<?xml version="1.0" encoding="UTF-8"?><rss xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:atom="http://www.w3.org/2005/Atom" version="2.0" xmlns:media="http://search.yahoo.com/mrss/"><channel><title><![CDATA[Chemistry Basics]]></title><description><![CDATA[A practical learning approach.]]></description><link>https://chemistrybasics.org/</link><image><url>http://chemistrybasics.org/favicon.png</url><title>Chemistry Basics</title><link>https://chemistrybasics.org/</link></image><generator>Ghost 3.40</generator><lastBuildDate>Fri, 10 Apr 2026 20:35:44 GMT</lastBuildDate><atom:link href="https://chemistrybasics.org/rss/" rel="self" type="application/rss+xml"/><ttl>60</ttl><item><title><![CDATA[Announcement]]></title><description><![CDATA[<p>Dear Reader</p><p>This is to inform you that I have not posted anything on this platform since a few days because I am out of my country. As soon as I am back, the posting will resume.</p>]]></description><link>https://chemistrybasics.org/announcement/</link><guid isPermaLink="false">6448fc336a348f57e3176219</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Wed, 26 Apr 2023 10:31:15 GMT</pubDate><content:encoded><![CDATA[<p>Dear Reader</p><p>This is to inform you that I have not posted anything on this platform since a few days because I am out of my country. As soon as I am back, the posting will resume.</p>]]></content:encoded></item><item><title><![CDATA[Problem # 57 Ice]]></title><description><![CDATA[<p>You leave in your fridge a glass bottle filled with water and well closed. After some hours you come to take the bottle out of the fridge. Surprise: the bottle has broken into pieces of ice and glass.</p><p><strong><em>Question</em></strong>: How do you explain that?</p>]]></description><link>https://chemistrybasics.org/problem-57-ice/</link><guid isPermaLink="false">64284d4d6a348f57e31761fe</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sat, 01 Apr 2023 15:32:44 GMT</pubDate><content:encoded><![CDATA[<p>You leave in your fridge a glass bottle filled with water and well closed. After some hours you come to take the bottle out of the fridge. Surprise: the bottle has broken into pieces of ice and glass.</p><p><strong><em>Question</em></strong>: How do you explain that?</p>]]></content:encoded></item><item><title><![CDATA[Problem # 56 Electrical conduction of aqueous solutions [Solved]]]></title><description><![CDATA[<p>You are given two aqueous solutions <strong>A</strong> and <strong>B</strong>. The only information you have is that <strong>A</strong> conducts electricity, <strong>B</strong> doesn’t.</p><p><strong><em>Question: </em></strong>Which of the two solutions contains potassium bromide (KBr) and which one contains ethanol (CH<sub>3</sub>CH<sub>2</sub>OH)? Explain/justify your answer.</p><p><strong><em> </em></strong><em><strong>Answer</strong></em>:</p><p>Solution A is</p>]]></description><link>https://chemistrybasics.org/problem-56-electrical-conduction-of-aqueous-solutions-solved/</link><guid isPermaLink="false">64204f8b6a348f57e31761e6</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sun, 26 Mar 2023 14:06:17 GMT</pubDate><content:encoded><![CDATA[<p>You are given two aqueous solutions <strong>A</strong> and <strong>B</strong>. The only information you have is that <strong>A</strong> conducts electricity, <strong>B</strong> doesn’t.</p><p><strong><em>Question: </em></strong>Which of the two solutions contains potassium bromide (KBr) and which one contains ethanol (CH<sub>3</sub>CH<sub>2</sub>OH)? Explain/justify your answer.</p><p><strong><em> </em></strong><em><strong>Answer</strong></em>:</p><p>Solution A is KBr(aq) solution; Solution B is CH<sub>3</sub>CH<sub>2</sub>OH(aq) solution.</p><p>Indeed, KBr is an ionic compound formed between a metal (K) and anon-metal (Br<sub>2</sub>); when it dissolves in water, it dissociates into its ions K<sup>+</sup> and Br<sup>-</sup>. The presence of ions allows KBr(aq) solution to conduct electricity. Such solutions are called “<strong>Electrolytes</strong>” or “<strong>Electrolytic solutions</strong>”. Hence <strong>A</strong> corresponds to KBr(aq) solution.</p><p>CH<sub>3</sub>CH<sub>2</sub>OH is a covalent compound (made of all non-metals); its dissolution in water does not involve the dissociation into ions. The absence of ions in the aqueous solution of CH<sub>3</sub>CH<sub>2</sub>OH explains why that solution does not conduct electricity. Such solutions are called “<strong>Non-electrolytes</strong>” or “<strong>Non-electrolytic solutions</strong>”. That is why the aqueous solution of ethanol doesn’t conduct electricity.</p>]]></content:encoded></item><item><title><![CDATA[Problem # 55 Why objects have different colors[Solved]]]></title><description><![CDATA[<p><em><strong>Question</strong></em>: Why a given object is colored orange, i.e. absorbs the blue light, whereas another object is red, i.e. absorbs the green light, while another reflects the whole white light to appear white, another black, another is transparent but colorless?</p><p><em><strong>Answer:</strong></em></p><p><strong>A/ Absorption, Transmission, Reflection of light</strong></p><p>As</p>]]></description><link>https://chemistrybasics.org/untitled-10/</link><guid isPermaLink="false">640e0f756a348f57e317619c</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sun, 12 Mar 2023 17:53:06 GMT</pubDate><content:encoded><![CDATA[<p><em><strong>Question</strong></em>: Why a given object is colored orange, i.e. absorbs the blue light, whereas another object is red, i.e. absorbs the green light, while another reflects the whole white light to appear white, another black, another is transparent but colorless?</p><p><em><strong>Answer:</strong></em></p><p><strong>A/ Absorption, Transmission, Reflection of light</strong></p><p>As we have seen previously, visible light is located in the region of light spectrum where all visible colors are located: λ= 700-400 nm. The lights outside that region, such as UV and IR, cannot be detected by our eyes.</p><p>We have also seen that the color of an object depends upon the light absorbed, reflected, or emitted by the object.</p><p>When the white light hits an object, a light of a given wavelength or frequency can match the energy needed by an electron to be excited. In this case that light (with a specific color) is absorbed by the object. The remaining other lights/colors that do not match the energy of excitation of the electrons are not absorbed.</p><p>Since the excitation energy differs from one chemical species to another due to their different electron structures, this explains why they absorb lights of different wavelength to be excited.</p><p><strong><em>Transparent objects</em></strong><strong>: </strong>a transparent object let pass light (whole white light or some lights); it can be colored or colorless.</p><p>- A colorless transparent object lets pass and transmit the whole white light (made of all colors), without absorption of any light color; ex: Water.</p><p>- A colored transparent object absorbs a light/color that matches the excitation energy of the electrons in the object, and let pass all the remaining lights/colors; the color perceived by our eyes is the color corresponding to the remaining or transmitted colors: the complementary color of the absorbed one; ex: KMnO<sub>4</sub> solution is red-purple because it absorbs yellow-green color.</p><p><strong><em>Opaque objects</em></strong><strong>: </strong>an opaque object is an object that is not transparent to light, this applies mainly to solids.</p><p>- A white object reflects the whole white light without absorbing any light color; ex: White paper.</p><p>- A colored but opaque object reflects a particular light color, and absorbs the remaining light colors of the white light; ex: The red colored car.</p><p>- A black and opaque object absorbs all light colors of the white light; ex: The black bag in which you keep your computer.</p><p><strong>B/ Emission of light</strong></p><p>Electrons in some chemical species such as metals and metal ions can be excited to higher energy levels by absorbing a light in the visible region, creating an excited state of the species. The excited state is not stable and tends to return to the ground state or non-excited state.</p><p>In the process, the energy absorbed for excitation is emitted back in form of corresponding light with the corresponding color <em>(you can revisit Problem 5)</em></p><p><strong><em>Emission flame:</em></strong> the principle described above is used to test and identify some metals and their ions by a technique called Flame test. In that technique, a sample is put in flame; the energy of the flame excites electrons in the sample to higher energy levels, excited state; but that excited state returns immediately to the ground state by emitting the energy previously absorbed. The color of the flame allows to identify the presence or not of a given metal <em>(see Figure7 of Problem 5)</em></p><p><strong><em>Electronic excitation can be caused by other means of energy such as electricity</em></strong>: the same principle is applied except that another kind of energy, such as electricity, is used. For example a sodium-vapor lamp is gas-discharge lamp where electrons of the gas metal (or any other gas such as He, Ar) are excited to higher energy levels, and when they come back to the ground state, they emit light of a characteristic color. Ex; Colors of the Traffic lights: the bright Yellow color is emitted by Sodium, Na.</p><figure class="kg-card kg-image-card"><img src="http://chemistrybasics.org/content/images/2023/03/image-1.png" class="kg-image" alt></figure><p><strong>                                                        T</strong>r<strong>affic Lights</strong></p><p><strong>                                                               </strong>(Pngtree)</p>]]></content:encoded></item><item><title><![CDATA[Problem #54[Solved]]]></title><description><![CDATA[<p><em><strong>Answer:</strong></em></p><p>1/ The change is caused by the phenomenon called “Evaporation”.</p><p>2/ No, because evaporation takes place at the surface of water (see Figure below), and since the two glasses have different surface (openings), the rates of water evaporation in the two glasses are different; therefore the remaining volumes of</p>]]></description><link>https://chemistrybasics.org/problem-54-solved/</link><guid isPermaLink="false">63ff2d866a348f57e3176156</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Wed, 01 Mar 2023 10:54:19 GMT</pubDate><content:encoded><![CDATA[<p><em><strong>Answer:</strong></em></p><p>1/ The change is caused by the phenomenon called “Evaporation”.</p><p>2/ No, because evaporation takes place at the surface of water (see Figure below), and since the two glasses have different surface (openings), the rates of water evaporation in the two glasses are different; therefore the remaining volumes of water are different.</p><p>3/ Yes, the rate of water evaporation in B is higher than in A, because water surface in B is larger than in A. This means that glass B has lost more water by evaporation than glass A and the water volume remaining in B is smaller than in A.</p><figure class="kg-card kg-image-card"><img src="http://chemistrybasics.org/content/images/2023/03/image.png" class="kg-image" alt></figure><p><strong>                                      Water Evaporation Principle</strong></p><p>                                                       (ResearchGate)</p>]]></content:encoded></item><item><title><![CDATA[Problem # 54]]></title><description><![CDATA[<p>You are given two glasses A and B of equal volumes; glass A is tall, has a small base; glass B is short with a larger base. You fill them with water and leave them in the laboratory. After a few days, you examine them and find that the levels</p>]]></description><link>https://chemistrybasics.org/problem-54/</link><guid isPermaLink="false">63f107926a348f57e317614a</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sat, 18 Feb 2023 17:18:58 GMT</pubDate><content:encoded><![CDATA[<p>You are given two glasses A and B of equal volumes; glass A is tall, has a small base; glass B is short with a larger base. You fill them with water and leave them in the laboratory. After a few days, you examine them and find that the levels of water are lower in both glasses than when you left them. Then you decide to measure the remaining volumes.</p><p><strong><em>Questions</em></strong>:</p><p>1/ What phenomenon caused that change?</p><p>2/ Do you expect to find equal remaining volumes in the two glasses? Justify your answer.</p><p>3/ Do you expect to find different remaining volumes in the two glasses? If yes, compare the remaining volumes V<sub>A</sub> and V<sub>B</sub>, and explain.</p>]]></content:encoded></item><item><title><![CDATA[Problem # 53 BP of water in a solution [Solved]]]></title><description><![CDATA[<p>You are at sea level and you are boiling water in which a spoon of sugar has been dissolved. <strong><em>Question</em>: </strong>Do you expect that solution to boil at: 100<sup>o</sup>C, &lt;100<sup>o</sup>C, or &gt;100<sup>o</sup>C? Justify your answer.</p><p><em><strong>Answer</strong></em>: Water in that solution, or any</p>]]></description><link>https://chemistrybasics.org/problem-53-bp-of-water-in-a-solution-solved/</link><guid isPermaLink="false">63e6819c6a348f57e317612e</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Fri, 10 Feb 2023 17:48:26 GMT</pubDate><content:encoded><![CDATA[<p>You are at sea level and you are boiling water in which a spoon of sugar has been dissolved. <strong><em>Question</em>: </strong>Do you expect that solution to boil at: 100<sup>o</sup>C, &lt;100<sup>o</sup>C, or &gt;100<sup>o</sup>C? Justify your answer.</p><p><em><strong>Answer</strong></em>: Water in that solution, or any aqueous solution of a non-volatile solute, will boil at a higher temperature than 100<sup>o</sup>C. As said in previous problems, the boiling temperature is the temperature at which the vapor pressure of the liquid equals the atmospheric pressure. But we know that the vapor pressure of the liquid is formed by the molecules coming from the surface of the liquid. In this case, the surface of the liquid solution contains molecules of water and molecules or ions from the solute; in other words the number of water molecules, or fraction of water molecules, on the surface of the liquid solution is smaller than on the surface of pure water. That is why it will require more heating or higher temperature than 100<sup>o</sup>C for the solution to form water vapor pressure equal to the atmospheric pressure.</p>]]></content:encoded></item><item><title><![CDATA[Problem #52 Boiling Temperature of Water[Solved]]]></title><description><![CDATA[<p>Answer:</p><p>Yes, there are some locations where the boiling temperature of water is higher than 100oC, and others where it is lower than 100oC.</p><p>Remember that the boiling temperature of a liquid is the temperature  at which the vapor pressure of the liquid is equal to the atmospheric pressure. </p><p>On</p>]]></description><link>https://chemistrybasics.org/problem-52-boiling-temperature-of-water-solved/</link><guid isPermaLink="false">63d5094b6a348f57e31760d3</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sat, 28 Jan 2023 12:57:33 GMT</pubDate><content:encoded><![CDATA[<p>Answer:</p><p>Yes, there are some locations where the boiling temperature of water is higher than 100oC, and others where it is lower than 100oC.</p><p>Remember that the boiling temperature of a liquid is the temperature  at which the vapor pressure of the liquid is equal to the atmospheric pressure. </p><p>On the other side, you know that the atmospheric pressure on a given point on the Earth is the force exerted by a column of atmospheric air on that point. The normal boiling points of different liquids that we find in the tables of data and textbooks are boiling points at Sea level, i.e. altitude 0 m.</p><p>If a location point is at higher altitudes, the column of air above that point is shorter than the column of air at Sea level; therefore the atmospheric pressure at that point is lower than at sea level and it will require less vapour pressure for water to boil. In other words, liquids at high altitudes than sea level  boil at a lower temperature than at sea level.</p><p>Examples: BP of water at the summit of Mount Kilimanjaro(~80.7oC), Mount Everest(68oC)</p><p>But if the location is below Sea level, the column of air above it is longer than at sea level, and the atmospheric pressure will be higher.</p><p>Example: I read in Google that the Eath's lowest land elevation point is at the Dead Sea shores, located at the border of Israel and Jordan with an altitude of 420 m below sea level; the boiling temperature of water at that location is ~101oC.</p>]]></content:encoded></item><item><title><![CDATA[Problem # 52: Boiling Temperature of water]]></title><description><![CDATA[<p><em>Question</em>: Answer, by Yes or No, if there may be a point/location on the Earth where the boiling point of water is lower or higher than 100<sup>o</sup>C, and justify your answer.</p>]]></description><link>https://chemistrybasics.org/problem-52-boiling-temperature-of-water/</link><guid isPermaLink="false">63cbafa96a348f57e31760ba</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sat, 21 Jan 2023 09:30:07 GMT</pubDate><content:encoded><![CDATA[<p><em>Question</em>: Answer, by Yes or No, if there may be a point/location on the Earth where the boiling point of water is lower or higher than 100<sup>o</sup>C, and justify your answer.</p>]]></content:encoded></item><item><title><![CDATA[Problem #51[Solved]]]></title><description><![CDATA[<p>You have two containers of the same volume. The container A has a fixed or constant volume, the container B has an expandable/expansible volume. You fill the two containers with air at rtp and close them tightly. You want to heat the air in both containers up to 50<sup></sup></p>]]></description><link>https://chemistrybasics.org/problem-51-solved/</link><guid isPermaLink="false">63c28aa26a348f57e31760a6</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sat, 14 Jan 2023 11:03:41 GMT</pubDate><content:encoded><![CDATA[<p>You have two containers of the same volume. The container A has a fixed or constant volume, the container B has an expandable/expansible volume. You fill the two containers with air at rtp and close them tightly. You want to heat the air in both containers up to 50<sup>o</sup>C.</p><p><strong><em>Question: </em></strong>Will it require the same amount of heat for the two containers, or different amounts of heat? Justify your answer.</p><p><strong><em>Answer</em></strong>: They will require different amounts of heat.</p><p>Because, in the fixed volume container A, the heat supplied will be fully utilized to increase the internal energy of the air, ΔU, or its temperature up to 50<sup>o</sup>C.</p><p>For the expansible/expandable container B, there are two actions: increasing the internal energy, ΔU,  or temperature of air up to 50<sup>o</sup>C; but also the volume expansion accomplishes work against the atmospheric pressure, w = -PΔV, where P = atm. pressure, and ΔV= volume expansion. (By convention the work done by a system on its surroundings has a negative sign because it results in a transfer of energy from a system to its surroundings).</p><p>In other words, in the expandable container B, a certain amount of internal energy is used to do the expansion work, not to increase the temperature; therefore, it will require more heat for the air in the container B to reach the same temperature of 50<sup>o</sup>C.</p><p>This explains why cooking in an under-pressure casserole is faster than in an open casserole.</p>]]></content:encoded></item><item><title><![CDATA[Problem # 50 Pressure [Solved]]]></title><description><![CDATA[<p><em><strong>Questions</strong></em>:(1) What’s the definition of pressure and more specifically the atmospheric pressure? (2) Explain how the atmospheric pressure at a given location can affect the performance of an athlete? (3) Do we need to indicate the pressure at which a chemical reaction takes place? Justify your answer. (4)</p>]]></description><link>https://chemistrybasics.org/untitled-8/</link><guid isPermaLink="false">63b026296a348f57e3176077</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sat, 31 Dec 2022 12:16:05 GMT</pubDate><content:encoded><![CDATA[<p><em><strong>Questions</strong></em>:(1) What’s the definition of pressure and more specifically the atmospheric pressure? (2) Explain how the atmospheric pressure at a given location can affect the performance of an athlete? (3) Do we need to indicate the pressure at which a chemical reaction takes place? Justify your answer. (4) Why usually we don’t feel that pressure on us?</p><p><em><strong>Answers:</strong></em></p><p>We live with many natural phenomena and take them as granted, without asking ourselves about their nature and how they work. One of such phenomena is the Atmospheric Pressure.</p><p>1/ A pressure is the force acting normally on a unit area of a surface or the ratio of force to area.</p><p>Imagine two persons of the same weight. One is wearing high heels shoes, another is wearing flat shoes. By accident, both of them walk on your feet; where do you feel more pain?</p><p>You feel more pain on the foot stepped on by the high heel shoe, because at that point of your foot, the pressure or the force acting on your foot is distributed on a small area, the surface of the high heel shoe. You feel less pain on the other foot, because the pressure or the force acting on that foot is distributed on a larger area of the flat shoe.</p><p>The atmospheric pressure is the force exerted by the weight of the air [composition of dry air: N<sub>2</sub>(78.08%), O<sub>2</sub>(20.95%), other gases (0.97%)] above any point of the Earth’s surface. It is expressed in atmosphere unit, <strong>atm</strong>., or Pascal unit, <strong>Pa</strong>:</p><p><strong>1 atm = 101,325 Pa = 760.0 mmHg</strong></p><p>The atmospheric Pressure is equal to 1 atm at sea level.</p><p>2/ As the altitude increases, the air become less dense than air nearer to the sea level; in other words the quantity of oxygen decreases with increasing altitude. An athlete, whose lungs and body are used to higher quantity of O<sub>2</sub> at lower altitude will experience difficulty at higher altitude where the oxygen is in lower quantity; this may affect his/her performance. That is why, in some cases, athletes need some acclimatization period.</p><p>3/ It is recommended to indicate the pressure and the temperature (if different from room temperature) at which a chemical reaction takes place. Because Temperature and Pressure are two factors that affect the speed of the reaction. Nevertheless the Pressure affects gaseous reactions more than reactions in liquid or solid state. That is why this recommendation is more critical for gaseous reactions.</p><p>4/ Two reasons can explain this:</p><p>(a)    Our body is hit by air molecules from all directions with the same force; that is why we do not feel it, because the forces in all directions are balanced. But if this balance is disturbed, for example by a lorry passing nearby, we are pushed away because of a higher pressure coming from the air displaced by the lorry on the other side.</p><p>(b)   The other reason is that the air within our bodies (in our lungs, stomach, ears, etc...) is exerting the same pressure outwards, so there is no difference of pressure and we feel nothing.</p><p>But this balance of pressure can be disturbed; for example when we change altitude quickly, for instance for people in an airplane that takes off or descends for landing, some people may feel pain in their ear drums (tympanic membrane) due to unbalanced pressure on both sides of the ear drum.</p>]]></content:encoded></item><item><title><![CDATA[Problem # 49Light [Solved]]]></title><description><![CDATA[<p><strong><em>Question: </em></strong>One daily phenomenon that we live with is Light. Give a scientific explanation of that phenomenon and the origin of colors.</p><p><strong><em>Answer:</em></strong></p><p>There are two combined origins or causes of the colors of objects we see: (i) the light, and (ii) the nature of the object.</p><p><strong>(1) </strong>Nature of</p>]]></description><link>https://chemistrybasics.org/untitled-6/</link><guid isPermaLink="false">639f75c56a348f57e3175ff8</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sun, 18 Dec 2022 20:43:25 GMT</pubDate><content:encoded><![CDATA[<p><strong><em>Question: </em></strong>One daily phenomenon that we live with is Light. Give a scientific explanation of that phenomenon and the origin of colors.</p><p><strong><em>Answer:</em></strong></p><p>There are two combined origins or causes of the colors of objects we see: (i) the light, and (ii) the nature of the object.</p><p><strong>(1) </strong>Nature of Light: light is a wave phenomenon as shown below:</p><figure class="kg-card kg-image-card kg-width-wide"><img src="http://chemistrybasics.org/content/images/2022/12/image.png" class="kg-image" alt></figure><figure class="kg-card kg-image-card"><img src="http://chemistrybasics.org/content/images/2022/12/image-1.png" class="kg-image" alt></figure><p>                                           (University of Hawai at Manoa)</p><p>A wave phenomenon is characterized by: (i) the wavelength, represented by the Greek letter lambda, λ, is the distance between two crests or two troughs; (b) the frequency or the number of wavelengths per second, represented by the Greek letter gamma, γ; (c) the amplitude or the height of the crest of the wave, and (d) its speed</p><p>Light, a wave phenomenon, is characterized by a speed, C:</p><p>C = 300,000 Km/s = 300,000,000 m/s: constant</p><p>C = λγ</p><p>λ =   C/γ ;  γ = C/λ</p><p>Since C is constant, the relation C = λγ, a constant, shows that longer λ corresponds to lower γ and vice-versa; shorter lambda corresponds to corresponds to higher gamma and vice-versa.</p><p>Light bears energy: E = hγ, where h = Planck constant = 6.626 x 10<sup>-34</sup> Js</p><p>As you can notice, since E is proportional to γ, short wavelength light bears higher energy than long wavelength light and vice-versa</p><p>White light such as Sunlight or light from an electrical bulb, is also a wavelength phenomenon; but in fact, it is a composite light made of many different lights of different wavelengths and frequencies. Those different lights are characterized, in the visible spectrum, by different colors that can be separated by a glass prism, light diffraction, as shown in the Figures below.</p><figure class="kg-card kg-image-card"><img src="http://chemistrybasics.org/content/images/2022/12/image-2.png" class="kg-image" alt></figure><p>                     <strong>Diffraction spectra of white light through a glass prism</strong></p><p><em>(clearinghouse.starnetlibraries.org/physics/145-let-s-see-light-in-a-new-way-diffraction-spectra.html)</em></p><p></p><figure class="kg-card kg-image-card"><img src="http://chemistrybasics.org/content/images/2022/12/image-3.png" class="kg-image" alt></figure><p>Light outside of the visible spectrum region, infrared and ultraviolet, cannot be detected by human eyes (but instruments that can detect light in the invisible region of spectrum exist).</p><p>According to their wavelength and frequency, the red light (λ = 740 - 625 nm) is the least energetic light, whereas violet light (λ = 435 - 380 nm) is the most energetic light of the visible spectrum.</p><p>(1)   The rainbow</p><p>The rainbow is the result of diffraction of the sunlight by droplets of water that play the role of a glass prism as shown in the figure representing the diffraction of light previously. You have certainly noticed that the rainbow appears when there is sunlight on a rainy day.</p><figure class="kg-card kg-image-card"><img src="http://chemistrybasics.org/content/images/2022/12/image-4.png" class="kg-image" alt></figure><p>                                                               <strong>Rainbow</strong></p><p>(2)    <strong>Colors of different objects</strong></p><p>The color of an object is the result of its interaction with the light.</p><p>- If an object reflects the whole sunlight, it will appear white because the whole white sunlight is reflected back; this is popularly known as white color, but in reality it is all colors together.</p><p>- If an object absorbs the whole white sunlight, it appears black; this is popularly known as black color, but in reality, black is the absence of colors or absence of light.</p><p>- If an object absorbs a color 1, it will appears as white light from which color 1 has been removed (white sunlight – color 1),  in this case the color of the object is called the complementary color of the absorbed color1.</p><p>- If an object is transparent to the sunlight and does not absorb any color, it appears transparent and colorless, because the entire sunlight is transmitted.</p><p>Here, we are not going to go into the explanation why some materials absorb some color others not.</p><p>Hereafter is a ring of colors representing different colors and their respective complementary colors. Colors in diagonal positions are complementary one another: ex: Yellow and Violet are complementary one another.</p><p></p><p>                                         <strong>Ring of Complementary Colors</strong></p><figure class="kg-card kg-image-card"><img src="http://chemistrybasics.org/content/images/2022/12/image-5.png" class="kg-image" alt></figure><p>                                                          (EasyEdit.pro)</p><p>If you are wearing a yellow shirt, according to the ring of colors above, this indicates that your shirt absorbs the violet and reflects the yellow color, its complementary color.</p><p>On the other side, the green color of the leaves of the trees, indicates that the leaves absorb the red and reflect the green color.</p><p>The violet solution of phenolphthalein in a basic solution indicates that the solution absorbs the yellow and lets the violet color pass.</p><p>The table below shows the respective complementary colors:</p><!--kg-card-begin: html--><table class="MsoNormalTable" border="1" cellspacing="0" cellpadding="0" style="margin-left:.75in;border-collapse:collapse;border:none;mso-border-alt:
 solid windowtext .5pt;mso-yfti-tbllook:1184;mso-padding-alt:0in 5.4pt 0in 5.4pt;
 mso-border-insideh:.5pt solid windowtext;mso-border-insidev:.5pt solid windowtext">
 <tbody><tr style="mso-yfti-irow:0;mso-yfti-firstrow:yes">
  <td width="168" valign="top" style="width:125.75pt;border:solid windowtext 1.0pt;
  mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Transmitted or Reflected Color (what we see)<o:p></o:p></span></p>
  </td>
  <td width="156" valign="top" style="width:116.8pt;border:solid windowtext 1.0pt;
  border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt:
  solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Absorbed Color (what we don’t see)<o:p></o:p></span></p>
  </td>
  <td width="228" valign="top" style="width:170.95pt;border:solid windowtext 1.0pt;
  border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt:
  solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Observation<o:p></o:p></span></p>
  </td>
 </tr>
 <tr style="mso-yfti-irow:1">
  <td width="168" valign="top" style="width:125.75pt;border:solid windowtext 1.0pt;
  border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt;
  padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Yellow<o:p></o:p></span></p>
  </td>
  <td width="156" valign="top" style="width:116.8pt;border-top:none;border-left:
  none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt;
  mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt;
  mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Purple/Violet<o:p></o:p></span></p>
  </td>
  <td width="228" rowspan="6" valign="top" style="width:170.95pt;border-top:none;
  border-left:none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt;
  mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt;
  mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">The two colors in the same row are called
  complementary one to the other, i.e. when one is absorbed, the other is
  transmitted or reflected, and vice-versa.<o:p></o:p></span></p>
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Ex: Yellow color is the complementary color of
  Purple and vice-versa, i.e. when Purple is absorbed, the Yellow color
  appears; and if Yellow color is absorbed, the Purple color appears.<o:p></o:p></span></p>
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">The yellow color of a flower is the reflected color,
  complementary to the purple color that has been absorbed.<o:p></o:p></span></p>
  </td>
 </tr>
 <tr style="mso-yfti-irow:2">
  <td width="168" valign="top" style="width:125.75pt;border:solid windowtext 1.0pt;
  border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt;
  padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Yellow-Orange<o:p></o:p></span></p>
  </td>
  <td width="156" valign="top" style="width:116.8pt;border-top:none;border-left:
  none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt;
  mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt;
  mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Blue-Purple<o:p></o:p></span></p>
  </td>
 </tr>
 <tr style="mso-yfti-irow:3">
  <td width="168" valign="top" style="width:125.75pt;border:solid windowtext 1.0pt;
  border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt;
  padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Orange<o:p></o:p></span></p>
  </td>
  <td width="156" valign="top" style="width:116.8pt;border-top:none;border-left:
  none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt;
  mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt;
  mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Blue<o:p></o:p></span></p>
  </td>
 </tr>
 <tr style="mso-yfti-irow:4">
  <td width="168" valign="top" style="width:125.75pt;border:solid windowtext 1.0pt;
  border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt;
  padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Red-Orange<o:p></o:p></span></p>
  </td>
  <td width="156" valign="top" style="width:116.8pt;border-top:none;border-left:
  none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt;
  mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt;
  mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Blue-Green<o:p></o:p></span></p>
  </td>
 </tr>
 <tr style="mso-yfti-irow:5">
  <td width="168" valign="top" style="width:125.75pt;border:solid windowtext 1.0pt;
  border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt;
  padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Red<o:p></o:p></span></p>
  </td>
  <td width="156" valign="top" style="width:116.8pt;border-top:none;border-left:
  none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt;
  mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt;
  mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Green<o:p></o:p></span></p>
  </td>
 </tr>
 <tr style="mso-yfti-irow:6;mso-yfti-lastrow:yes">
  <td width="168" valign="top" style="width:125.75pt;border:solid windowtext 1.0pt;
  border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt;
  padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Red-Purple<o:p></o:p></span></p>
  </td>
  <td width="156" valign="top" style="width:116.8pt;border-top:none;border-left:
  none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt;
  mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt;
  mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt">
  <p class="MsoNormal"><span style="font-size:12.0pt;line-height:107%;font-family:
  &quot;Times New Roman&quot;,serif">Yellow-Green<o:p></o:p></span></p>
  </td>
 </tr>
</tbody></table><!--kg-card-end: html-->]]></content:encoded></item><item><title><![CDATA[Announcement: Scientific explanations of some Natural Phenomena]]></title><description><![CDATA[<p>In the Introduction of this Series of Problems and Solutions, Chemistry has been defined as the Science related to the Study and Observation of Matter that we find in our environment and interact with in our daily life; the study of the interaction between matters themselves and interaction between matter</p>]]></description><link>https://chemistrybasics.org/untitled-5/</link><guid isPermaLink="false">638cd3986a348f57e3175fdc</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sun, 04 Dec 2022 17:09:30 GMT</pubDate><content:encoded><![CDATA[<p>In the Introduction of this Series of Problems and Solutions, Chemistry has been defined as the Science related to the Study and Observation of Matter that we find in our environment and interact with in our daily life; the study of the interaction between matters themselves and interaction between matter and energy, etc.…. Every day of our life we observe different natural or non-natural phenomena; as someone who is interested about Science and particularly Chemistry and its relationship with our environment, you must be interested about the scientific explanation of those phenomena.</p><p>In the next series of problems and questions, a certain number of common natural phenomena will be explained.</p>]]></content:encoded></item><item><title><![CDATA[Problem # 48Exercise [Solved]]]></title><description><![CDATA[<p><em>Question</em>: How much energy is lost (produced) or gained when Co-60 undergoes a beta decay? Mass of   Co-60 is 59.9338 g, Mass of Ni-60 is 59.9308 g.</p><p><em>Solution</em>: </p><p>m = 59.9308 g - 59.9338 g = -0.003 g = -0.003 x 10<sup>-3</sup> kg</p><p>E = mc<sup>2</sup></p>]]></description><link>https://chemistrybasics.org/problem-48exercise-solved/</link><guid isPermaLink="false">6381e1ce6a348f57e3175fb4</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Sat, 26 Nov 2022 10:00:20 GMT</pubDate><content:encoded><![CDATA[<p><em>Question</em>: How much energy is lost (produced) or gained when Co-60 undergoes a beta decay? Mass of   Co-60 is 59.9338 g, Mass of Ni-60 is 59.9308 g.</p><p><em>Solution</em>: </p><p>m = 59.9308 g - 59.9338 g = -0.003 g = -0.003 x 10<sup>-3</sup> kg</p><p>E = mc<sup>2</sup>, where m = expressed in Kg, and c = 3.00 x 10<sup>8</sup> m/s</p><p>E = -0.003 x 10<sup>-3</sup> kg (3.00 x 10<sup>8</sup>m/s)<sup>2</sup> = -3 x 10<sup>-6</sup> kg x 9.00 x 10<sup>16</sup> kg.m<sup>2</sup>/s<sup>2 </sup>=       -27 x 10<sup>10</sup> kg.m<sup>2</sup>/s<sup>2</sup> = <strong>- 2.7 x 10<sup>11</sup>kg.m<sup>2</sup>/s<sup>2</sup></strong> = <strong>-2.7 x 10<sup>11</sup>J = Energy produced</strong> = <strong>-6.45 x 10<sup>10</sup> calories</strong></p><p>(1J = 0.2388 calorie; 1calorie = 4.1868 J)</p><p><strong><em>N.B</em></strong>: <em>Notice the huge amount of energy produced by just 0.003 g of matter!</em></p>]]></content:encoded></item><item><title><![CDATA[Issue # 47 Origin of Nuclear Energy?]]></title><description><![CDATA[<p>A short reminding on ordinary chemical reactions. In a chemical reaction, atoms involved in the reaction do not change, only the way they are combined changes; in other words, only the chemical bonds change during a chemical reaction. The difference between the bond energies in the compounds and the bond</p>]]></description><link>https://chemistrybasics.org/issue-47-origin-of-nuclear-energy/</link><guid isPermaLink="false">636e6f326a348f57e3175f9b</guid><dc:creator><![CDATA[Daniel Iyamuremye]]></dc:creator><pubDate>Fri, 11 Nov 2022 15:55:27 GMT</pubDate><content:encoded><![CDATA[<p>A short reminding on ordinary chemical reactions. In a chemical reaction, atoms involved in the reaction do not change, only the way they are combined changes; in other words, only the chemical bonds change during a chemical reaction. The difference between the bond energies in the compounds and the bond energies in the reactant explains why some reactions are exothermic, other endothermic. Since only chemical bonds are concerned, this justifies the Lavoisier’s mass conservation law.</p><p>In 1930s, Scientists discovered that the masses of nuclei combined are always less than these nucleons individually.</p><p><em>Example</em>: Helium nucleus (2 p + 2 n)</p><p>Mass of 2 protons = 2 x 1.00728 amu = 2.01456 amu</p><p>Mass of 2 neutrons = 2 x 1.00867 amu = 2.01734 amu</p><p><strong>Expected total mass of nucleons: 4.03190 amu</strong></p><p><strong>The observed mass of He-4: 4.0150 amu, causing a <u>mass defect</u> of 0.0304 amu</strong></p><p>The origin of that mass defect is due to some of the mass converted into binding energy which binds the nucleons together in the nucleus (Einstein’s Principle of mass-energy equivalence:      E = mc<sup>2</sup>, where m= mass, c = velocity of light).</p><p>This explains the origin of high energy generated during Nuclear Fusion and Nuclear Fission reactions, where a part of mass (mass defect) is converted into energy.</p><p>Nuclear Fission: n + <sup>235</sup>U → <sup>236</sup>U → <sup>92</sup>Kr + <sup>141</sup>Ba + 3n</p><p>Nuclear Fusion: <sup>2</sup>D + <sup>3</sup>T → <sup>4</sup>He + n</p><p>Although the above nuclear equations seem to show the conservation of mass, in real terms, there is a mass defect in the products, mass defect that has been converted into energy.</p><p>The term c<sup>2</sup> shows that a small mass loss can cause a large energy loss (or energy release). For example, combustion of one mole of methane (16 g), loses 9.9 x 10<sup>-9</sup> g, which is negligible.</p><p>In nuclear reaction, this mass change are much greater, 50,000 times greater (4.95 x 10<sup>-4</sup>g) than methane combustion.</p><p><em>Example</em>: <sup>238</sup><sub>92</sub>U  →  <sup>234</sup><sub>90</sub>Th  +  <sup>4</sup><sub>2</sub>He</p><p><sup>238</sup><sub>92</sub>U = 238.0003 amu; <sup>234</sup><sub>90</sub>Th = 233.9942 amu, and <sup>4</sup><sub>2</sub>He = 4.0015 amu</p><p>The mass defect : 233.9942 g + 4.0015 g – 238.0003 g = -0.0046 g</p><p>E = mc<sup>2</sup> = (-0.0046 x 10<sup>-3</sup>kg)(3.00 x 10<sup>8</sup> m/s)<sup>2</sup> = -0.0414 x 10<sup>13</sup> kg.m<sup>2</sup>/s<sup>2</sup>=         -4.14 x 10<sup>11</sup> kg m<sup>2</sup>/s<sup>2</sup> = -4.14 x 10<sup>11</sup>J</p>]]></content:encoded></item></channel></rss>