Answer:
1. <u>No, you cannot calculate the solubility of X in water at 26ºC.</u>
Explanation:
You cannot calculate the solubility of X in <em>water at 26 degrees Celsius </em>because you do not know whether the solution formed by dissolving the crystals in 3.00 liters of water is saturaed or not.
The only way to determine the solubility of the compound X is by dissolving the crystals in certain (measured) amount of water and making sure that some crystals remain undissolved, as a solid on the bottom of the beaker.
Next, you should filter the solution to remove the undissolved crystals. Then, weigh the solution, evaporate, wash, dry, and weigh the crystals.
Then you have the mass of the crystals dissolved and the mass of the solution which will let you calculate the mass of pure water, and then the solubility.
Answer: GeH4 (Germanium(IV) Hydride)
Explanation:
A Binary molecular compound Hydrogen and a Group 4A element which is more more acidic than SiH4 in aqueous solution is GeH4.
The pKa of GeH4;
= 25
Whilst that of SiH4
= 35
The lesser the pKa the higher the Ka which means more acidic.
I believe the answer to this is A.
Answer:
6.4 × 10^-10 M
Explanation:
The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).
CaF2 will dissociate as follows:
CaF2 ⇌Ca2+ + 2F-
1 mole of Calcium ion (x)
2 moles of fluorine ion (2x)
NaF will also dissociate as follows:
NaF ⇌ Na+ + F-
Where Na+ = 0.25M
F- = 0.25M
The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M
Ksp = {Ca2+}{F-}^2
Ksp = {x}{0.25}^2
4.0 × 10^-11 = 0.25^2 × x
4.0 × 10^-11 = 0.0625x
x = 4.0 × 10^-11 ÷ 6.25 × 10^-2
x = 4/6.25 × 10^ (-11+2)
x = 0.64 × 10^-9
x = 6.4 × 10^-10
Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M
