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Lead carbonate and lead iodide are insoluble. Which 2 soluble salts could be use in the...
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The key to answering this question is to know the solubility rules. Let's look first at finding a "carrier" for the lead. Since compounds with the nitrate ion are almost always soluble, lead nitrate will make the perfect choice for preparing both of these compounds.
Find a carbonate compound and a chloride compound that are soluble (see table linked to below). Compounds with alkali metals (Na, K, etc) and ammonium (NH4+) are generally soluble so they make a great choice as a "carrier" for the anion.
Start by determining the formulas for the compounds, then assigning them as s (solid) or aq (aqueous) and balance the equation. Write it out in words just as you would read it out loud (including the coefficients and phases).
For the ionic equation, you need to clarify whether is an ionic equation or net ionic equation. For an ionic equation, you will write everything marked as "(aq)" as ions. Compounds that are solids remain written as the compound. If you are looking for a net ionic equation, cancel out ions that are the same on both sides (i.e Na+(aq) on the reactant and product side would be cancelled out) as these are spectator ions and are not undergoing any changes.
Posted by mlsiasebs on February 5, 2012 at 11:59 AM (Answer #1)
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