5ml of mixture of CH4 and C2H4 need 13.2 ml of oxygen for complete combustion. calculate the volume of each gas in the mixture based on the law of gaseous volume.

Explanation:

CH₄ + 2O₂ = CO₂ + 2H₂O

C₂H₄ + 3O₂ = 2CO₂ + 2H₂O

Let volume of CH₄ be V , so volume of C₂H₄ will be 5 - V

CH₄    +    2O₂   =    CO₂   +   2H₂O

1 vol           2 vol

V                2 V

V vol of CH₄   will require 2 V of O₂

C₂H₄     +    3O₂    =    2CO₂   +   2H₂O

5 - V       3( 5 - V )

5 - V vol of C₂H₄ will require 3 ( 5 - V ) vol of oxygen .

Total oxygen required = 2 V + 3 ( 5 - V )

2 V + 3 ( 5 - V ) = 13.2

2 V + 15 - 3V = 13.2

V = 1.8 ml

vol of CH₄ = 1.8

vol of C₂H₄ = 5 - 1.8 = 3.2 ml.

heat or electricity in the atmosphere- c.

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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