The solubility equilibrium of :
[tex] CaCrO_{4}(aq)<===>Ca^{2+}(aq) + CrO_{4}^{2-}(aq)\\
Q_{sp}=[Ca^{2+}][CrO_{4}^{2-}]\\
= (0.0200 M)(0.0300 M) \\
= 0.0006
Ksp (0.00071) > Qsp (0.0006). So, <u>no precipitate would form</u>.
On a class trip to the coast, you and your lab partner find a rock in the spot marked in red on the diagram. You show your teach
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Answer:
The molar mass of the vapor is 43.83 g/mol
Explanation:
Given volume of gas = V = 247.3 mL = 0.2473 L
Temperature = T = 100 = 373 K
Pressure of the gas = P = 745 mmHg (1 atm = 760 mmHg)
Mass of vapor = 0.347 g
Assuming molar mass of gas to be M g/mol
The ideal gas equation is shown below
The molar mass of the vapor comes out to be 43.834 g/mol