### HighFour Chemistry Round 5 Category D: Grades 11 – 12 Tuesday

```HighFour Chemistry Category D: Grades 11 – 12 Round 5 Tuesday, January 10, 2017 The use of calculator is required. Answer #1: Explanation: Answer #2: Solution: ethylbenzene Ethylbenzene is an organic compound that is formed (along with hydrochloric acid) with benzene and ethyl chloride. It is a highly flammable liquid and is an important component in the manufacture of styrene. 1.58 molal To calculate the molality of the aqueous solution, we apply the formula \$%&'()*
m = . First, we calculate the number of moles from the +, -. /-012\$3
molarity: n/-0532 = 1.55
9-0
: /-0\$
1 L = 1.55 moles To solve for the mass of the solvent in kilogram, we determine the mass of the solution and the solute (MW of NaOH = 40 g/mol): m/-0\$ = 1.04
m/-0532 = 1.55 mol
9:
1000 mL = 1040 grams ,
The mass of the solvent is: Therefore, the molality of the solution is: 40
9-0
= 62 grams m/-012\$3 = 1040 g − 62 g = 978 g m = ,
(1.55 moles)
978 g M +,
MNNN ,
= 1.58 molal HighFour Chemistry Category D: Grades 11 – 12 Round 5 Tuesday, January 10, 2017 The use of calculator is required. Answer #3: Solution: pH = 0.086 The pH of the solution can be solved using the formula pH = -­‐‑log [H+]. First, we determine the concentration of the solution. The MW of HCl is 36.46 g/mol: HCl = H V =
V
= 0.82
750 mL MNNN 9:
2.56 seconds Determination of the time is simply a direct substitution of values into the equation hinted. M
t=
:
XY.ZY ,
M :
The pH of the solution is pH = − log 0.82 = 0.086 Answer #5: Solution: 9-0
M 9-0
22.3 g \)
−
M
M
\]
N.XN _
−
M
N.YN _
aM
0.65 M ∙ s aM
=
k
t = 2.56 seconds 3.20 M Solving the problem requires 3 steps. First, we determine the mass of the solution: ,
m/-0\$ = 0.984
1000 mL = 984 g Then we solve for the mass of the solute: m/-0532 = 0.15 984 g = 147.6 g 9:
We can now solve the molarity of the ethanol solution (MW = 46.08 g/mol): M=
147.6 g M 9-0
ZY.Ne ,
1 L soln
= 3.20 𝑀 HighFour Chemistry Category D: Grades 11 – 12 Round 5 Tuesday, January 10, 2017 The use of calculator is required. Answer #6: Solution: Answer #7: Solution: Kc = 0.29 The equilibrium constant Kc for the synthesis of ammonia is as follows: [NHX ]
𝐾h =
k
m Nj l Hj l
(0.25 M)
𝐾h =
k
m = 0.29 (0.11 M)l (1.91 M)l
44.50 ppm 9n// /-0532
Use the formula ppm = 9n// -. /-053o-\$ x 10Y . The concentration of the solution in ppm is: ppm = Answer #8: Explanation: Answer #9: Solution: 44.5 mg M ,
MNNN 9,
1000g + 44.5 mg M ,
x 10Y = 44.50 MNNN 9,
sulfur dioxide Sulfur dioxide is a dangerous primary pollutant which can harm people, plants and materials. It is produced as a secondary pollutant by the oxidation of hydrogen sulfide: 2H2S (g) + 3O2 (g) ⟶ 2SO2 (g) + 2H2O (l) pOH = 7.04 To solve for the pOH of the solution, use the formula pH + pOH = 14 𝑝𝑂𝐻 = 14 − 6.96 = 7.04 HighFour Chemistry Category D: Grades 11 – 12 Round 5 Tuesday, January 10, 2017 The use of calculator is required. Answer #10: Explanation: 2-­‐bromo-­‐2-­‐chloropropane In naming organic compounds, follow these steps: 1) Identify the longest straight chain of carbon atom, which is 3 carbon atoms, and assign the stem of the name: prop-­‐ 2) Identify the functional group: alkane 3) Identify the side chain and functional group: bromo-­‐ and chloro-­‐ Note that if there are difference groups on the same atom, they are put in alphabetical order. Therefore, the name of the molecule is 2-­‐bromo-­‐2-­‐
chloropropane. Answer #11: Solution: Answer #12: Explanation: Answer #13: Explanation: Answer #14: Solution: 218.2 kJ/mol The decomposition of hydrogen gas is the formation of 2 moles of atomic hydrogen. The standard enthalpy of formation of atomic hydrogen is (436.4 kJ ÷ 2 mols) = 218.2 kJ/mol. Kevlar This plastic is also being used in military armor, fishing rods, and running shoes. It is a condensation polymer and has a structure similar to nylon-­‐
6,6 with the carbon chain replaced by benzene. Kp = 0.255 No solution is required in solving for the Kp for the reaction since the concentration of the solid (CaCO3) does not appear in the equilibrium constant expression. Therefore, 𝐾v = 𝑃xyj = 0.255 89 Joules The change in energy of the gas is simply the difference between the work done and the energy (heat) released. Therefore, the net change in energy is 122 J – 33 J = 89 J. HighFour Chemistry Category D: Grades 11 – 12 Round 5 Tuesday, January 10, 2017 The use of calculator is required. Answer #15: Explanation: Answer #16: Solution: fluorine and chlorine Fluorine is the lightest halogen and exists as a pale yellow gas; chlorine is the second-­‐lightest halogen and exists as a yellow-­‐green gas at room temperature. 1.37 femtometer Before using the formula to solve for the wavelength, the constants are as follows (all can be found in the Chemistry Data Booklet): Planck’s constant, h = 6.63 x 10aXZ J ∙ s mass of proton = 1.6726 x 10-­‐‑27 kg The wavelength is therefore: λ=
Answer #17: Solution: 6.63 x 10−34 J ∙ s
1.6726 x 10
−27
λ = 1.37x10aM• m kg
𝑚
2.90x108
𝑠
M .9
M€MN•k‚ 9
[Table 2] [Table 4] = 1.37x10aM• m = 1.37 fm 30.52 g/mol The molar mass of the gas can be calculated using the ideal gas equation: :∙n39
PV = nRT. Use the universal gas constant, R = 0.08205 which can 9-0∙ˆ
be found in Table 2 of the Chemistry Data Booklet. L∙atm
6.90 g 0.08205 35 + 273 K
mRT
mol∙K
MW =
=
1atm
PV
720 torr (6.03 L)
,
MW = 30.52 9-0
760torr
HighFour Chemistry Category D: Grades 11 – 12 Round 5 Tuesday, January 10, 2017 The use of calculator is required. Answer #18: Solution: 359.57 L of O2 To solve for the volume of O2, we break down the solution into three parts. First, solve for the number of moles of ethanol (MW = 46.08 g/mol): n = 227 g M 9-0
ZY.Ne ,
= 4.93 mols The moles of oxygen is therefore: The volume of oxygen is: n Oj = 4.93 mol ethanol X 9-0/ ‹l
= 14.79 mols M 9-0 23Œn\$-0
:∙n39
14.79 mols 0.08205 35 + 273 K
nRT
9-0∙ˆ
V=
=
Mn39
P
790 mmHg •YN99Ž,
Answer #19: Solution: Answer #20: Solution: V = 359.57 L pH = 4.87 In solving the pH of the solution, use the formula pH = -­‐‑log [H+]. pH = −log (1.35x10a• ) pH = 4.87 Kc = 1.19 To solve for the Kc, use the formula 𝐾v = 𝐾h (0.0821 ∙ T)∆\$ . The value of Δn can be obtained from the chemical reaction: Δn = 2 – (3 + 1) = -­‐2 The equilibrium constant is therefore: 𝐾v
4.2 x 10aZ
𝐾h = =
(0.0821 ∙ T)∆\$
0.0821 ∙ 648
𝐾h = 1.19 aj
```