Determine the molar solubility for Cr(OH)3 Ksp = 6.3x10^-31 ....
1 answer:
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Answer:
Explanation:
*Since the titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7. On interpolation, we will obtain that 9.50mL and 9.82 mL of HCl is required to completely neutralized the given Ca(OH)2 solution.
*pH at the equivalence point =7
we know that pH + pOH = 14
Hence pOH= 14-7=7
pOH= -log(OH-)
The concentration of OH-= 10-pH= 1X10-7 M
One reason for the low solubility may be the higher reaction temperature, Another reason is the common ion effect.
Answer:
Elemental gold to have a Face-centered cubic structure.
Explanation:
From the information given:
Radius of gold = 144 pm
Its density = 19.32 g/cm³
Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:
a = 407 pm
In a unit cell, Volume (V) = a³
V = (407 pm)³
V = 6.74 × 10⁷ pm³
V = 6.74 × 10⁻²³ cm³
Recall that:
Net no. of an atom in an FCC unit cell = 4
Thus;
density d = 19.41 g/cm³
Similarly; For a body-centered cubic structure
where;
r = 144
a = 332.56 pm
In a unit cell, Volume V = a³
V = (332.56 pm)³
V = 3.68 × 10⁷ pm³
V 3.68 × 10⁻²³ cm³
Recall that:
Net no. of atoms in BCC cell = 2
∴
density =17.78 g/cm³
From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.
This makes the elemental gold to have a Face-centered cubic structure.