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Chem 104 M1 to M6 Exam-with 100%

verified answers-

Module 1: Question 1 In the reaction of gaseous N 2 O 5 to yield NO 2 gas and O 2 gas as shown below, the following data table is obtained: → 4 NO 2 (g)

+ O

2 (g)

1. Using the [O 2 ] data from the table, show the calculation of the instantaneous rate early in the reaction (0 secs to 300 sec).
2. Using the [O 2 ] data from the table, show the calculation of the instantaneous rate late in the reaction (2400 secs to 3000 secs).
3. Explain the relative values of the early instantaneous rate and the late instantaneous rate. Your Answer:
4. rate = (0.014 - 0) / (300 - 0) = 4.67 x 10 -5^ mol/Ls

2 N

2

O

5 (g) Data Table # Time (sec) [N 2 O 5 ]^ [O 2 ] 0 0.300 M 0 300 0.272 M 0.014 M 600 0.224 M 0.038 M 900 0.204 M 0.048 M 1200 0.186 M 0.057 M 1800 0.156 M 0.072 M 2400 0.134 M 0.083 M 3000 0.120 M 0.090 M

2. rate = (0.090 - 0.083) / (3000 - 2400) = 1.167 x 10 -5^ mol/Ls
3. The late instantaneous rate is smaller than the early instantaneous rate.

Question 2 The following rate data was obtained for the hypothetical reaction: A + B → X + Y Experiment # [A] [B] rate 1 0.50 0.50 2. 2 1.00 0.50 8. 3 1.00 1.00 64.

1. Determine the reaction order with respect to [A].
2. Determine the reaction order with respect to [B].
3. Write the rate law in the form rate = k [A]n^ [B]m^ (filling in the correct exponents).
4. Show the calculation of the rate constant, k. Your Answer: rate = k [A]x^ [B]y rate 1 / rate 2 = k [0.50]x^ [0.50]y^ / k [1.00]x^ [0.50]y 2.0 / 8.0 = [0.50]x^ / [1.00]x 0.25 = 0.5x x = 2 rate 2 / rate 3 = k [1.00]x^ [0.50]y^ / k [1.00]x^ [1.00]y 8.0 / 64.0 = [0.50]y^ / [1.00]y 0.125 = 0. y y = 3 rate = k [A]^2 [B]^3 2.0 = k [0.50]^2 [0.50]^3 k = 64 Question 3 ln [A] - ln [A] 0 = - k t 0.693 = k t1/ An ancient sample of paper was found to contain 19.8 % 14 C content as compared to a present-day sample. The t1/2 for 14 C is 5720 yrs. Show the calculation of the decay constant (k) and the age of the paper.

Your Answer: 0.693 = k t1/ 0.693 = k (5720) k = 1.21 x 10 - ln [A] - ln [A] 0 = - k t ln 19.8 - ln 100 = - 1.21 x 10-4^ t t = 13, 384 years Question 4 Using the potential energy diagram below, state whether the reaction described by the diagram is endothermic or exothermic and spontaneous or nonspontaneous, being sure to explain your answer. Your Answer: The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous because it has relatively large Eact. Question 5 Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H 2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of H 2 O and corresponding amounts of CO, H 2 , and CH 4. CO (^) (g) + 3 H 2 (g) CH4 (g) + H 2 O (^) (g) Your Answer: 0.309 mole of H 2 O formed = 0.309 mole of CH 4 formed

0.309 mole of H 2 O formed = 0.800 - 0.309 = 0.491 mole CO 0.309 mole of H 2 O formed = 3 x 0.309 mole H 2 reacted = 2.40 - (3 x 0.309) = 1.473 mole H 2 [CO] = 0.491 mole / 8.00 L = 6.1375 x 10-2^ M [H2] = 1.473 mole / 8.00 L = 18.4125 x 10-2^ M [CH4] = 0.309 mole / 8.00 L = 3.8625 x 10-2^ M [H2O] = 0.309 mole / 8.00 L = 3.8625 x 10-2^ M Kc = [3.8625 x 10 -2] [3.8625 x 10-2] / [6.1375 x 10 -2] [18.4125 x 10-2]^3 Kc = 3. Question 6 Explain the terms substrate and active site in regard to an enzyme. Your Answer: Enzymes are large protein molecules that act as catalysts and increases the rate of a reaction. The catalysts act only on one type of substance to cause one type of reaction and this is called a substrate. Active sites are enzymes that are spherical in shape together ith a group of atoms on a surface of protein. These active sites bind with the substrate causing it to undergo a reaction. The substrate and active site forms a complex that creates a new pathway with lower activation energy and thus speed up a reaction. Question 7 The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of predominately reactants, predominately products or significant amounts of both products and reactants. Be sure to explain your answer. 2 O 3 (g) 3 O 2 (g) Kc = 2.54 x 10^12 Your Answer: Kc = 2.54 x 1012 is very large. Thus, the equilibrium mixture is made up of predominantly products. Question 8 The equilibrium reaction below has the following equilibrium mixture concentrations:

H 2 O = [0.0380], CH 4 = [0.0380], CO = [0.0620] and H 2 = [0.186] with Kc = 3.62. If the concentration of CO at equilibrium is increased to [0.200], how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. CO (^) (g) + 3 H 2 (g) CH4 (g) + H 2 O (^) (g) Your Answer: Kc = 3. Qc = [0.200] [0.0380] / [0.0620] [0.186]^3 Qc = 19. Qc is greater than Kc and thus the equilibrium will shift to the left towards the direction of the reactants. Kc = 3.62 when H 2 O = [0.0380], CH 4 = [0.0380], CO = [0.0620] and H 2 = [0.186 M] When CO = [0.200], Q = [0.0380] [0.0380] = 1. [0.200] [0.186]^3 The reaction must shift briefly in the forward direction to decrease the [CO] to come back to equilibrium.This is in agreement with Qc < Kc which also predicts the reaction will proceed to the right. new CO concentration is 0. Question 9 The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is decreased from 6.00 liters to 2.00 liters, how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. CO (^) (g) + 3 H 2 (g) CH4 (g) + H 2 O (^) (g) Your Answer: Initial volume = 6.00 L Final volume = 2.00 L Volume is reduced by

Pressure is inversely proportional to volume, thus pressure is tripled and concentration of gases is also tripled. Final concentration = Initial concentration x 3

Qc = [3CH 4 ] [3H 2 O] / [3CO] [3H 2 ]^3 Qc = Kc / 9 Qc is lesser than Kc and thus the equilibrium will shift to the right towards the direction of the products. When volume decreases from 6.00 to 2.00, the pressure triples and the concentration of all gases (CO, H 2 , CH 4 , and H 2 O) triples so: (at equilibrium) Qc =Kc = [CH 4 ] [H 2 O] = 3. [CO] [H 2 ]^3 (volume 1/3 = pressure tripled = conc tripled) Qc = [3 CH 4 ] [3 H 2 O] = Kc [3 CO] [3 H 2 ]^3 The reaction must shift briefly in the direction that decreases the pressure by going toward the side with the lesser moles of gas (forward direction : 4 moles of gas yields 2 moles of gas) to come back to equilibrium. This is in agreement with Qc < Kc: the reaction will proceed to the right (in the direction of the products).

• Qc should be used to confirm your answer - for what reason will it shift this way? Question 10 The equilibrium reaction below has the Kc = 3.93 at 25oC. If the temperature of the system at equilibrium is decreased to 10 oC, how and for what reason will the equilibrium shift? Also show and explain how and why the Kc value will change. CO (^) (g) + 3 H2 (g) CH 4 (g) + H 2 O (^) (g) ∆H^0 = -206.2 kJ Your Answer: Temperature is decreased in the reaction at equilibrium that has a - ∆H^0 , thus reaction shifts in the forward direction. The reaction shift in the direction that produces some heatIn this reaction, the concentration of the product increases and the concentration of the reactants decreases and thus increases the value of Kc.

4 4 4 2 4 3 3 4 Module 2: Question 1 Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula and explain your answer. (1) Cr(OH) 3 (2) HAsO 4 (3) CoCO 3 Your Answer:

1. base - contains Cr metal + OH polyatomic group
2. acid - contains H + polyatomic group (AsO 4 )
3. salt - contains Co metal + polyatomic group (CO 3 ) Question 2 For the Brønsted-Lowry acid base reactions shown below, list the stronger acid, stronger base, weaker acid and weaker base in the answer blanks provided: NH + + H PO - NH + H PO Stronger acid: Stronger base: Weaker acid: Weaker base: Your Answer: Stronger acid: H 3 PO 4 Stronger base: NH 3 Weaker acid: NH + Weaker base: H 2 PO

Question 3 For the compounds shown below, choose the stronger acid, explain why it is the stronger acid and write the formula of its conjugate base in the answer blank:

4 HClO 4 or HBrO 4 or HIO 4 Stronger acid is: Explanation: Formula of conjugate base of the stronger acid: Your Answer: Stronger acid is: HClO 4 > HBrO 4 > HIO 4 Cl is more electronegative than Br so HClO 4 is stronger acid than HBrO 4 and so is Br to I, Br is more electronegative than I so HBrO 4 is stronger than HIO 4. Stronger acid is: HClO 4 Explanation: Cl has the higher electronegativity (of Cl, Br or I) which makes the H-O bond of HClO 4 most polar and most likely to form H+ Formula of conjugate base of stronger acid: conjugate base of HClO 4 is ClO - Question 4 Show the calculation of the [H

] and pH of a 0.00350 M solution of the strong acid H 2 SO 4. Your Answer: [H+] = 2 x [H 2 SO 4 ] = 2 x (0.00350] [H+] = = 0.007 M pH = - log [H+] = - log ( 0.007) pH = 2. Question 5 In the titration of 15.0 mL of H 2 SO 4 of unknown concentration, the phenolphthalein indicator present in the colorless solution turns pink when 26.4 mL of 0.130 M NaOH is added. Show the calculation of the molarity of the H 2 SO 4.

2 3 2 a Your Answer: (Ma x mLa)/1000 x Sa/Sb = (Mb x mLb)/ (Ma x 15.0)/1000 x 1/2 = (0.130 x 26.4)/ MH2SO4 = 0.458 M ( H 2 SO 4 is the acid and Na OH is the base) Ma x mLa / 1000 x Sa / Sb = Mb x mLb / 1000 Ma x 15.0 / 1000 x 2 / 1 = 0.130 x 26.4 / 1000 MH2SO4 = 0.114 M ratio: 2 / 1 Question 6 Show the calculation of the [H+], pH and % ionization for 0.645 M acetic acid (HC 2 H 3 O 2 ) HC 2 H 3 O 2 H+^ + C H O -^ K = 1.8 x 10- Your Answer: HC 2 H 3 O 2 H+^ + C 2 H 3 O 2 - 0.645 0 0 -x +x +x 0.645-x x x Ka = 1.8 x 10- Ka = [C 2 H 3 O 2 - ] [H+] / [HC 2 H 3 O 2 ] 1.8 x 10-5^ = (x) (x) / (0.645-x) x^2 = 1.8 x 10 - (0.645) x = 3.41x10- =[H+]

2 3 2 2 3 2 4 pH = - log [H+] = - log (3.41x10-3) pH = 2. % ionization = ([C H O - ]/ [HC H O ]) x 100 % ionization = (3.41x10-3^ / 0.645) x 100 % ionization = 0.53% Question 7 Predict and explain whether a solution of LiF is acidic, basic or neutral. Your Answer: Basic since Li hydrolyzes to form strong base [LiOH]. LiF: Basic since F-^ hydrolyzes to form a weak acid (HF) and OH- Question 8 Show calculation of the pH of a buffer prepared by mixing 0.100 M NH 4 Cl and 0.0750 M NH 3. NH 3 + H 2 O (^) (liq) NH +^ + OH-^ Ka = 1.8 x 10- Your Answer: NH 3 + H 2 O (^) (liq) NH 4 +^ + OH- 0.0750 0.100 0 -x +x +x 0.0750-x 0.100+x x Ka = 1.8 x 10- Ka = [NH 4 +] [OH-] / [NH 3 ] 1.8 x 10-5^ = (0.100+x) (x) / (0.0750-x)

2 3 2 a 2 3 2 1.8 x 10-5^ = (0.100) (x) / (0.0750) x = 1.35x10-5^ = [OH-] pOH = - log [OH-] = - log (1.35x10-5) pOH = 4. pH = 14 - pOH = 14 - 4. pH = 9. Question 9 Show the calculation of the pH of a solution obtained by adding 0.0100 mole of OH-^ to a buffer of pH 4.74 which originally contains 0.100 M NaC 2 H 3 O 2 and 0.100 M HC 2 H 3 O 2. HC 2 H 3 O 2 H+^ + C H O -^ K = 1.8 x 10 - Your Answer: HC 2 H 3 O 2 H+^ + C H O - 0.100 0 0. +0.0100 mole OH

0.090 0 0. -x +x +x 0.090-x x 0.110+x Ka = 1.8 x 10- Ka = [H+] [C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] 1.8 x 10 -5^ = (x) (0.110+x) / (0.090-x ) 1.8 x 10-5^ = (x) (0.110) / (0.090) x = 1.47x10-5^ = [H+] pH = - log [H

] = - log (1.47x

) pH = 4. pH went up slightly fron 4.74 to 4.83 when OH-^ was added.

Question 10 List and explain what type of acid and base are involved in the titration curve below and select an appropriate indicator to be used in this titration from the list below. pH Range of Indicator Color Change Methyl violet 0.0 - 1. Phenolphthalein 8.0 - 9. Alizarin yellow R 10.1 - 12. Your Answer: Titration of a strong acid with a strong base because the location of the equivalence point is at pH 7. The rapid rise occurs approximately between pH 3 and pH 11. Indicator that can be used is limited to phenolphthalein because its range of color change is at 8.0 - 9.8. Methyl violet and Alizarin yellow R cannot be used for this titration. Module 3: Question 1 Show the calculation of the molar solubility (mol/L) of BaF 2 , Ksp of BaF 2 = 1.0 x 10

• . Your Answer: Ksp = [Ba+2] [F-]^2 = 1.0x10-

initial 1.0x10-6^ = [s] [2s]^2 1.0x10-6^ = 4s^3 s = 6.3x10-3^ mol/L Question 2 Show the calculation of the Ksp of MnS if the solubility of MnS is 0.0001375 g/100 ml. MW of MnS = 87. MnS (s) Mn+2^ (aq) + S-2^ (aq) Your Answer: molar solubility of MnS = (0.0001375 g/100 ml) x (1000 mL) x (1 mol/87.01 g MnS) molar solubility of MnS = 1.580x10- Ksp = [Mn+2] [S-2] Ksp = (1.580x

• ) (1.580x - ) = 2.496x - Question 3 Show the calculations and predict if a precipitate of PbI 2 will form if 1.0 x 10-4^ M Pb(NO 3 ) 2 is mixed with 1.0 x 10-3^ M NaI? The Ksp of PbI 2 = 7.1 x 10-9. Your Answer: Ksp = [Pb+2] [I-]^2 = 7.1 x 10 - Qc = [Pb+2]initial [I-]^2 = [1.0 x 10-4] [ 1.0 x 10-3]^2 Qc = 1.0x10- Qc (1.0x10-10) < Ksp (7.1 x 10-9) thus reaction should go in the forward direction (solution) and PbI 2 will not precipitate. Question 4 State and explain which has greater entropy: 1 mole of N 2 gas at 1 atm or 1 mole of N 2 gas at 2 atm. Your Answer:

1 mole of N 2 gas at 1 atm has greater entropy than 1 mole of N 2 gas at 2 atm because the increase in pressure will cause compression of the particles of nitrogen gas at 2 atm into a smaller space and decreases the entropy. Question 5 Explain whether this physical reaction results in an increase or decrease in entropy. H 2 O (s) → H 2 O (l) Your Answer: The reaction is from solid H 2 O to liquid H 2 O and liquid has greater entropy because it has more available motions than solid. Thus, reaction from solid to liquid increases the entropy. Question 6 Determine ΔS^0 for the following reaction using the data given. Explain whether this value agrees with prediction of the ΔS 0 value. ΔS 0 C 2 H 6 = 229.5 J/mole, ΔS 0 O 2 = 205.5 J/mole, ΔS 0 CO 2 = 213. J/mole, ΔS^0 H 2 O = 188.7 J/mole 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) Your Answer: ΔS 0 = ΔS 0 products - ΔS 0 reactants ΔS 0 = [4 ΔS 0 CO 2 + 6 ΔS 0 H 2 O] - [2 ΔS 0 C 2 H 6 + 7 ΔS 0 O 2 ] ΔS^0 = [(4x213.7) + (6x188.7)] - [(2x229.5) + (7x205.5)] ΔS^0 = 89.5 J/mol Reaction has a higher entropy and system is more disorder because ΔS^0 is positive. Question 7 Determine ΔG^0 for the following reaction using the data given and use it to predict the spontaneity of the reaction. ΔGf^0 C 2 H 6 = - 32.9 kJ/mole, ΔGf^0 CO 2 = -394.4 kJ/mole, ΔGf^0 H 2 O = - 228.6 kJ/mole 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) Your Answer:

ΔGf^0 = ΔGf^0 products - ΔGf^0 reactants ΔGf 0 = [(4 ΔGf 0 CO 2 ) + (6 ΔGf 0 H 2 O)] - [(2ΔGf 0 C 2 H 6 ) + (7 ΔGf 0 O 2 )] ΔGf^0 = [(4x-394.4) + (6x-228.6)] - [(2x-32.9) + (7x0)] ΔGf 0 = -2883.4 kJ/mol Reaction is spontaneous since ΔG is a larger negative number than -20 kJ. Question 8 Using the values of ΔH^0 and ΔS^0 given, determine the value of ΔG^0 for the reaction below at 25 oC. ΔH^0 = 311.4 kJ /mole and ΔS^0 = 232.6 J /mole. Explain whether this reaction is spontaneous as written. C 2 H 6 (g) → C 2 H 2 (g) + 2 H 2 (g) What would be the value of ΔG^0 for the reaction at 900oC? Explain whether this reaction is spontaneous as written at 900 oC. Your Answer: ΔG^0 = ΔH^0 - TΔS^0 ΔG^0 = 311.4 kJ/mol - (25+273K)(232.6/1000 kJ/mol) ΔG^0 = 242.1 kJ/mol Reaction is nonspontaneous at 25 o C since ΔG is a larger positive number than +20 kJ. ΔG^0 = ΔH^0 - TΔS^0 ΔG^0 = 311.4 kJ/mol - (900+273K)(232.6/1000 kJ/mol) ΔG^0 = 38.56 kJ/mol Reaction is nonspontaneous at 900 oC since ΔG is a larger positive number than +20 kJ. Question 9 The formation of Mn from MnS is nonspontaneous as written in reaction 1 below but can be accomplished by coupling it with the oxidation of S to SO 2 (shown in reaction 2 below). Show the overall spontaneous reaction resulting from the coupling of these two reactions and calculate the ΔG^0 of the overall reaction. (1) MnS (s) → Mn (s) + S (g) ΔG^0 = + 214.2 kJ (2) S (s) + O 2 (g) → SO 2 (g) ΔG^0 = - 300.1 kJ

Your Answer: MnS (s) + O 2 (g) → Mn (s) + SO 2 (g) ΔG^0 = -85.9 kJ Question 10 Using the table and images (A - F) below, predict, and explain by doing a Qc vs Ksp calculation, which tube shows the result when colorless 1 x 10 -6^ M AgNO 3 solution was added to colorless 1 x 10 -6^ M NaBr solution. A B C D E F Substance Formula Color K sp Copper (II) carbonate CuCO 3 Blue 1.4 x^10

Lead (II) chromate PbCrO 4 Yellow (^) 2.8 x 10 - Silver bromide Your Answer: AgBr White (^) 5.0 x 10 - Ksp = [Ag+] [Br-] = 5 .0 x 10- Qc = [Ag+]initial [Br-]initial = [1.0 x 10-6] [ 1.0 x 10 -6] Qc = 1.0x10- Qc (1.0x10-12) < Ksp (5 .0 x 10-6) thus the reaction should go in the forward direction (solution) and AgBr will not precipitate. Image B shows the result of when colorless AgNO 3 solution was added to colorless NaBr solution because it does not precipitate, its color is still colorless.

4. E (^) cell = E - E Qc = (1 x 10 -6) x (1.0 x 10 -6) = 1 x 10 - Ksp for AgBr = 5.0 x 10

Since Qc > Ksp a white precipitate will form = E Module 4: Question 1 For the cell described by the following cell diagram. Cr (s) | Cr+3^ (aq, 1 M) || Al+3^ **(aq, 1 M) | Al (s) Al

• 3e

→ Al E 0 = - 1.66 v Cr+3**^ + 3e-^ → Cr E^0 = - 0.73 v (1) The anode half reaction is (2) The cathode half reaction is (3) The overall cell reaction is (4) Show the calculation for the total cell potential (5) State and explain whether the cell is voltaic or electrolytic Your Answer:

1. Cr (s) → Cr

(aq) + 3e

2. Al+3^ (aq) + 3e-^ → Al (s)
3. Cr (s) + Al+3^ (aq) → Cr+3^ (aq) + Al (s) 0 0 0 cathode anode E^0 cell = -1.66 - (-0.73) E^0 cell = -0.93 v
4. Since the total cell potential is negative, cell will not occur spontaneously and would have to be electrolytic. Question 2 In the following reaction, identify the following and explain your answers Zn + 2 H+^ + 2 Cl-^ → Zn+2^ + 2 Cl-^ + H 2

(1) the material undergoing oxidation (2) the material undergoing reduction (3) the oxidizing agent (4) the reducing agent Your Answer: Zn + 2 H+^ + 2 Cl-^ → Zn+2^ + 2 Cl-^ + H 2 loses e gains e

1. Zn
2. H

3. H+
4. Zn Question 3 Complete the following two statements: A. The lead-acid battery is called a battery because it is rechargeable. B. The reactants are fed into a battery. Your Answer: A. secondary or storage B. fuel cell Question 4 For the voltaic cell described by the cell reaction below, (1) write the cell diagram, (2) identify which specie is being reduced, (3) which specie is being oxidized, (4) identify the anode and (5) identify the cathode. Al + Cu+1^ (0.30 M) → Al+3^ (0.10 M) + Cu Your Answer:
5. Al (s) I Al

(aq, 0.10 M) II Cu

(aq, 0.30 M) I Cu (s)

2. Cu

is reduced to Cu

3. Al is oxidized to Al+
4. Al is the anode

cell cell cell cell cell cathode anode

5. Cu is the cathode Question 5 For the voltaic cell described by the following cell reaction: Ca + Ag+1^ (0.20 M) → Ca+2^ (0.10 M) + Ag Ca+2^ + 2e-^ → Ca E^0 = - 2.76 v Ag+1^ + e-^ → Ag E^0 = 0.80 v (1) calculate the value of Q (2) using the given standard reduction potentials, show the calculation of the Ecell using the Nernst equation [Ecell = E 0

• (0.0592/n x log Q)]. Your Answer:

1. Q = [ Ca + ] / [ Ag + ] 2 = 0.10 / (0.20) 2 = 2.

E^0 = 0.80 - (-2.76) v = 3.56 v Ecell = E^0 – (0.0592/n x log Q) Ecell = 3.56 - (0.0592/2 x log 3.56) Ecell = 3.56 - (0.0592/2 x 0.551) Ecell = 3.544 v (1) calculate the value of Q Q = Product ion / Reactant ion = [Ca+2] / [Ag+1] = (0.10) / (0.20) = 0. (2) using the given standard reduction potentials, show the calculation of the Ecell using the Nernst equation [Ecell = E^0 – (0.0592/n x log Q)]. E^0 = E - E = 0.80 - (- 2.76) = 3.56 v Ecell = 3.56 - (0.0592/2 x log 0.5) = 3.56 - (0.0592/2 x - 0.30) = 3.56 + 0.0089 = 3.5689 v (0.10) / (0.20) = 0.

Question 6 Answer the following multiple choice questions:

1. The ground state of has an electron configuration of 1s^1. A. Hydrogen B. Nitrogen C. Carbon D. Sodium E. Iron
2. According to the course materials, which of the following planets has a significant amount of hydrogen in the atmosphere due to its large gravitational pull? A. Venus B. Mars C. Earth D. Jupiter E. Uranus
3. Which of the following is known as protium? A. 1 P B. 1 H C. 2 H D. 3 H E. 2 P
4. Which of the following is a mixture of metals? A. Mineral B. Ore C. Alloy D. Metallurgy E. None of the above Question 7 Answer the following multiple choice questions:
5. Smelting is a type of processing of metals. A. Electrometallurgy B. Hydrometallurgy C. Pyrometallurgy

D. Thermometallurgy E. None of the above

2. Which of the following indicates that metals can be pounded into shape? A. Ductile B. Malleable C. Luster D. Electrical conductivity E. Thermal conductivity
3. Which of the following is not an alkali metal? A. Mg B. Na C. Li D. K E. All of the above are
4. What gas is evolved when alkali metals react with water? A. Br 2 B. O 2 C. Cl 2 D. N 2 E. None of the above Question 8 Answer the following multiple choice questions:
5. Which alkali compound is used in water softening? A. MgF 2 B. BaCO 3 C. Na 2 CO 3 D. LiNO 3 E. All of the above
6. Gunpowder contains which alkali compound? A. KNO 3 B. KCl C. NaF

D. NaNO 3 E. LiNO 3

3. Which compound is the crucial ingredient in milk of magnesia? A. CaCO 3 B. MgCl 2 C. Mg(OH) 2 D. Ca(OH) 2 E. CaO
4. When magnesium burns in air, what is one compound produced? A. MgH 2 B. MgI 2 C. Mg 2 N 3 D. Mg 2 P 3 E. None of the above Question 9 Answer the following multiple choice questions:
5. Tin and lead tend to form compounds with and charges. A. +1, +2 B. +2, +3 C. +2, +4 D. -2, -4 E. -1, -2
6. Which of the following is not a characteristic of nonmetals? A. Brittle B. Poor electrical conductors C. Covalent bonding D. Forms only – charges E. None of the above
7. Carbon exists as two allotropes, and. A. graphite; ozone B. Coal; oil C. Carbon monoxide; carbon dioxide

D. Oil; graphite E. Graphite; diamond

4. The solid form of is called dry ice and is non-toxic. A. Baking soda B. Carbon monoxide C. Baking powder D. Carbon dioxide E. Graphite Question 10 Answer the following multiple choice questions:
5. The nonmetal forms the azide ion, which is used in airbags of cars. A. Nitrogen B. Carbon C. Oxygen D. Phosphorus E. Chlorine
6. The following allotrope of phosphorus is used in incendiary bombs. A. Calcium phosphate B. Tetraphosphorus hexaoxide C. White phosphorus D. Red phosphorus E. None of the above
7. The following nonmetal has six valence electrons as a neutral atom. A. phosphorus B. chlorine C. Sulfur D. Both A and B E. Both B and C
8. The two common oxides of sulfur are: A. SO 2 and SO 3 B. SO 3 and SO 4 C. SO 2 and SO 4 D. SO and SO 2 E. SO 3 and SO 5

Module 5: Question 1 Hexene is a hydrocarbon containing (#) carbons and a (type) bond. Your Answer: 6 C=C double Question 2 The compound below is a type of hydrocarbon called a/an. Your Answer: aromatic hydrocarbon Question 3 The compound below is called (#s)-dimethyl. Your Answer: 1,4 cyclohexane Question 4 The compound below contains a/an functional group. Your Answer: carboxylic acid Question 5 An ester contains a bond with a/an attached to the C. Your Answer: carbonyl (C=O) OR (alkoxy group)