After black and white photographic film has been developed, unreacted silver bromide is removed by reaction with sodium thiosulfate.

Question 19 [Group 7 > reactions of halide ions]

After black and white photographic film has been developed, unreacted silver bromide is removed by reaction with sodium thiosulfate.

AgBr + 2Na2S2O3 → 4Na⁺ + Br ^– + [Ag(S2O3)2]^3–

What is the function of the thiosulfate ion?

A to make the silver ions soluble

B to oxidise the silver ions

C to reduce the bromine

D to reduce the silver ions

Reference: Past Exam Paper – Chemistry (9701) March 2016 Paper 12 Q19

Solution:

     Answer:A
 

     Firstly, if confused work out oxidation changes for both silver and bromine to see if oxidation or reduction has taken place. Neither oxidation or reduction takes place and so automatically the only feasible answer is A.

   Secondly, AgBr is a precipitate, which you should be able to recall from qualitative tests and other chapters from inorganic chemistry. We see that Ag is then incorporated into this complex on the products side of the equation. The formula is enclosed in square brackets and we also observe a charge of -3. Automatically we see that this complex is in ionic form and therefore silver has gone from an insoluble solid to a soluble ion.

A rod PQ is attached at P to a vertical wall, as shown in Fig. 3.1. The length of the rod is 1.60 m. The weight W of the rod acts 0.64 m from P.

Question 42

A rod PQ is attached at P to a vertical wall, as shown in Fig. 3.1.

The length of the rod is 1.60 m. The weight W of the rod acts 0.64 m from P. The rod is kept horizontal and in equilibrium by a wire attached to Q and to the wall at R. The wire provides a force F on the rod of 44 N at 30° to the horizontal.

(a) Determine

(i) the vertical component of F,          [1]

(ii) the horizontal component of F.     [1]

(b) By taking moments about P, determine the weight W of the rod.           [2]

(c) Explain why the wall must exert a force on the rod at P.             [1]

(d) On Fig. 3.1, draw an arrow to represent the force acting on the rod at P. Label your arrow with the letter S.             [1]

Reference: Past Exam Paper – June 2015 Paper 22 Q3

Solution:

(a)

(i) Vertical component = 44 sin 30° = 22 N

(ii) Horizontal component = 44 cos 30° = 38(.1) N

(b)

{The vertical component of F acts at a distance of (0.64 + 0.96 =) 1.60m from the pivot.

Clockwise moment = Anti-clockwise moment}

W × 0.64 = 22 × 1.60

W = 55 N

(c) F has a horizontal component (not balanced by W)

{The horizontal component of F acts on the wall and this is not balanced by W. From Newton’s 3^rd law, there should be an equal and opposite force exerted by the wall on the rod.}

OR F has 38 N acting horizontally

OR 38 N acts on wall

OR vertical component of F does not balance W

OR F and W do not make a closed triangle of forces

(d) Line from P in the direction towards the point on wire vertically above W and direction up

{This force, along with F and W should form a system that is in equilibrium. For equilibrium, the resultant force and the resultant moment should be zero. If all the 3 forces pass through the same point, the resultant moment would be zero.

Moment = Force × perpendicular distance from line of action of force to pivot.

In this case, the pivot is that point where all the 3 forces pass. As they pass on the point, the ‘distance …’ is zero, and thus, the moment at that point is zero.}

Solid carbon dioxide, CO2, is similar to solid iodine, I2, in its structure and properties. Carbon is in Group 14. Silica, SiO2, is a Group 14 compound.

Question 8 [Chemical Bonding >3.2 Covalent bonding and co-ordinate (dative covalent) bonding including shapes of simple molecules]

Solid carbon dioxide, CO2, is similar to solid iodine, I2, in its structure and properties. Carbon is in Group 14. Silica, SiO2, is a Group 14 compound.

Which statement about solid CO2 and solid SiO2 is correct?

A Both solids exist in a lattice structure.

B Both solids have a simple molecular structure.

C Both solids have atoms joined by single covalent bonds.

D Both solids change spontaneously to gas at s.t.p.

Reference: Past Exam Paper – Chemistry (9701) March 2016 Paper 12 Q8

Solution:

     Answer: A

     Option A:  “It should be known that solid I₂ has a simple molecular lattice, therefore it can be deduced from the information in the question that solid CO₂ does too”

     Option B can be discredited since we know that SiO₂ has a similar structure to diamond, i.e, a giant covalent structure, not simple molecular.

    Option C can be eliminated. While SiO₂ has four single bonds, CO₂ has two double bonds and is linear rather than tetrahedral like SiO₂.

    Option D can be eliminated since at stp, SiO₂ is a solid and will not change spontaneously (naturally) into a gas.