Answer:
9.1
Explanation:
Step 1: Calculate the basic dissociation constant of propionate ion (Kb)
Sodium propionate is a strong electrolyte that dissociates according to the following equation.
NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻
Propionate is the conjugate base of propionic acid according to the following equation.
C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻
We can calculate Kb for propionate using the following expression.
Ka × Kb = Kw
Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰
Step 2: Calculate the concentration of OH⁻
The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.
[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M
Step 3: Calculate the concentration of H⁺
We will use the following expression.
Kw = [H⁺] × [OH⁻]
[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M
Step 4: Calculate the pH of the solution
We will use the definition of pH.
pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1
N₂+3H₂⇒ 2NH₃
m(NH₃)=1250+225*2=1700 grams
N₂ is the limiting <span>reagent.
1250 grams are</span><span> left when the maximum amount of ammonia is formed.</span>
Q = mct
-Q= energy in Joules
-m = mass in grams
-c= specific heat capacity in J/g degree C
-t = delta temperature in degrees Celsius
So,
Q = m c t
Q = (7 grams)(0.448J/g C)(750 C - 25 C)
Q = 2273.6 J
Your final answer = 2273.6 Joules
Answer:
3. 75.0%
Explanation:
2 ClO2(g) + F2(g) → 2 FClO2(g)
First order with respect to ClO2 and F2.
This means the rate equation is given as;
Rate = k [ClO2][F2]
When the initial concentrations of ClO2 and F2 are equal?
Let's assume an initial value of 1 for both reactants, so rate equation is given as;
Rate = k * 1 * 1 = k
The rate after 25% of the F2 has reacted is what percent of the initial rate?
The concentration left of F2 is 75% ( 100% - 25%) = 0.75
Concentration of ClO2 remains 1.
So rate equation is given as;
Rate = k * 1 * 0.75 = 0.75 k
Comparing 0.75k and k.
This means our answer is;
3. 75.0%
