1. 12 L = 12 dm³
2. 3.18 g
<h3>Further explanation</h3>
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:
the ratio of the universe's helium to hydrogen by mass is 2:5 approximately.
Explanation:
To solve this question, we assign arbitrary values to these elements in tandem to the data given in the question.
nH, Number of Hydrogen atoms = X = 100
nHe, Number of Helium atoms = 0.1X = Y = 10
From the data given, by mass:
if the mass of Hydrogen atom = 1g = 100g
therefore, the mass of a Helium atom = 4g (1*4) = 40g
Mass ratio of Helium to Hydrogen = 40:100 = 4:10 = 2:5 approximately
Answer:
Zero (0)
Explanation:
Since Na^+ has a charge of +1 and Cl^- has a charge of —1 in NaCl
Then the net = +1 —1 = 0
