J / mol K

c) two points

At boiling pt, [delta]G° = 0 and thus,

T = [delta]H° / [delta]S° = 3.08 x 104 / 92.9 = 332 K

d) one point

Vapor pressure = 0.281 atm

Note added for student's benefit: this comes directly from the fact that Kp = PBr2.

3.)

a) three points

From Hess's law:

[delta]H°f = [4 (393.5) + 4 (205.85)] - 2183.5 kJ = - 533.8 kJ

b) one point

4 C(s) + 4 H2(g) + O2(g) ---> C3H7COOH

c) two points

[delta]S°f butyric acid = S° butyric acid - 4 S° carbon - 4 S° H2 - S° O2

[delta]S°f butyric acid = [226.3 - 4(5.69) - 4(130.6) - 205] joules/K = - 523.9 joules/K

Note: If part c was based on the equation in part b, and was done correctly, credit was given for part c even if part b was wrong.

d) three points

[delta]G°f butyric acid = [delta]H°f - T [delta]S°f

[delta]G°f = [533.9 - (298) (- 0.524)] kJ

[delta]G°f= - 377.7 kJ

Note: For each of the problems, a maximum of one point was subtracted for gross misuse of significant figures and. or for a mathematical error if correct principles were used.

Chapter 18: Electrochemistry

1.)

a) two points

No. moles S2O32¯ = (0.224 M) (0.0373 L) = 8.39 x 10¯3 mol

1 mole I2 = 1/2 mole S2O32¯

moles I2 = 4.20 x 10¯3 mol

b) three points

Phydrogen = Ptotal - Pwater

= (752 - 24) mm Hg = 728 mm Hg

1 mole H2 = 1 mole I2

PV = nRT

V = [(4.2 x 10¯3) (0.0821 L atm) (298K)] / [(728/760 atm) ( mol-K)] = 0.107 L

c) one point

At anode: 2 I¯ ---> I2 + 2e¯

d) three points

4.20 x 10¯3 mole I2 = 8.40 x 10¯3 mole e¯

(8.40 x 10¯3 F) (96500 coul / F) = 811 coul

amperes = 811 coul / 180 sec

2.)

a) any two parts = 1 point; all three parts = 2 points

Sodium is softest of the three.

Na added to H2O leads to gas and base.

Na + H2O ---> H2 + NaOH

b) any two parts = 1 point; all three parts = 2 points

Magnesium reacts with HCl

Mg + 2H+ ---> Mg2+ + H2

Redn. potentials: Mg = - 2.37 V; Ag = 0.80 V

Mg, but not Ag, reacts with HCl.

c) unbalanced equation = 1 point; balancing adds another point

Ag + 4 H+ + NO3¯ ---> 3 Ag+ + NO + 2 H2O

OR

Ag + 2 H+ + NO3¯ ---> Ag+ + NO2 + H2O

d) two points

A white precipitaion forms.

Ag+ + Cl¯ ---> AgCl(s)

3.)