2

AFAIK, such a reaction has no name. However, if one were so inclined, one could call it a replacement reaction, because the $\ce{MgCO3}$ is being converted into $\ce{Mg(OH)2}$. To calculate the "equilibrium point," which I take to mean the concentration of all the ions at equilibrium, we would need to know the $k_\text{sp}$ of both $\ce{MgCO3}$ and $\ce{Mg(...


2

Yes, the calcium ion could lead to precipitation. The solubility of $\ce{CaCO3}$ in distilled water is about 15 mg/L, which is about 0.15 mM calcium ion if there is no other source of carbonate. The solubility constant for $\ce{CaF2}$ is about $4\times 10^{-11}$, which means that we can only have 0.5 mM fluoride ions before precipitation will start. That's ...


1

If you look at the graph of the reactivity with relation to temperature it increases proportionally until it reaches its activation energy and then it proceeds to decrease. For exothermic reactions since the activation energy is met early the reactivity decreases when more temperature is provided. While for endothermic reactions since they require a higher ...


1

Consider three compositions: A. 2NaF + CaCO3 B. CaF2 + Na2CO3, and C. NaF + 0.5 CaCO3 + 0.5 CaF2 + 0.5 Na2CO3. Using data from the CRC Handbook (62nd ed), the heats of formation of A and B are respectively 560.47 and 560.6 kcal, so there is little driving force to make a reaction go to completion. Note that A should be near neutral pH, but B ...


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