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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...

Pressure in the fluid

       Fluid refers to gases, liquids or anything that flows. When we say pressure in fluid, we are referring to the kind of pressure which fluids exert. So, what is pressure?                                               Pressure  Pressure is a force or vertical force per unit area acting on a surface. It has the following units         I.             Newton per square metre (Nm -2 )       II.             Pascal (Pa)     III.             Bar 1 Bar = 10 5 Nm -2 = 10 5 Pa Pressure is a...

Motion

        Motion is a change of position of a body or object with time                            Types of motion There are four types of motion which include         I.             Vibratory motion       II.             Translational motion     III.             Rotational motion   IV.             Random motion                           Vibratory motion   This is a to-and-fro motion of a body at a fixed ...

Laws of Logarithms

    Log (AB) = Log A + Log B   Log (A/B) = Log A – Log B   Log (A x ) = X Log A                                             Example Question Simple the following without using tables 1.     Log 16 ÷Log 4 2.     Log 24 – Log6÷ Log 16                           Solution 1)    Log 16÷Log 4   = Log 2x2x2x2÷Log 2x2   = Log 2 4 ÷Log 2 2 = 4Log 2÷2Log2 = 2 2)    Log 24 – Log6÷ Log 16                   Solution Log 24 – Log 6÷ Log 16   Log (8x3) – Log (2x3) ÷ Log (2x2x2x...

Logarithm Theory 2

  1.       Given that Log 2 = 0.30101, evaluate the following, i)         Log 16 Ii)   Log 128                          Solution i)           Log 16   = Log 2 4   =    4 Log 2 Substitute 0 . 30101 to Log 2   4         (0.30101) = 1.20404   ii)       Log 128          = log 2 7 =   7 Log 2 = 7 (0.30101) = 2.10707 2.       Evaluate without any tables 3Log2 +Log20 – log1.6                Solution 3Log2+Log20-Log 1.6 = Log2 3 +Log20-Log16/10 = Log{8x20x10/16} = Log 100 = Log 10 2 = 2Log10 =...

Logarithms

     Logarithm to a certain base of a number is equal to the power to which the base must be raised to give the number.                   M = X a                   Log x M = a                  4 = 2 2                    Log 2 4 = 2 Evaluate the following logarithms 1.     Log 10   10000 2.     Log 4    1/16 3.     Log 100   0.00001                                        ...

The Angle of Elevation and depression

       The foot of a ladder is 6m from the base of an electric pole, and the top of the ladder rests against the pole at a point 8m above the ground. How long is the ladder?                                              Solution                                                                                                     ...