HELP PLZ!CHEMISTRY may/june 2018 paper 22.
i need explanations and steps for these.
thank you so much!!

1. 12 L = 12 dm³

2. 3.18 g

Given

1. Reaction

K₂CO₃+2HNO₃⇒ 2KNO₃+H₂O+CO₂

69 g K₂CO₃

2. 0.03 mol/L Na₂CO₃

Required

1. volume of CO₂

2. mass Na₂CO₃

Solution

1. mol K₂CO₃(MW=138 g/mol) :

= 69 : 138

= 0.5

mol ratio of K₂CO₃ : CO₂ = 1 : 1, so mol CO₂ = 0.5

Assume at RTP(25 C, 1 atm) 1 mol gas = 24 L, so volume CO₂ :

= 0.5 x 24 L

= 12 L

2. M Na₂CO₃ = 0.03 M

Volume = 1 L

mol Na₂CO₃ :

= M x V

= 0.03 x 1

= 0.03 moles

Mass Na₂CO₃(MW=106 g/mol) :

= mol x MW

= 0.03 x 106

= 3.18 g

on passing the small ampere of electric current through the tube made of ar and hg of cfl generates uv rays which excites the fluorescent and thus emits the high radiation by consumption of just 5- 12 w in comparison to incandescent light which requires more power such as around 40 - 55 w to lighten.

moreover the life span of fluorescent or cfl is around 8000 h whereas the life span of incandescent is around 1200 h

hence, the building installed with incandescent lights bears more electricity bill.