Answer:
Answer:
B. It's dissociation is a reversible reaction
Explanation:
NH3 is a weak alkali that does not dissociate fully into its solution. Only parts of the ammonia takes part in the dissociation process.
NH3 + H20 —> NH4+ + OH-
This dissociation is reversible which means the reactants can be formed from the product gotten from the dissociation
It has a high pH due to its basic nature. It also has a Low concentration of H+ ions and not all the OH- ions are released.
Answer:
Total mass of original sample = 0.38 g
Explanation:
Percentage of chloride = 74.5%
Mass of AgCl precipitate = 1.115 g
Mass of original sample = ?
Solution:
Mass of chloride:
1.115 g/ 143.3 g/mol × 35.5 g/mol
0.0078 g × 35.5 = 0.28 g
Total mass of compound:
Total mass = mass of chloride / %of Cl × 100%
Total mass = 0.28 g/ 74.5% × 100%
Total mass = 0.0038 g× 100
Total mass = 0.38 g
The rinsing of mouth is to remove food particles and minimize mucous
C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
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
