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Sistema Internacional de Unidades, Densidad, Temperatura, Materia y Energía — Transcript

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0:01Hello, welcome to Academia Internet. We
0:05are going to start the chemistry course
0:06. By way of introduction, we will talk
0:09about the International System of Units
0:11, density, temperature, and some matter
0:13and energy. So, the International
0:16System of Units. The International
0:18System of Units arises from a need to
0:22bring order to measurements on the
0:25planet, establishing base standards on
0:28which to make comparisons. Remember
0:32that is what the act of measuring is
0:34about, comparing by taking one as a
0:36base, as a parameter. The international
0:41system is formed by base units,
0:42supplementary units, and derived units.
0:46It also includes the use of multiple
0:49and submultiple prefixes. We are going
0:51to study all of that. What are the base
0:54units? And what does that "base units"
0:56mean? They are those that are not
0:58derived from others defined based on
1:03natural and invariable physical
1:05phenomena. There are seven. Length,
1:09meter, symbol lowercase m. It is
1:12important that you learn this with its
1:14symbols. Mass, kilogram, kg, symbol in
1:18lowercase, watch out, time. Second, a
1:21lowercase s, electric current intensity
1:24, ampere. And here I already have an
1:27uppercase A. When the name of the units
1:30refers to a researcher, the symbol has
1:33an uppercase letter. Thermodynamic
1:36temperature. Name of the unit. Kelvin,
1:40uppercase K, is the researcher's name
1:44or surname. Luminous intensity, candela
1:48, symbol lowercase cd, and amount of
1:50substance, mole. Okay, those are the
1:55base physical magnitudes, the
1:57fundamental ones. Derived units are
2:03those formed by algebraically combining
2:06the former, the base units, the ones we
2:09just explained. Well, here we have the
2:13derived units that do not have their
2:16own name. For example, surface or area,
2:19square meter, symbol m². Product of
2:23two dimensions. Volume, cubic meter,
2:25product of three dimensions. That is
2:27what this means. M cubed. Density,
2:30which we will talk about later,
2:32kilogram per cubic meter. Velocity,
2:35meter per second, and so on, right?
2:38Angular velocity that we saw in physics
2:41, radians per second, acceleration,
2:43meters per second squared, angular
2:45acceleration, radian per second squared
2:47, molar concentration. Mole per cubic
2:52meter and current density, ampere per
2:56square meter. There it is. We also have
3:00derived units with their own name and
3:02symbol. For example, frequency, hertz,
3:05the researcher's surname, therefore,
3:08symbol uppercase H, lowercase z, and
3:11expression of the base or derived unit.
3:16Look, what relationship does it have
3:17with the base unit? It is second to the
3:19power of -1. Its reciprocal, force,
3:24Newton, the famous scientist, symbol N,
3:32kilograms, meters per second squared,
3:36and so on. Pressure and tension: pascal
3:40, the Frenchman, right? Uppercase P,
3:45lowercase a, Newton per square meter.
3:49Work, energy, quantity of heat: joule.
3:53Here we have its equivalence with base
3:55units: power, joule over time; quantity
3:58of electricity, uppercase C; electrical
4:01capacitance, Farad, uppercase F;
4:03electrical resistance, and it has this
4:06little symbol which is the Greek letter
4:08omega, volts divided by amperes. Well,
4:15multiples and submultiples. Remember
4:19that the International System does not
4:22only include base units and
4:24supplementary and derived units, but
4:26also the use of multiples and
4:29submultiples. The prefix system, here
4:33we have it. Multiple that multiplies
4:38and submultiple that divides. Deca
4:41multiplies the unit by 10. For example,
4:44if we are talking about meters:
4:46decameter, hectometer, kilometer, and
4:49so on. Look how practical it is. Symbol
4:52dk is da, hecto lowercase h, kilo
4:55lowercase k, and then the others are
4:58uppercase. Mega, giga, tera, peta, exa.
5:03In such a way that in chemistry or in
5:05science in general, every time you see,
5:07for example, an uppercase M, it means
5:0910 to the sixth. You can choose to
5:11write the uppercase M, put 10 to the
5:14sixth, giga, 10 to the ninth, and so on
5:17. Submultiples: centi, milli, micro,
5:20nano, pico, femto, atto. Symbols are
5:25all lowercase: d, c, m, micro has this,
5:28which is a Greek letter, nano, pico,
5:33femto, and atto. What do they mean?
5:37Factor 10 to the -1. Factor 10 to the
5:40-2, 10 to the -3, 10 to the -6, 10 to
5:44the -9, and so on; the equivalent is
5:47also usually worked with factors,
5:50obviously. Look how cumbersome it is to
5:52write so many zeros, or here, it's much
5:54more practical to put 10 to the 18th.
6:01Here I present to you conversion
6:02factors and constants, the most common
6:04ones, the ones you have to know. Or at
6:09least some of them, the most important
6:12ones. For example, this one is very
6:14important and it is not there. It is an
6:17Angstrom. You write an uppercase A, and
6:20to differentiate it from ampere, they
6:22have put a little circle on it; one
6:24Angstrom is 10 to the -8 cm. It is
6:28important that you know this, for
6:30example, that 1 meter is equivalent to
6:323.2 feet. Just feet. Or which is the
6:38same, one foot is equivalent to 30.48
6:41cm. It is important to know these
6:43equivalences. The same as one foot
6:46equals 12 inches, therefore, one inch
6:49is equivalent to 2.54 cm. We obtained
6:54it by dividing 30.48 by 12. This
6:57Western system, right? English system.
7:00Inch, approximately 2.5 centimeters.
7:03One yard, three feet, 0.9 m. One
7:07nautical mile, 1852 m. One land mile,
7:11differences, 1609. Units of mass.
7:15Important to remember this. One pound,
7:1716 ounces. One ounce, 28.3 g. One
7:22metric ton, 1,000 kg, 10 to the third.
7:261 kg, 2.205. Volume units, one barrel,
7:3142 gallons. liters, 1 m³, 1 ml, 1 cm³
7:43, 1 dm³, 1,000 cm³ of course, and the
7:48same for pressure units, energy units,
7:53look, one calorie, 4.184 J. Hm. We will
7:58see this anyway. We will surely see
8:01each one of these as we advance in the
8:03chemistry and physics course. Constants
8:06. These constants are important.
8:08Capital C is the speed of light
8:11constant, which you can express as 3*10
8:16to the 5th km per second, or its
8:21equivalent in meters per second, 3*10
8:26to the power of 8. meters per second,
8:31Planck's constant, Avogadro's number,
8:38and the universal gas constant.
8:42Temperature. Temperature is an
8:47arbitrarily determined parameter that
8:49indicates the average energy of a body.
8:52It is what is known as cold or hot.
8:55General formula, look. Notice the
8:59relationship between degrees Celsius
9:02and Kelvin, the same as Fahrenheit and
9:04Rankine. Both have the same denominator
9:09, and here they have the same
9:10denominator. They differ, for example,
9:16Kelvin from Celsius by 273. Note that
9:20this is temperature variation; it is
9:22not the same as this. The variation of
9:25Celsius degrees with respect to
9:27Fahrenheit degrees is 1.8. Of course,
9:30the variation from Celsius to Kelvin is
9:32the same, and this right here with the
9:37Rankine scale. We are going to see
9:41exercises where we will explain in
9:42detail what that means later.
9:45Thermometric scale. Look, these are
9:52absolute and the others relative.
10:04Boiling point of water, 100 ° C. In
10:08Fahrenheit it reaches 212, in Kelvin
10:10373. Rankine 672. Freezing point of
10:15water. We know that in degrees Celsius,
10:19zero is our reference; this is for the
10:22Western world, 32 ° F, Kelvin 273
10:25scientifically. Rankine, now obsolete,
10:28492. Finally, absolute zero. It is
10:35called absolute zero when there is a
10:37cessation of molecular activity. There
10:43it is. density, the ratio of the mass
10:52and volume of bodies. Therefore, it is
10:55a derived magnitude. absolute density,
10:59mass, volume. Here we have the
11:01different units that relate mass and
11:04those that relate volume. Relative
11:07density is when we compare the density,
11:11whether of a solid or a liquid, with
11:14respect to the density of water. That
11:18is relative density. We find the ratio.
11:22The density of water is 1 g per ml.
11:26density of the solid divided by the
11:27density of water, density of the liquid
11:29divided by the density of water. For
11:31gases, relative density takes a similar
11:34form, but in this case we compare it,
11:36we find the ratio with respect to the
11:39density of air. The density of air is
11:42greater than the density of water. The
11:45density of air is 1.293 g per liter.
11:52Well, the density of oil is a known
11:55value, 0.8. It is important to know,
11:58grams per milliliter, and the density
12:01of mercury is 13.6 g per milliliter.
12:08Mixtures. How is the density of a
12:12mixture found? It is a quotient of the
12:16masses over the volumes. We calculate
12:20the sum and then we divide. If they had
12:25equal volumes, then I just add the
12:28densities and divide by the number of
12:33substances. There we go. Matter and
12:38energy. What is meant by matter? Matter
12:42is everything that occupies a place in
12:44space. It has mass and therefore volume
12:46. According to Einstein, matter is
12:49condensed energy and energy is
12:52dispersed matter. This idea is the
12:56modern conception regarding matter and
12:59energy. So, matter and energy are the
13:07same, except one represents condensed
13:09energy and the other is dispersed
13:12matter. Properties of matter can be
13:18divided into two broad categories:
13:21general or extensive properties and
13:23particular or intensive properties.
13:27Extensive ones depend on mass. That is
13:32the difference from intensive ones,
13:34which do not depend on mass. For
13:38example, those that depend on mass,
13:41which are the general ones: inertia,
13:43indestructibility, impenetrability,
13:46extension, gravity, and divisibility,
13:48which obviously depend on mass. Those
13:52that do not depend on mass, for example
13:55: elasticity, porosity, malleability,
13:57sheets, ductility, ease of making wires
14:00, right? Flexibility, hardness, which
14:03is resistance to scratching,
14:05conductivity, viscosity—which has
14:08nothing to do with mass—and tenacity.
14:12Very well. States of matter: solid,
14:18liquid, and gaseous. To the three
14:21classics, we can add plasma. We will
14:26talk about all of them in detail over
14:29the course of the classes we will have
14:32in chemistry. We will say now that in
14:36the solid state, the forces of cohesion
14:38are greater than the forces of
14:40repulsion. It has a defined shape, an
14:43invariable volume, and also an
14:45invariable mass. In the liquid state,
14:50the forces of cohesion are equal to the
14:53forces of repulsion. Undefined shape;
14:56it adapts to the container that holds
14:58it. The volume is invariable, and so is
15:01the mass. And gaseous, as expected, the
15:05force of repulsion is greater than the
15:07force of cohesion. That determines an
15:12undefined shape, an invariable volume,
15:16and also an invariable mass. The plasma
15:21state is a system found at high
15:23temperatures, consisting of ions and
15:26subatomic particles; it includes the
15:28sun, stars, and Earth's core. We can
15:35add colloid, which is a dispersion
15:37phenomenon. We will see that later on.
15:40It has two phases, one dispersed and,
15:43of course, the dispersant. Also, its
15:47characteristic is Brownian motion. To
15:50identify them, the Tyndall effect is
15:52applied. Classic examples. Gelatin,
15:55flan, egg white. We will see this later
15:59, don't worry. Phase changes. We have
16:05the three classic states of matter:
16:07solid, liquid, and gaseous. So, if I go
16:12from solid to liquid, it's fusion. If I
16:18go from liquid to gas, it's
16:20vaporization. From gas to solid, it's
16:25deposition. From solid to gas, it's
16:27sublimation. It's also called direct
16:30sublimation, and deposition is called
16:31reverse sublimation. From liquid to
16:33solid, it's solidification. From gas to
16:38liquid, it's liquefaction. A classic
16:44example, liquefied gas, right? Here we
16:49have some examples of sublimation.
16:51Classic dry ice, naphthalene. And well,
16:56volatilization evaporates without
16:59boiling; for example, acetone, benzene,
17:02and vaporization, which is the
17:05evaporation that occurs on the surface.
17:09Example, seawater. What do we
17:16understand by energy? It is anything
17:19capable of producing work. It is also
17:24defined, as we already said at the
17:26beginning, remember? Dispersed matter
17:29according to Einstein's theory. Types:
17:33mechanical energy, electrical energy,
17:35chemical energy, radiant energy, light
17:38energy, and atomic energy. We have
17:41Einstein's law of conservation of mass,
17:43which established two equations. The
17:45first equation: energy equals mass
17:47times the speed of light squared. We
17:50already knew that this lowercase 'c'
17:51represents the speed of light in
17:53physics. 3*10 to the 5th km per second.
17:57Using our prefixes, we realize that a
17:59kilometer is equivalent to 10 cubed
18:01meters. Therefore, 3*10 to the 8th m/
18:05second, and 1 m is equivalent—let's
18:09put the equivalence here—1 km is 10³
18:14meters, and of course 1 m is 10² cm.
18:21So, if I want to go from kilometers to
18:23meters, I simply replace it; instead of
18:25putting it here, I will put 10 to the
18:263rd, which is what that means. And the
18:31same for meters to centimeters. The
18:34units of energy: ergs and joules. Here
18:39we have the second equation. Let's
18:41present it. Mass in motion equals mass
18:45at rest, also initial mass, final mass,
18:48in quotation marks, final velocity, C
18:53is the speed of light, that ratio
18:56squared. Well, Surely in some exercise
19:01we will see the direct application of
19:03this property. Let's review a little
19:07bit what mixtures and combinations mean
19:11. Mixtures are those whose components
19:13are in any proportion, do not undergo
19:16changes in their properties, there is
19:18no chemical reaction and they can be
19:21separated, and I think this is the most
19:24important characteristic, by physical
19:26methods. Classic examples, seawater,
19:30brass, which is an alloy, right?
19:34Petroleum. Some of them can be
19:39separated by physical methods, such as
19:41centrifugation. Mixture system phases
19:48liquid, sol, gas, gaseous, colloid.
19:55Regarding their components and
19:57constituents, we have that the
19:58components can be elements or compounds
20:00. elements like copper, for example,
20:04and compounds like H2O. And the
20:07constituents represent the types of
20:09atoms in the mixture. For example, we
20:13have this mixture. So, its constituents
20:15are the types of atoms: hydrogen,
20:18oxygen, sodium, and chlorine. Let's
20:23look at combinations. They are those
20:25whose components are in defined
20:27proportions. There I already have a
20:30difference from mixtures. whose
20:32components are found in any proportion.
20:34Here they must be in defined and fixed
20:37amounts, where chemical reactions occur
20:40. Watch out for that detail, thus
20:43forming products that are new
20:45substances, they are only separated by
20:49chemical means. That is the basic
20:51difference. They are separated by
20:53chemical means and thus form new
20:55substances. For example, the combustion
20:57of paper. The paper turns to ash and,
21:00well, it is a totally different
21:02substance. From ash I can never again
21:05reconstruct the paper, at least by
21:09physical means. Let's go with the
21:13application problems to put into
21:14practice everything we have learned.
21:17They ask me, how many do not correspond
21:20to base units of the international
21:22system? We have talked about the
21:24international system which established
21:26precisely base units, supplementary
21:28units, and derived units. Also the use
21:31of multiple and sub-multiple prefixes.
21:33The base units are those that do not
21:36decompose into others, right? Since
21:39they are defined according to physical,
21:41natural, and invariable phenomena.
21:43Which ones are those? You must remember
21:46it is length, mass, time, the main ones
21:49, electric current, thermodynamic
21:51temperature, luminous intensity, and
21:53amount of substance. There we have them
21:59. Therefore, which are the ones that do
22:02not correspond? The ones that do not
22:04correspond are acceleration. Another
22:07one that does not correspond is volume.
22:10How many do not correspond? Two. The
22:13others do. Do you remember the symbols
22:15for each one? Length is the meter, mass
22:19, kilogram, time, S, electric current
22:24the ampere uppercase A, thermodynamic
22:29temperature Kelvin uppercase K Luminous
22:35intensity, candela lowercase cd, amount
22:38of substance mole. Well, let's go with
22:44the next one. Which is the incorrect
22:47equivalence? How do we figure this out?
22:53What do liters equal? That is the first
22:56thing I need to know. 1 L equals, let's
23:02put it this way, 1 L is equal to 1
23:07cubic decimeter. Okay, that is what it
23:11means. However, I know that 1 m is
23:18equivalent to 10 dm. We know that. from
23:29our table of multiples and submultiples
23:31prefixes, right? I invite you to look
23:33at the table, but since I want the cube
23:37, look, you cube both sides. So, 1 m³
23:44is equal to 10 cubed cubic decimeters,
23:48obviously, but 1 cubic decimeter equals
23:531 liter. Therefore, we say that 1 cubic
23:58meter is equal to 10 cubed. Instead of
24:03putting this, we already know it is the
24:05same as liters. Okay. That is the
24:09equivalence. 1 cubic meter, 10 cubed
24:11liters, or 1000 L. So, the first one is
24:14false and is the solution. The others
24:18we know by simply looking at the table.
24:21This letter, which is the micron,
24:26equals 10 to the -6. Okay, there it is.
24:33This A with the little circle on top is
24:35the Angstrom, a very common measure
24:37that means 10 to the -10. The others
24:41are also known equivalences. 10 yards
24:44is approximately 30 feet. And just as
24:47we explained, 1 cubic decimeter is the
24:49same as 1 L; it is a basic, fundamental
24:51equivalence you must know. And then,
24:55how do you convert 1 cubic meter to
24:57liters? The same goes for centimeters
25:00with decimeters. I invite you to do it
25:01later on your own. Let's keep
25:04practicing. How many microseconds are
25:09in an hour? I recommend that for
25:15conversions you use the unit factor
25:17method. It is very practical and very
25:20simple, for which you need to have the
25:23equivalences. You get the equivalences
25:25from the table. There are other values
25:27that you have to know. For example,
25:30what do I mean by equivalences? I know
25:33that one hour is equivalent to 3600
25:40seconds. Yes. Those are the
25:43equivalences with the main unit, in
25:45this case, seconds. Now I know that
25:48this little symbol here, the micro, is
25:53equivalent to 10 to the -6. Like that.
25:59Therefore, if we are talking about
26:00seconds here, it also has to be seconds
26:02here. Those are my equivalences. Now
26:06let's see what the technique consists
26:07of. You start with the information you
26:11are given, you want to convert it to
26:14microseconds, therefore, you say one
26:17hour, you write it here and then you
26:19are going to multiply by a unit factor,
26:22that is, by an equivalence. Okay? This
26:26first one says that one hour is equal
26:29to 3600 seconds. So, I put one hour
26:31here and in the numerator, I put 3600.
26:35I always put the initial given data or
26:38what I want to cancel in the
26:40denominator. Then, since I already have
26:44seconds, okay? Because look, here I
26:47have canceled this with this, so only
26:49seconds will remain. I am going to
26:51convert from seconds to micros, which
26:52is what they are asking for. I do the
26:54same, I multiply by one because they
26:57are equal. What do I put underneath? 10
27:01to the 6th seconds and above I put one
27:09microsecond. This way, look, I cancel
27:14seconds and I have the solution. What
27:19is the solution? I multiply in the
27:22numerator I only have 3600*1, therefore
27:28I can put 3600 here and in the
27:30denominator I am left with 10 to the
27:326th. Regarding my units, the only unit
27:36left is microseconds. There it is. That
27:39could be an answer. Now I am going to
27:43give it the form they want. Look, they
27:46have 36. I am going to make 36 appear.
27:51So that 3600 I can write as 36*10².
27:59Here I have 10 to the 6th microseconds.
28:04You apply elementary algebra 36*10 to
28:07the 2nd. I move this 6 to the numerator
28:10as 10 to the -6. So you are left with
28:1536*10 to the 8th. microseconds. I think
28:22now you have an answer. There it is.
28:26Okay. That is the famous unit factor
28:29method. You write your equivalencies
28:32and then put your equivalencies as a
28:34fraction. Since they are equal, it is
28:37as if you were multiplying by one,
28:39keeping in mind that in the denominator
28:42you will put precisely what you are
28:44what you want to eliminate, okay? In
28:48that way, I guarantee that I eliminate
28:50the units I do not want and am left
28:52with those I will use, which is what
28:54they are asking for. Let's see more
28:57examples. For example, they ask me to
29:01convert 18 kg per liters over hours to
29:04grams per milliliter per minute. The
29:08first thing I do then is establish my
29:09equivalencies from kilograms to grams,
29:11from liters to milliliters, and from
29:13hours to minutes. the equivalencies I
29:15have on paper in my table or those I
29:17know. Of course, 1 kg is 1000 g. You
29:22know that. I am going to put my
29:23equivalency here. 1 kg is equal to 10
29:26cubed grams. That will be for my
29:28prefixes. Then I also know that 1 liter
29:31, how many milliliters is it equivalent
29:34to? 1 L is equivalent to 10 cubed
29:40milliliters. We put it like this, okay?
29:48And then one hour is equivalent to 60
29:53minutes. These are all units that you
29:56know. Alright, now we are going to use
29:59our unit factor strategy. We put 18
30:01here kg per liter divided by h. Look,
30:10for the first one. I know that 1 kg is
30:14equal to 10 cubed. I'm going to put
30:16kilogram at the bottom. Why are you
30:17putting it at the bottom? Because I
30:19want to eliminate it with the one above
30:21, since I want grams to appear in the
30:23numerator. There it is. Let's see, the
30:27next one. The next one is similar. 1 L.
30:31I want milliliters to appear above and
30:33I have liters in the numerator, look.
30:36Therefore, I'm going to put liters here
30:38. That's how I realize, 1 L and in the
30:41numerator I put 10 cubed milliliters.
30:48Now the last one. Watch out, I want
30:54minutes to appear in the denominator
30:56now and I have hours in the denominator
30:59. So, hour is now going to appear in
31:01the numerator. I put one hour here and
31:05below I put 60 minutes. Done. Once that
31:11is done, what follows is simply to
31:14multiply, multiply, and divide. Look, I
31:18cancel like this, kilogram with
31:20kilogram, I cancel hour with hour, I
31:22cancel liter with liter and I will be
31:23left with the units I want. Let's solve
31:27it. In the numerator I have 18 by these
31:32, I combine them and I get 10 to the
31:34sixth. I add the exponents and in the
31:36denominator instead of putting 60 I put
31:386*10. Always express it with powers of
31:41base 10. Here I have, uh, grams per
31:47milliliter left over minute. That is
31:50already converted. Well, 18/6 gives me
31:543. Since this is 10 to the 1, it would
31:57have to be subtracted, right? Because
31:59they have the same base and they are
32:01being divided. So I will be left with 3
32:03*10 to the fifth. I'll put 6-1. This
32:10finally gives me 3*10 to the fifth
32:14units gram per milliliter divided by
32:20minute. Solution. There it is. It would
32:27have to be the answer, letter E. Let's
32:35move on to the next one. I have a kind
32:39of equation. They tell me to calculate
32:41the value of R in cubic centimeters
32:43from the following expression. Look at
32:45how R is. And I have a root. So, I have
32:47to square it to get rid of the root
32:50here and here. I will be left, of
32:53course, with R squared cm squared here.
32:56All of this is equal. The root goes
33:00away with the exponent 27 meters cubed
33:04per liter per centimeter. All of this
33:08divided by R. Now I can do the
33:09following. I multiply in this way. I
33:13will be left with R cubed and
33:15centimeter squared goes over to
33:16multiply there. It is equal to 27 m
33:20cubed per liter per cubic centimeter.
33:27Sure, because this one hooks up with
33:28the one here. Centimeter times
33:30centimeter gives me cubic centimeter.
33:31Now come the equivalencies. For example
33:35, 1 m is equal to 10 cm. But if I want
33:41the cube, well, I raise it to the cube.
33:43So I have here 1 m³ is equal to 10 to
33:47the sixth cubic cm. That is what I am
33:51going to put here instead of cubic
33:53meter. And I already knew that 1 L is
33:57equivalent to 10 to the third cubic cm.
34:04Ready. Well, I am going to replace
34:07these two. Therefore, you have
34:09something like this. R³ is equal to 27
34:13. Instead of putting cubic meters, you
34:16are going to put 10 to the sixth cubic
34:18cm, all cubic centimeters, of course.
34:21Instead of liters, you are going to put
34:2410 to the third cubic cm times cubic
34:28centimeters. What am I left with? I am
34:36left with r³ is equal to 27*look 10 to
34:43the 9th cm 3 times 3 is 9. Ready. Since
34:53they ask for R, you are going to take
34:55the cube root here and cube root over
34:57there, since R is raised to the cube.
34:59Look, this way, I cancel like this and
35:06I am left with R here. I take this cube
35:09root of everything. cube root of 27 is
35:123 times, uh, taking the cube root of 10
35:15to the 9th is as if I divided these
35:18exponents, therefore I will be left
35:20with 10 to the 3rd and here I also
35:23divide that in roots. In algebra we
35:27have seen it. There it is. That is the
35:31answer. So they ask me for it in cubic
35:32centimeters. Here is the solution 3*10
35:34to the third. Answer is letter C.
35:38applying our uh conversion ideas that
35:41we had explained at the beginning.
35:44Problem six, we leave it as homework.
35:52There we leave you two additional
35:53exercises. Now a little bit of
35:57temperature. They tell us that a
36:01student has a fever and his temperature
36:03indicates 38ºC. How much will it
36:06indicate on his thermometer? in degrees
36:09Fahrenheit. You have to remember your
36:13equivalencies. Everything is a matter
36:14of equivalencies, right? Degrees
36:16Celsius over 5 was equal to degrees
36:24Fahrenheit-32 over 9. But the complete
36:29scale, how was it? You added those,
36:32those are the relative ones, the
36:34absolute ones, right? Kelvin and
36:36Rankine. Here we put Kelvin-273. Watch
36:44out. And the other was Rankine, which
36:47is already in disuse. We only use it to
36:49do some exercises. Okay, I think we
36:55didn't put its name, right? Rankine.
37:00And this is Kelvin. Here we are going
37:07to use these two. They tell me it is at
37:1038ºC. So I replace, I put 38 here/5 is
37:15equal to Fahrenheit-32/9. I want to
37:19find Fahrenheit. Therefore, I am left
37:23here with 38/5. This 9 goes over to
37:27multiply. times 9 and then add 32.
37:34First, obviously, I do this
37:35multiplication. This is equal to
37:37Fahrenheit. And from here we get that
37:40it is 1004. Careful, though, in degrees
37:52Fahrenheit. Answer. Here they put
37:57Celsius, it must be Fahrenheit. Okay.
38:07Number two. At what temperature on the
38:10Celsius scale is the Fahrenheit reading
38:12equal to 2.6 times the Celsius reading?
38:16I start the same way as the previous
38:19ones with the equivalence: degrees
38:22Celsius over 5; they ask me to relate
38:25degrees Celsius with degrees Fahrenheit
38:28-32/9. But they are telling me that one
38:31will be 2.6 times the reading of the
38:33other. Therefore, you say, let's see,
38:37degrees Celsius, degrees Fahrenheit.
38:43Suppose that degrees Celsius is x. Then
38:45they tell me that Fahrenheit is equal
38:47to 2.6 times that. There it is. Well,
38:51now I am going to replace the values I
38:52have put there. So, instead of degrees
38:55Celsius I will put x over 5 is equal to
38:582.6 -32/9. Then, what comes next? A
39:05simple operation comes next. I multiply
39:07in this direction. 9x 5*2.6 x-32.
39:15distributive property. Here I have 9x,
39:18uh, here we get 13 x-160. 160 positive
39:26over here, 13x-9x 4x Well, 160/4 from
39:40here you discover that x must be 40.
39:45Since the question refers to the
39:47Celsius scale, 40ºC, solution simply
39:56by applying our equivalences. Let's go
39:59with the next one. A new scale in
40:04degrees X is constructed in which the
40:07temperature at the freezing and boiling
40:10points of water are -10º X and 110º
40:13X. Calculate what a reading of -20ºC
40:16is equivalent to on the X scale. This
40:22type of exercise is solved using
40:25proportions. Let's make a little sketch
40:29. Degrees X we compare it with degrees
40:34Celsius, since that is what the
40:36exercise mentions. Look, there we have
40:42our little diagram. They say that the
40:46boiling point is 110. And the freezing
40:53point is -10. Let's put -10 here. But
40:57you know that boiling in degrees
40:59Celsius is equivalent to how much, and
41:05freezing of water, of course, is
41:14equivalent to zero and then 100.
41:23Remember that these are data we already
41:25know. the boiling point of water at
41:27100ºC and the freezing point of water
41:29at 0ºC. Those are the correspondences
41:33on the X scale. But then they are
41:35asking me what a reading of -20ºC is
41:37equivalent to. -20 is around here,
41:39right? So we put -20 over here. And
41:43since we don't know this, I'm going to
41:46call it "a" or a "T" for temperature.
41:50Well, we are going to use proportions,
41:53right? which is the master of
41:55proportions. What does that mean about
41:57proportions? Remember that thermometers
42:01are graduated according to a scale, and
42:04if I have two points, I already have
42:06the scale's proportion. That's what
42:09it's about. In other words, the
42:10difference will ultimately be the same.
42:12For example, look, I can subtract this
42:14one with this one, and then this one
42:16with this one; it will be the same as
42:18if I subtract, for example, this one
42:20with this one and then this one with
42:22this one. For example, 110 minus -10,
42:30okay, over -10-t will be equal to the
42:35same thing I do on the Celsius scale,
42:41that is, 100 minus 0 and then 0 minus
42:46-20. The same proportion will be
42:49established. In the numerator, I will
42:52be left with 110. Minus times minus is
42:55plus and in the denominator -10-t,
42:59which is what I want; here I got 100 in
43:01the numerator and in the denominator,
43:03look, minus times minus is plus, so I'm
43:06just left with 20. Let's keep working.
43:09I'll go over here. In the numerator, I
43:13will be left with 120. Then in the
43:17denominator -10-t, and here I got
43:19100/20 equals 5. All of this moves over
43:22to multiply. So I have 120 here; 5 when
43:25multiplying -10 gives me -50-5 t. Since
43:30it's multiplying both. I'll move this
43:32to add to the other side. So I have 120
43:35+ 50 equals -5 t. I got 170 equals -5 t
43:41here. Then t must be equal to 170/-5.
43:49Okay. T is equal, let's write it this
43:56way, 170/-5 equals t. Therefore, from
44:04here you discover that t is equal to
44:05-34. That is the equivalent in degrees
44:10x. Your answer. We have established
44:15proportions. It is said. Working with
44:19this method is quite practical because
44:21I could have subtracted, for example,
44:23this one with this one and then this
44:25one with this one, but what I do on one
44:26side I would have to do on the other. I
44:29chose that for the practicality of
44:30working with zero. Look, 100 minus 0
44:33and then 0 minus 20. But I could have
44:36worked 100 with this one and then like
44:39that. Then, I would have done the same
44:41here. This here. And then these, I
44:45think I told you, but, eh, if I worked
44:49like that and like that, I would have
44:50had the variable in the numerator and
44:52in the denominator. So I tried to
44:54choose what was most practical. First
44:57these with these and then the
44:59difference that exists from here to
45:01here, since that difference will remain
45:03constant because we had said that this
45:06is a calibrated measurement that
45:11maintains proportion. By establishing
45:14two points or having two points, I can
45:16already find any other. Anyway, that's
45:19how you solve these exercises. Is it
45:21clear now? Well, here are a few more
45:24for you to practice or tell me how it
45:26went. Let's move on to density. What do
45:31we have there? Do you remember your
45:33absolute density formula? Yes, let's
45:36start with that. Density is equal to
45:39mass divided by volume. They are asking
45:43me for the mass. How many grams are in
45:45400 ml of ethyl alcohol? That
45:47milliliter is, of course, a unit of
45:49volume. They have given me the volume,
45:51which is 400 ml, and they even gave me
45:58the density of 0.8 g/ml. We are in luck
46:02because it has the same unit,
46:04milliliters. So you say, well, since I
46:06want to find the mass, I rearrange it;
46:09mass is equal to density times volume.
46:13Therefore, we write the appropriate
46:15values. Instead of density, I put 0.8.
46:17I won't put the units because I've
46:19already checked that they're correct.
46:21Times 400. Therefore, the mass has to
46:24be 320. Obviously, it will come out in
46:28grams. Done, solution. The next one is
46:37a density problem, but with a mixture.
46:40Two liquids are mixed: liquid A, with a
46:42density of 1 g/ml, with liquid B, which
46:45has a density of 2 g/ml, in a
46:46volumetric ratio of 3 to 2. Find the
46:49density of the mixture. You have to
46:50remember this. In this problem, density
46:58one is 1 g/ml. Volume one. There is an
47:05interesting detail here. And density
47:07two is 2 g/ml, and volume two says they
47:12are in a ratio of 3 to 2. So volume one
47:16can be like 3K and volume two can be
47:19like 2K. That is what the ratio means.
47:23With these data, I am going to
47:25substitute. So I write it in the
47:27following way. Density of the mixture 1
47:34*3K + 2*2K over 3K + 2K. Okay, since
47:45that was the volumetric ratio. Density
47:48of the mixture. I do the math; I am
47:51left with 3K + 4K. Then, in the
47:54denominator, 3K + 2K is 5K. 3 + 4, 7. K
47:59, and here I have 5K. Notice that this
48:03constant can be canceled. So I am left
48:05with 7/5. 7/5 is 1.4. That is the
48:10solution. The density of the mixture
48:13should be 1.4 g/ml, since that is the
48:18unit. There it is. Solution. Applying
48:23this idea that we had seen in the
48:26theoretical section. The most notable
48:29thing here is perhaps the volume ratio.
48:32I wrote them in that way. Liquid A is
48:37mixed with water such that the
48:39resulting density is 1.5 g per cm³ in
48:42a volume of 1 L. Then, 100 cm³ of A is
48:45removed and the same amount of water is
48:48added. As a result, the density
48:51decreases to 1.25 g per cm³. Find the
48:55density of liquid A in grams per cm³.
48:57If they ask for the density of liquid A
49:00, density of A, I need to have the mass
49:05of A, of course, divided by the volume.
49:12Here we have the mixture. This is in 1
49:15L. 1 L has 1000 cm³. Therefore, if you
49:25apply your formula, density equals mass
49:28divided by volume, the density is 1.50.
49:33It's like the initial density. Let's
49:35call it density one. Substituting, we
49:38obtain that 1.50 is equal to the mass.
49:48That mass will be determined by both
49:51the water and liquid A. So, let's put
49:54mass one here and here I have 1000 cm³
50:02. We multiply and obtain that 1500 g is
50:10like the mass one. But, what happened
50:13there? 100 cm³ of A is removed and the
50:16same amount of water is added. As a
50:19result, there is a density of 1.25. So
50:22we now have a density two. Density two,
50:28which will be equal to m2, of course,
50:33divided by volume. The volume will not
50:37vary; it remains 1000 cm³. We perform
50:40the same operation. Instead of density
50:41two, we put 1.25. This is equal to mass
50:47two, volume 1000 cm³. We move this to
50:54multiply. So I am left here with 1250 g
50:57. But there is a detail regarding mass
51:00two. What happens with mass two? What
51:02does mass two mean? It means mass one
51:10minus the 100 cm³ that will represent
51:18a mass. For example, the mass of A
51:20already appeared and you added 100 of
51:23water. Those 100 cm³ represent 100 g
51:27of water. Well, the initial mass you
51:33already knew was 1500. Therefore, we
51:39will replace 1250 is equal to 1500 +
51:46100 minus the mass of A. I'll put the
51:50mass of A over here. I have 1600-1250.
51:57I get that the mass of A is 350 grams.
52:07We already have the mass, which is 350.
52:13We do our substitution, 350 g. But we
52:18are working with a volume of how much?
52:22With a volume of 100 cm³. since that
52:27is what we used to find the mass, and
52:35that's it. We divide and obtain that
52:37the density of A is equal to 3.5 grams
52:46per cm³. Answer. Well, we wanted to
52:56review a bit of the density part, also
52:58temperature, unit conversions. Let's
53:05look a little bit at matter and energy.
53:09While I leave you a couple of
53:10additional exercises here. Matter and
53:16energy. The property of matter that
53:18determines the degree of resistance to
53:20scratching is we had discussed that, I
53:25think, uh, hardness was not put here.
53:33That is the solution. Let's see some
53:36exercises. Uh, the third one, right?
53:41Determine the energy in Joules released
53:43when exploding a small 200 g uranium
53:46reagent. How do we solve that? It is an
53:49application problem. We have to use
53:52Einstein's very famous formula. Energy
53:56is equal to mass times the speed of
54:01light squared. Keeping units in mind,
54:07remember that they are asking me for
54:10Joules, so I need to convert the mass
54:13from grams to kilograms and work with
54:16meters per second. The unit of C. How
54:20would C be now? C has to be the speed
54:24of light. We express it like this. 3*10
54:28to the power of 8 meters per second and
54:33200 g for the mass. Instead of 200 g we
54:39divide by 1000. We already knew that.
54:430.2 kg. Now, let's substitute. So,
54:48energy, instead of m I put 0.2. I
54:53already have the correct unit. Instead
54:55of C I put 3*10 to the 8 only. All of
55:00this squared. Energy 0.2 and then here
55:099*10 to the 16. When we multiply we get
55:199*0.2 1.8*10 to the 16 Joules. There it
55:33is solution B. Simply applying the
55:40formula. I think we have another
55:43application one here. I leave the
55:46others for you to practice. Here it is.
55:58Problem number four. What will be the
56:00mass of the products of the reaction if
56:03the grams of uranium-235 undergo
56:04nuclear fission and produce 1.5*10 to
56:07the 14 ergs of radiant energy?
56:09Releasing thermal energy. Well, we are
56:13going to apply this formula, but they
56:19are talking to me here about ergs and
56:25ergs relates to what? Ergs relates to
56:30centimeters and with grams. The idea is
56:37this. You have uranium here -235
56:47undergoes nuclear fission. Fission, a
56:49breakdown and it produces a large
56:53amount of radiant energy releasing, of
56:58course, thermal energy. It says to me,
57:01"What will be the mass of the products?
57:02" Here I had 2 g, but there is a piece
57:07of that energy that transformed into
57:09radiant, which is precisely what we are
57:12going to find. It interconverted. Once
57:16we find that with this formula, what
57:18was made isothermal, there it is, with
57:23that we find it and then we subtract
57:25from what there was, which was 2 g. So,
57:30you say the energy value is 1.5*10 to
57:35the 14th ergs. This is equal to the
57:40mass. The mass that was lost and
57:45converted into energy by C². How much
57:49will it be. Since we are working in
57:53centimeters, you say C is 3*10 to the
57:5710th cm. Let's write this better here.
58:05Like this. Cm per second. Velocity. 3*
58:1010 to the 10th. The formula says I have
58:17to square this. Therefore, I have here
58:201.5*10 to the 14th. This is equal to
58:24mass times 9. This affects both, times
58:2610 to the 20th. Move it to divide 1.5*
58:3110 to the 14th/9*10 to the 20th. We
58:38perform the division and we are left
58:40with. 1.67. *10. Look, here I have 14,
58:49here I have 20. Subtracting 14-20 gives
58:51-6. A tiny bit of mass. In grams, of
58:56course. What comes next? The difference
59:01. The mass of the product, therefore,
59:07will be 2 minus what became radiant.
59:101.67*10 to the -6 th. What does this 10
59:14to the -6 th mean? That I move to the
59:17left six spaces, look, like this. In
59:20other words, an incredibly small
59:21decimal number, almost nothing has been
59:23converted. Therefore, my answer is 1,
59:27approximately 1.99 g. We can leave it
59:30like that. Well, there it is. So it is
59:36important that we know how to recognize
59:39this formula and use it. There we have
59:43two applications, one direct and
59:44another where we were asked for the
59:46mass of the products. Anyway, guys,
59:52here we leave you more so that you can
59:53practice later. See you. See you soon.

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