WEBVTT
1
00:00:00.240 --> 00:00:03.600 A:middle L:90%
in a mixture oven or that contains both mag unease
2
00:00:03.799 --> 00:00:09.820 A:middle L:90%
and oxide that are two different ion compounds. We
3
00:00:09.820 --> 00:00:14.080 A:middle L:90%
can use our concept of percent composition to determine how
4
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many grams of this mixture RMN Oh, and how
5
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many grams air em into 03 So we know the
6
00:00:21.710 --> 00:00:27.510 A:middle L:90%
ratio of manganese oxygen is one to 1.42 So if
7
00:00:27.510 --> 00:00:36.109 A:middle L:90%
we look at our total mass, we can determine
8
00:00:36.119 --> 00:00:47.219 A:middle L:90%
how much of that is oxygen. Oxygen is 1.42
9
00:00:47.219 --> 00:00:50.689 A:middle L:90%
parts out of the hole, which is one plus
10
00:00:50.689 --> 00:01:02.429 A:middle L:90%
1.42 or there should be 318.21 grams oxygen. We
11
00:01:02.429 --> 00:01:07.769 A:middle L:90%
can also find how much of this his mag unease
12
00:01:07.939 --> 00:01:11.090 A:middle L:90%
mag unease is one part of the whole where one
13
00:01:11.090 --> 00:01:21.230 A:middle L:90%
divided by one plus 1.42 and so there are 224.9
14
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grams of mm. We then look at dividing up
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00:01:26.349 --> 00:01:29.120 A:middle L:90%
between the two compounds. We know that our two
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compounds M n O. And mn two 03 So
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00:01:36.409 --> 00:01:47.549 A:middle L:90%
there are a total of three amends and four oxygen's
18
00:01:48.040 --> 00:01:51.930 A:middle L:90%
. If we consider m n o, we know
19
00:01:51.930 --> 00:01:55.200 A:middle L:90%
that it has 1/3 of the Mangga knees because it
20
00:01:55.200 --> 00:02:09.860 A:middle L:90%
has one out of the three and it has 1/4
21
00:02:09.860 --> 00:02:13.699 A:middle L:90%
of the oxygen because it has one of the four
22
00:02:13.710 --> 00:02:29.449 A:middle L:90%
oxygen's total, or 154 0.2 grams of this mixture
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00:02:29.840 --> 00:02:31.729 A:middle L:90%
, his M N o. You can do the
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00:02:31.729 --> 00:02:36.219 A:middle L:90%
same for em. In 203 You see that there
25
00:02:36.229 --> 00:02:44.919 A:middle L:90%
are 2/3 of the mag unease present, and 3/4
26
00:02:44.930 --> 00:02:58.599 A:middle L:90%
of the oxygen or this compound is 388 0.5 grams
27
00:02:58.610 --> 00:03:07.419 A:middle L:90%
of the total mixture. Look at the ratio of
28
00:03:07.419 --> 00:03:14.539 A:middle L:90%
the ions. M n 20 is made up of
29
00:03:14.539 --> 00:03:20.349 A:middle L:90%
the MN two plus I am and M in 203
30
00:03:21.000 --> 00:03:23.550 A:middle L:90%
is made up of the mn three plus ion.
31
00:03:23.939 --> 00:03:27.960 A:middle L:90%
So using our masses, we can find out how
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00:03:27.960 --> 00:03:31.550 A:middle L:90%
many moles of each eye on is present. First
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00:03:31.550 --> 00:03:35.770 A:middle L:90%
, we changed two moles by dividing by the molar
34
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Mass and then noticing that there's one ah of mn
35
00:03:54.969 --> 00:04:06.680 A:middle L:90%
two plus ions for everyone. Moul of compound or
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00:04:06.680 --> 00:04:12.990 A:middle L:90%
there are about two moles of mn two plus ions
37
00:04:15.439 --> 00:04:27.379 A:middle L:90%
and using our mass for him. In 203 we
38
00:04:27.379 --> 00:04:30.149 A:middle L:90%
changed two moles by dividing by the molar Mass.
39
00:04:35.680 --> 00:04:39.350 A:middle L:90%
And we see in this equation there are two moles
40
00:04:40.439 --> 00:04:45.350 A:middle L:90%
of mn three plus for every mole of compound.
41
00:04:48.839 --> 00:04:57.300 A:middle L:90%
So this gives us five moles of MN three plus
42
00:04:57.300 --> 00:05:00.180 A:middle L:90%
, so our ratio is a 5 to 2 ratio
43
--> A:middle L:90%
.