To get the molarity you need to follow this equation
moles of solute
Molarity (M = -----------------------
Liters of solution
But before you apply that equation you need to find the moles of solute and the liters of solution. Follow this equation
Na2SO4 + BaCl2 = BaSO4 + 2 NaCl
Solution
Moles of BaSO4 = 5.28 g
---------------
233.43 g / mol
= 0.0226 moles
Moles of NaSO4 = 0.0226
0.0226 mole
Molarity = -----------------
0.250 L
= 0.0905 mol / L
So the answer is 0.0905 mol / L
Answer: A
Explanation: Beta particles have a charge of -1
Answer:

Explanation:
Hello there!
In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

Then, we solve for specific heat of the metallic alloy to obtain:

Thereby, we plug in the given data to obtain:

Regards!
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, 
Explanation:
no lo c crack solo es para poder iniciar la sección y no c ingel Jaksjs