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!
Concentration can be expressed in different forms: molarity, molality, normality, percentage, part per million and many more. For molality, it is a unit of concentration expressed as moles of solute per kilogram of solvent. Therefore,
0.3 = moles solute/0.10 kg solvent
moles solute = 0.03 moles
Anytime an atom shares or transfers electrons a bond is formed, sharing= covalent and transfer= ionic
A) -0.5(9.8)*t^2 = -25(t-2) - 0.5(9.8)(t-2)^2
-4.9t^2 = -25t + 50 - 4.9(t^2-4t+4)
0 = -25t+50+19.6t - 19.6
5.4t = 30.4
t = 5.62962963 s
b) h = -4.9(5.62962963)^2
h = -155.2943759
the building is 155.2943759 m high
c) speed 0of first stone
= at
= 9.8*5.62962963
= 55.17037037 m/s
speed of second stone
= v + at
= 25+9.8*3.62962963
= 60.57037037 m/s
The oxidation state of the compound Mn (ClO4)3 is to be determined in this problem. For oxygen, the charge is 2-, the total considering its number of atoms is -24. Mn has a charge of +3. TO compute for Mn, we must achieve zero charge overall hence 3+3x-24=0 where x is the Cl charge. Cl charge, x is +7.
