Determine the molar solubility (in m to two decimal places) of solid cobalt(ii) hydroxide ksp=5.92x10^-15 when dissolved in:

a) water (at 25 degree c)

b) an aqueous solution containing 0.1 m naoh (at 25 degree c)

beta decay is the answer

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

i think it is true since you can test the answer and can use observations to complete it.

6.4 cm (max) - 25 cm (min)