F321 Updated Unofficial Mark Scheme Watch

its mee
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(Original post by Mr van der WAALs)
An updated version of the F321 mark scheme ~ Friday 21st May 2010

F321 ~ Unofficial Mark Scheme (Updated version)

Answers to Question 1:

a) 50 68 50 ~ proton, neutrons and electrons of Tin (Sn) [1]

b) The weighted mean mass; of an atom of an element compared; to 1/12th of the carbon-12 isotope; [3]

c) 120 Sn has 2 more neutrons than 118 Sn [1]

d) 2.08kg = 2080/118.7 =17.52
17.52 x 6.02 x 10^23 =1.05 x10^25 [2]

e) 78.8% of Sn and 21.2% O2; therefore SnO2 [2]

Answers to Question 2:

a) H+ and SO4 2- [2]

b) bubbles/fizzing of gas (CO2); solid carbonate dissolves;

K2CO3 + H2SO4 = K2SO4 + CO2 + H2O; [3]

c) 2.47 x 10^-3 [1]

d) 4.94 x 10^-3 [1]

e) Since 2.00g of soda was used to make a 250cm^3 solution; then 25cm^3 was titrated against 24.70cm^3 of 0.100mol dm^3 H2SO4;

The calculation for the % mass was:

moles of NaOH ~ n= 2.00/40 = 0.05mol
moles of the amount titrated ~ n= (24.7 x 0.1)/1000 = 2.47 x 10^-3

2:1 reaction, hence; 4.94 x 10^-3 moles used in 25cm^-3
Therefore x10 gives 250cm^-3 = 0.0494

Then; 0.0494/0.05 = 0.988

0.988 x 100 = 98.8% [3]

Answers to Question 3:

a) 4s then 3d [1]

b) Orbital~ where two electrons 'spin'; opposite directions; eq [1]

c) 11 electrons in p-orbitals [1]

d) Ca 2+ = 18 electrons [1]

e) 2nd, 3rd, 10th and 11th [2]

f) Al (2+) (g) = Al (3+) (g) + e- [2]

Answers to Question 4:

a) purple/violet; orange [2]

b) displacement between Br and I; eq [1]

c) Cl simultanously oxidised and reduced;

From 0 to +1 in HClO;
From 0 to -1 in HCl; [3]

ii) Cl kills bacteria; eq [1]

d) thermal decomposition [1]

e) Calculation involving a carbonate; find volume of CO2;

I cannot really remember what my answer was; something like 17.4dm3 [2]

f) Thermal stability increases down the group; eq [1]

Answers to Question 5:

a) Lithium has giant metallic structure; diagram etc;

Fixed cations; surrounded by delocalised electrons; tightly packed/strong attractions; eq ~ high boiling point [3]

b) Fluorine covalently bonded; 8 electrons for each one [ 3 x lone pair]; one shared pair; eq [1]

c) LiF ~ Ionic structure ~ Li electron in outer shell completes the incomplete shell to leave Li with a full shell; and a + charge []; F with a full shell (show by dots and crosses); and a - charge []; eq [2]

d) Fails to conduct when solid; ions are fixed; conducts when molten; ions are delocalised; [2]

e) 2B + 3[F2] = 2[BF3] [1]

f) Trigonal planar; 120 degrees; 3 bonded pairs and no lone pairs; eq ~ repulsion theory explained; eq [4]

g) Permeanent dipoles arise due to difference in electronegativity; F very electronegative; electrons pulled closer; eq [3]

h) Atomic radii decrease across period 2; increased proton number; therefore increased nuclear charge; electrons added to same shell; similar shielding; hence atomic radii decrease; shells/electrons pulled closer/more strongly to nucleus; ionisation increases; eq [4]



heyy thanks for the uploading the mark scheme i was just doing this past paper and on question 3(a) where you have to complete the sub-shell the paper already has 1s 2s 2p 3s 3p 4s then you have to fill out the next 2? i was just wondering do you still put 4s before 3d?
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Contrad!ction.
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(Original post by its mee)
heyy thanks for the uploading the mark scheme i was just doing this past paper and on question 3(a) where you have to complete the sub-shell the paper already has 1s 2s 2p 3s 3p 4s then you have to fill out the next 2? i was just wondering do you still put 4s before 3d?
1s 2s 2p 3s 3p 3d 4s etc is the way you should write the configuration in the exam but they're filled as 1s 2s 2p 3s 3p 4s 3d (normally).
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