Hw Help M.9a
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Hw Help M.9a
Hello! I was just working on M.9a (Copper(II) nitrate reacts with sodium hydroxide to produce a precipitate of light blue copper(II) hydroxide. (a) Write the net ionic equation for the reaction.) Can someone please walking me through step by step on how they write chemical equations like this? Also, what does the (II) in copper equate to? Thank you for all the help!
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Re: Hw Help M.9a
Hello! To be honest with you we haven't really learned it so most of what I am going to tell you is based on my prior knowledge and experience.
That being said, the (II) part of copper refers to the oxidation state it is in. Because copper is a transition metal in the d-block, there's a tendency for metals in the same block to lose their outer electrons thus changing their charges. In this case, copper (II) means Copper has a charge of +2, so Cu+2.
Ionic chemicals equations are easy once you learn the names, they're basically the charged version of the regular chemical equations. In this case, because copper (II) is specified to have a +2 charge, you would then include this as: Cu+2. Then we found out it reacted with sodium hydroxide, the sodium is not really important in this case because we already have a cation, the copper, and we know hydroxide have a -1 charge so: OH-. Put them together you get: Cu+2 + OH- = Cu(OH)2. On the product side, you have the charges crisscross that's why only 1 Cu but 2 OH. Hope it helps! :)
That being said, the (II) part of copper refers to the oxidation state it is in. Because copper is a transition metal in the d-block, there's a tendency for metals in the same block to lose their outer electrons thus changing their charges. In this case, copper (II) means Copper has a charge of +2, so Cu+2.
Ionic chemicals equations are easy once you learn the names, they're basically the charged version of the regular chemical equations. In this case, because copper (II) is specified to have a +2 charge, you would then include this as: Cu+2. Then we found out it reacted with sodium hydroxide, the sodium is not really important in this case because we already have a cation, the copper, and we know hydroxide have a -1 charge so: OH-. Put them together you get: Cu+2 + OH- = Cu(OH)2. On the product side, you have the charges crisscross that's why only 1 Cu but 2 OH. Hope it helps! :)
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