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All right, guys.
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We're going to be doing problem number 70 in chapter 5 of chemistry, central science.
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So we're going to be looking at these four compounds right here, hydrogen peroxide, calcium carbonate, p -o -c -o -c -l -3, e and c2h5 -o -h.
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We're going to write the balanced equation for the formation of these compounds, and then we're going to look up the delta h value.
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So let's start off hydrogen peroxide.
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So that's going to be h2 plus o2 yields h2 o2 -2.
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So we have two h's on the left, two h's on the right, two o's on the left, two o's on the right.
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So that means our equation is balanced.
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And if you look at appendix c, you can find the delta h formation of these compounds, and that is going to be negative 136.
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1 and 0 kilojoules per kilojoules.
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Now let's look at calcium carbonate.
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So we're going to add calcium.
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We're going to add our carbon.
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And now we're going to add the oxygen.
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Now, this will produce our carbonate.
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Now, let's see what we have here.
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We have one calcium on the left, one calcium on the right, one carbon on the right.
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And we have two oxygens left, but three on the right.
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So remember when we're talking about the standard entope of formation, we're actually producing one mole of our product.
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So we have to create a scenario.
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We're only producing one mole of calcium carbonate.
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So instead of having a define an integer coefficient, we're going to be using a fraction.
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So we're going to be using three halves of oxygen.
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So, oh, you break up o2, ooh, and half, you have individual oxygen atom.
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You multiply that by three.
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That gives you three oxygen, which will be used for carbonate.
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Now, we're going to go to p .o .o .c .l .3...