Answer:
3.62x10⁻⁷ = Kb
Explanation:
The acid equilibrium of a weak acid, HX, is:
HX + H₂O ⇄ X⁻ + H₃O⁺
Where Ka = [X⁻] [H₃O⁺] / [HX]
And basic equilibrium of the conjugate base, is:
X⁻ + H₂O ⇄ OH⁻ + HX
Where Kb = [OH⁻] [HX] / [X⁻]
To convert Ka to Kb we must use water equilibrium:
2H₂O ⇄ H₃O⁺ + OH⁻
Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]
Thus, we can obtain:
Kw = Ka*Kb
Solving for Kb:
Kw / Ka = Kb
1x10⁻¹⁴ / 2.76x10⁻⁸ =
3.62x10⁻⁷ = Kb
Answer: The statement (B) is not true about chemical reactions.
Explanation:
A chemical reaction rate is affected by the several factors few of which are temperature, concentration of reactants, surface area etc.
In a chemical reaction, if temperature is increased then the rate of reaction will increase because it will increase the average kinetic energy of the reactant molecules. Thus, large number of molecules will have minimum energy required for an effective collision.
It is known that increasing the amount of reactants will increase the rate of reaction.
Therefore, rate of reaction will change if concentration or temperature is changed.
Hence, the statement (B) is not true about chemical reactions.
Answer:
The answer to your question is below
Explanation:
Data
mass of CaCO₃ = 155 g
mass of HCl = 250 g
mass of CaCl₂ = 142 g
reactants = CaCO₃ + HCl
products = CaCl₂ + CO₂ + H₂O
1.- Balanced chemical reaction
CaCO₃ + 2HCl ⇒ CaCl₂ + CO₂ + H₂O
2.- Limiting reactant
molar mass of CaCO₃ = 40 + 12 + 48 = 100 g
molar mass of HCl = 2[1 + 35.5 ] = 73 g
theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37
experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62
As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃
3.-
Calculate the molar mass of CaCl₂
CaCl₂ = 40 + 71 = 111 g
100 g of CaCO₃ ------------------ 111 g of CaCl₂
155 g of CaCO₃ ----------------- x
x = (155 x 111) / 100
x = 17205 / 100
x = 172.05 g of CaCl₂
4.- percent yield
Percent yield = 142 / 172.05 x 100 = 82.5 %
5.- Excess reactant
100 g of CaCO₃ -------------------- 73 g of HCl
155 g of caCO₃ ------------------- x
x = (155 x 73)/100
x = 133.15 g
Mass of HCl = 250 - 133.15
= 136.9 g
