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Hi there.
00:03
In this question, we are working with double replacement reactions.
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We know it's a double replacement reaction because in a double replacement reaction, we react to aquaous ionic compounds, and we form an insoluble product.
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We precipitate in this case.
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So in a double replacement, those two aquis ionic compounds react, and they exchange ions.
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The positive ion from the first combines with the negative ion from the second, and the positive ion from the second combines with the negative ion from the first.
00:41
Okay, so for this, we want to write, first of all, a balanced chemical equation.
00:49
Well, what we have reacting are potassium iodide, so that's going to be k -i, and it is aquaous.
00:58
It's reacting with silver nitrate.
01:02
Silver nitrate has the formula ag, and then nitrate is n -o -3.
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We have a one positive and a one negative charge, so it's simply a .g.
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N -o -3.
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When these exchange ions, we get the precipitate silver iodide.
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So agi is our precipitate.
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That is a solid.
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Our other product is going to be the potassium with the nitrate.
01:33
That remains aquaous.
01:36
Okay, looking at this equation, we see it's already balanced as written.
01:40
We have one potassium on each side, one iodine on each side, one silver, and one nitrate ion.
01:49
So this is our balanced chemical equation.
01:53
Next, we want to write the complete ionic equation.
01:58
So that means we need to write everything as ions here.
02:00
So we're going to split these things apart into the ions, because that's the way they really are in solution.
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So potassium iodide is really potassium.
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Ions in solution and iodide ions in solution.
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The silver nitrate is really silver ions and nitrate.
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And nitrate ions.
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Okay...