This is an incomplete question, here is a complete question.
Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.
CaCO₃, Ksp = 8.7 × 10⁻⁹
Answer : The solubility of CaCO₃ is, 
Explanation :
As we know that CaCO₃ dissociates to give 
ion and 
ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ca^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
Let solubility of CaCO₃ be, 's'
Therefore, the solubility of CaCO₃ is,
Answer:
319.8 m/min
Explanation:
533 cm/s
We can convert 533 cm/s to m/min by doing the following:
First, we shall convert 533 cm/s to m/s. This can be obtained as illustrated below:
Recall:
100 cm/s = 1 m/s
Therefore,
533 cm/s = 533 cm/s /100 cm/s × 1 m/s
533 cm/s = 5.33 m/s
Finally, we shall convert 5.33 m/s to m/min. This can be obtained as follow:
1 m/s = 60 m/min
Therefore,
5.33 m/s = 5.33 m/s / 1 m/s × 60 m/min
5.33 m/s = 319.8 m/min
Therefore, 533 cm/s is equivalent to 319.8 m/min
Answer:
It can't be done.
Explanation:
If you have only 5.4 g of oxygen, the most lithium oxide you can get is 7.7 g.
Only 2.3 g of lithium will react. and the other 22.3 g of lithium will not be used.
Answer:
Oxygen's atomic weight is 16.00 amu. 1 mole of oxygen is 6.02 x 1023 atoms of oxygen 1 amu = 1.661 x 10-24g What is the molar mass (g/mole) of oxygen? Molar mass (in grams) is always equal to the atomic weight of the atom! Molar mass (in grams) is always equal to the atomic weight of the atom!
