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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3

Work Energy and Power

  Work ; This is the product of force and the displacement in the direction of the force. Work done = force x distance moved in the direction of force The unit of work done is Joule (J) How to Calculate Work Done Example Questions 1.         If a car of mass 80kg moved a distance of 5m. calculate the work done by the car (g =10m/s 2 )   Solution Mass = 80kg Force = 80 x 10 = 800N Distance = 5m Work done = 800 x 5 = 4000J 2.        A boy pulls a load of 300N to a distance of 6m. Calculate the work done   Solution Force = 300N Distance = 6m Work done = 300 X 6 = 1800J Energy : This is the ability to do work. It is also measured in Joule (J). Energy has many forms which include: ·          Heat energy ·          Light energy ·          Chemical energy ·          Electrical energy ·          Atomic energy ·          Solar energy ·          Mechanical energy We are going to be studying mechanical energy. Mechanical energy is divided i

Titration of acid and base (calculation)

 Titration is the method employed in volumetric analysis. In this method, an unknown concentration of a liquid substance is titrated against another of known concentration. We are going to learn how to calculate the concentration of the substance using the question below. A is a dilute tetraoxosulphate (VI) acid. B contains 2g of sodium hydroxide per 250 cm3 of solution. 25cm 3 of B requires 23cm 3 of A for complete neutralization. Calculate i.                      Concentration of solution B in mol/dm 3 ii.                    Concentration of A in mol/dm 3 iii.                   The number of hydrogen ions in 1dm 3 of solution A (Na=23, H=1, S=32, O=16) Equation of reaction H 2 SO 4(aq)   +  2NaOH (aq) → Na 2 SO 4(aq)   + 2H 2 O (l) Solution i.                      Concentration of solution B in mol/dm 3 250cm 3    = 2g 1000cm 3 = 2/250 x1000 = 8g Molar mass of NaOH = 40g/mol Concentration in mol/dm 3    = 8/40   = 0.2mol/dm 3 Or C = n/V Wher

Graham’s Law of Diffusion calculation

 Graham’s law states that any gas's diffusion rate at constant temperature and pressure is directly proportional to the square root of its relative molecular mass or vapour density. R 1 /R 2 = √ m 2 / √ m 1                    or t 2 /t 1 = √ m 2 / √ m 1 Example Questions 1.        If 200cm 3 of hydrogen diffused through a porous pot in 30 seconds. How long will it take 400cm 3 of oxygen to diffuse through the same pot? (H = 1, O = 16) Solution Before solving the problem, the first thing is to find how long it will take an equal volume (400cm 3 ) of hydrogen to diffuse through the pot. This is because we are comparing oxygen and hydrogen. Since 200cm 3 of hydrogen = 30s 400cm 3 will = 30/200 X 400 = 60s         t 2 /t 1 = √ m 2 / √ m 1         .t 1 = 60s          .t 2 = ??          M 1 = 2           M 2 = 32                60/.t 2 = √ 2/ √ 32                .t 2 2  = 60x60 x 16                    = 57600             .t 2 = √ 57600 = 240se