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

answer: at room temperature (roughly 20 degrees c), you can dissolve 203.9 grams (or 200 grams) of sugar in 100 ml of water. as water temperature increases, the amount of sugar that will dissolve increases, up to about 500 grams of sugar in 100 ml water at 100 degrees c.

explanation:

the answer is the