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Identifying Oxyacids

Jules Bruno
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Now we're gonna say the second type. They're called oxy acids. And from the name Oxy, you could tell that it has one particular element in their oxygen. So we're gonna say oxy acids are acids that contain H plus I on. Of course, they contain oxygen, and they contained some type of nonmetal. Okay, So the difference between binary and Oxy is that Oxy has oxygen's involved. Binary does not. Something else they both share in common is that they usually do not possess any medals. Again, we're gonna see usually when we have a metal involved, that compound will be a base. Now, we're gonna say that oxy acids are formed when we hydrate non metal oxides. So in this example, here are non metal oxide would be carbon dioxide here and here would be sulfur try oxide. Okay, So non metal oxide just means you have a non metal connected to oxygen. Now all that's gonna happen is the water is going to attach itself to this whole compound. And what we get are two types off oxy acids and their oxy acids. Because both have h both have oxygen. Remember that H is really age plus they both have oxygen, and they both have some type of nonmetal involved. And that's what makes them both oxy acids. So here we would have carbonic acid. And here we have sulfuric acid. And again, we'll learn how to name these particular types of acids later on. But for right now, just realize, how do we create oxy acids? We just add H plus to some type of non metal oxide Co two s +02 s, +03 Those types of compounds, uh, P +03 Those are the kinds of compounds you add water to to make your new types of oxy acids. Now that you guys have seen the two types of assets binary and oxy, I want you guys to try to answer the next following question here. I say which the following compound or compounds cannot be classified as an acid. I gave you just brief description on on both, but try to use your best judgment to determine which one could be in which ones could not be. Then, once you're done with that click on the explanation button, you'll see a video of me explaining how to approach this problem. Guys, good luck