Question

Part III-Long Answer Questions - (approx.. 45 min) Answer these on looseleaf paper in full. Show all of your work including equations, logic, significant figures and units. Long Answer PROBLEM 100 marks): When backpacking in the wilderness, hikers often boil water to sterilize it. Suppose you are hiking and need to boil 25.0 L of water for your group, but only 15% of energy released by fuel can be used for heat, how many litres (L) of fuel (C-H16) do you need to bring? Here is the data you'll need: > Density of fuel (C7H16) -o.780 g/mL >Standard AH r (CH16)--224.4 kJ/mol >The Combustion of the fuel produces carbon dioxide and water vapour > AH°r values can be found in your text/workbook. > Assume you only need to bring the water from 25.0 °C to 100.0 °C, you do not need to consider heat of vaporization See Questions 21 -25 for hints on solving this problem Long Answer PROBLEM 2 (5 marks) a) Give the complete ground state electron configuration (do not use the rare gas notation) of the following elements and state the number of unpaired electrons present in each: F,Ag. As b) What are the condensed electron configurations for two common oxidation states of Lead? c) Give two examples of elements or ions that are isoelectronic with each of the lead ions from part b)

Answer 1

Let the volume of fuel be V mL

density of the fuel = 0.780 g / mL

mass of the fuel = ( volume of fuel ) $\times$ ( density of the fuel )

= 0.780 V g

molar mass of C₇H₁₆ = 7 $\times$ 12 + 16 $\times$ 1

= 100 g / mol

number of moles of C₇H₁₆ =

masso ftheC7H16 molarmassofC7H16

=

0. 780У 100

= 7.8 V $\times$ 10^-3 moles

$\Delta$H⁰_f of the fuel = -224.4 kJ / mol

energy released from combustion of fuel = number of moles of fuel $\times$$\Delta$H⁰_f of the fuel

= 7.8 V $\times$ 10^-3 moles $\times$ 224.4 kJ / mol

= 1.75 V kJ

since only 15 % of energy is useful

The amount of Usable energy =

15 100 energyreleasedfromcombustionof fuel

=

15 × 1.75VkJ 100

= 0.2625 V kJ

Specific heat of water = 4.186 J / g °C

density of water = 1 kg / L

volume of water = 25.0 L

Change in temparature ( $\Delta$T ) = Final temparature - initial temparature

= 100⁰C - 25⁰ C

= 75⁰C

amount of energy required to raise the temparature of water from  25⁰ C to 100⁰C =volume of water $\times$ density of water $\times$ Specific heat of water $\times$ Change in temparature

= 25.0 L$\times$ 1 kg / L$\times$ 4.186 J / g °C  $\times$ 75⁰C

= 7848.75 kJ

amount of energy required to raise the temparature of water from  25⁰ C to 100⁰C =The amount of Usable energy

7848.75 kJ = 0.2625 V kJ

V7848.75 0.2625

V = 29900 mL

V = 29.9L

2)

ground state electronic configuration of F is 1s² 2s² 2p⁵

ground state electronic configuration of Ag is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰

ground state electronic configuration of As is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p³

b)

The two oxidation states of lead are +2 and +4

electronic configuration of Pb⁺² is (Xe) 4f¹⁴, 5d¹⁰, 6s²

electronic configuration of Pb⁺⁴ is (Xe) 4f^14, 5d¹⁰

c)

Mercury ( Hg ) has an atomic number of 80 and is iso electronic with Pb⁺²

Platinum ( Pt ) has an atomic number of 78 and is iso electronic with Pb⁺⁴