Answer:
Explanation:
Hello there!
In this case, given the solubilization of cadmium (II) hydroxide:
The solubility product can be set up as follows:
![Ksp=[Cd^{2+}][OH^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BCd%5E%7B2%2B%7D%5D%5BOH%5E-%5D%5E2)
Now, since we know the concentration of cadmium (II) ions at equilibrium and the mole ratio of these ions to the hydroxide ions is 1:2, we infer that the concentration of the latter at equilibrium is 3.5x10⁻⁵ M. In such a way, the resulting Ksp turns out to be:
Regards!
The energy release when dissolving 1 mol of NaOH in water is 445.1 kJ
the mass of NaOH to be dissolved is 32.0 g
The number of NaOH moles in 32.0 g - 32.0 g / 40 g/mol = 0.8 mol
the energy released whilst dissolving 1 mol of NaOH - 445.1 kJ
when dissolving 0.8 mol - the energy released is 445.1 kJ/mol x 0.8 mol
therefore heat released is - 356.08 kJ
answer is -356.08 kJ
Answer:
57.8kg
Explanation:
Potential energy is given by:
Where U is potential energy, m is mass, g is acceleration due to gravity, and h is height. Using this equation:
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
