Answer:
The enthalpy change in the the reaction is -47.014 kJ/mol.
Explanation:
Volume of water in calorimeter = 22.0 mL
Density of water = 1.00 g/mL
Mass of the water in calorimeter = m
Mass of substance X = 2.50 g
Mass of the solution = M = 2.50 g + 22 g = 24.50 g
Heat released during the reaction be Q
Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C
Specific heat of the solution is equal to that of water :
c = 4.18J/(g°C)
Heat released during the reaction is equal to the heat absorbed by the water or solution.
Heat released during the reaction =-1.433 kJ
Moles of substance X= 
The enthalpy change, ΔH, for this reaction per mole of X:
Group 2
Period 3
Alkaline Earth Metal
When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,
cC ⇔ aA + bB
Ksp = {[A^a][B^b]}
Now, for the this problem, the reaction is
BaSO₄ ⇔ Ba²⁺ + SO₄²⁻
The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.
1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M
When we can get the Kinetic energy from this formula KE= 1/2 M V^2and we can get the potential energy from this formula PE = M g H
we can set that the kinetic energy at the bottom of the fall equals the potential energy at the top so, KE = PE
1/2 MV^2 = M g H
1/2 V^2 = g H
when V is the velocity, g is an acceleration of gravitational force (9.8 m^2/s) and H is the height of the fall (8 m).
∴ v^2 = 2 * 9.8 * 8 = 156.8
∴ v= √156.8 = 12.5 m/s
