Paper 1-2017

Question 1

Enthalpy lattice formation

Answer 1

Solution= -lattice formation +lattice hydration

Question 2

Why does enthalpy of lattice formation based on perfect ionic model give smaller numerical value than calculated value?

Answer 2

-Value calculated contains covalent character
-Therefore forces holding lattice are stronger

Question 3

Indicate presence of iodide ions in aq. sol

Answer 3

AgNo3
Yellow ppt forms

Question 4

CaOH2 was a saturated solution. Students added same moles of MgOH2 to water and stirred till no more solid dissolved. Predict if PH of MgOH2 is larger, smaller or same as CaOH2?

Answer 4

-Smaller
-MgOH2 is less/sparingly soluble

Question 5

Test tube reaction to distinguish sodium oxide and phosphorus oxide?

Answer 5

React with water and add:
-UI: Sodium- Blue + phosphorus-Red
-PH: Sodium-Ph12-14 + Phosphorus-1-2

Question 6

Sulfur trioxide with KOH

Answer 6

SO3 + 2KOH—> K2SO4 +H2O

Question 7

Excess MgO with H3PO4?

Answer 7

3MgO+ 2H3PO4—> Mg3(PO4)2+ 3H2O

Question 8

Type of boning between oxygen and hydrogen in water

Answer 8

Covalent

Question 9

Why cl- ion not considered ligand? (Outside the complex)

Answer 9

Cl- doesn’t done lone pairs or form co ordinate bonds go Cu2+

Question 10

Reagent D and fed to solution A to form yellow green solution.

Answer 10

-Reagent D: HCl
Equation:
[Cu(H2O)6]2+ + 4Cl- —> [CuCl4]2- +6H2O

Question 11

Why calorimeter can’t be used to determine conc. of solution of [CuCl2]-?

Answer 11

-Has full 3d subshell
-Colourless therefore can’t absorb frequencies of visible light.

Question 12

Distinguish between aq. sol. of KNO3 and K2SO4?

Answer 12

Reagent: BaCl2/ Ba(OH)2
Obs. KNO3: colourless sol/ no change
Obs. K2SO4: White ppt

Question 13

Distinguish aq. sol of MgCl2 and AlCl3?

Answer 13

R: NaOH
Obs with MgCl2: white ppt
Obs with AlCl3: white ppt in excess NaOH

Question 14

Deduce optimum conditions for industrial conversion of methane and steam into gaseous products?

Answer 14

1. Yield increases as temp increases. After certain temp yield no longer increases. Yield decreases as pressure increases.
   2.High temps=high costs. After certain temp yield no longer increases therefore no gain in using really high temps. Optimum temp: 780-880 celcius
2. Low pressure may be too slow so comprise pressure requires. Optimum- 1000-2000KPa.

Question 15

Weakest reducing agent

Answer 15

Strongest=most negative on right hand side. Therefore weakest is most positive on RHS

Question 16

What can oxidise copper?

Answer 16

-HNO3 as E0 for HNO3> E0 for Cu

Question 17

Why is enthalpy change for reaction close to 0?

Answer 17

-6Fe-Fe bonds broken then made. Same no. of bonds broken then made.

Question 18

Type of isomerism in Fe(ethanedioate)3 ligand sub?

Answer 18

-Optical isomerism
-90° bond angle