I think this is what you wanted, so good luck!
Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
Answer:
D
Explanation:
Hopefully this helps you!
Answer: 4.21×10⁻⁸
Explanation:
1) Assume a general equation for the ionization of the weak acid:
Let HA be the weak acid, then the ionization equation is:
HA ⇄ H⁺ + A⁻
2) Then, the expression for the ionization constant is:
Ka = [H⁺][A⁻] / [HA]
There, [H⁺] = [A⁻], and [HA] = 0.150 M (data given)
3) So, you need to determine [H⁺] which you do from the pH.
By definition, pH = - log [H⁺]
And from the data given pH = 4.1
⇒ 4.10 = - log [H⁺] ⇒ [H⁺] = antilog (- 4.10) = 7.94×10⁻⁵
4) Now you have all the values to calculate the expression for Ka:
ka = 7.94×10⁻⁵ × 7.94×10⁻⁵ / 0.150 = 4.21×10⁻⁸
