The acid dissociation constant is 1.3 × 10^-3.
<h3>What is acid-dissociation constant?</h3>
The acid-dissociation constant is a constant that shows the extent of dissociation of an acid in solution. We have to set up the reaction equation as shown below;
Let the acid be HA;
HA + H2O ⇄ H3O^+ + A^-
since the pH of the solution is 2.57 then;
[H3O^+] = Antilog(-pH) = Antilog(-2.57) = 2.7 × 10^-3
We can see that; [H3O^+] = [A^-] so;
Ka = (2.7 × 10^-3)^2/(5.5 × 10^–3)
Ka = 1.3 × 10^-3
Learn more about acid-dissociation constant: brainly.com/question/9728159
Because if they are not balanced, then the amount of reaction is not equal to the amount of the product. You must have the same ratio of atoms on both sides of the equation because we are not losing moles of each element.
The third one, it releases H+ions into a solution.
When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three phases changes from steam to liquid and from liquid to ice (solid) :
when Q = M*C*ΔT
Q is the heat in J
and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g
C is the specific heat J/g.°C
ΔT is the change in temperature
Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice * ΔT ice)]
= 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]
∴Q = 7444.8 J
and when we know that the heat of fusion for water = 334J/g
and heat of vaporization for water = 2260J/g
∴Q for the two phases changes = M * (2260+334)
= 18 * (2260+334)
= 46692 J
∴ Q total = 7444.8 + 46692 = 54136.8 J
