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, 
Step 1 - Discovering the ionic formula of Chromium (III) Carbonate
Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

Step 2 - Finding the molar mass of the substance
To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.
The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

The molar mass will be thus:

Step 3 - Finding the percent composition of carbon
As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

The percent composition of Carbon is thus 12.7 %.
Answer:
Explanation:
Molar mass of sodium hydroxide is = 100
any compund with its molar mass dissovled in 1L itres solution gives rise to 1M solution
40 g ------> 1L ---->1M
Xg ------>1L------>2M
X g= 40*1*2/1*1 =80 gram
Answer:
Nonpoint pollution is harder to control
Explanation:
Nonpoint pollution is harder to control because they include runoff from farms, golf courses, construction sites, roads. Whereas the single origin of the pollution is harder to trace.
Answer:
5 moles of Fe
Explanation:
The equation of the reaction is;
2 Al(s) + Fe 2O 3(s) --> 2Fe (s) + Al 2O 3 (s)
Now;
1 mole of Fe2O3 require 2 moles of Al
3 moles of Fe2O3 requires 3 × 2/1 = 6 moles of Al
Hence Al is the limiting reactant.
If 2 moles of Al yields 2 moles of Fe
5 moles of Al yields 5 × 2/2 = 5 moles of Fe