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NAME : EKO NEVRIANSYAH

CLASS : REGULER B '16

STUDY PROGRAM : CHEMISTRY EDUCATION

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New Post!

  • DataPrint telah memasuki tahun kedelapan. Setelah sukses mengadakan program beasiswa di tahun 2011 hingga 2017, maka DataPrint kembali membuat program beasiswa bagi penggunanya yang berstatus pelajar dan mahasiswa. Hingga saat ini lebih dari ribuan beasiswa telah diberikan bagi penggunanya.

    Di tahun 2018 sebanyak 400 beasiswa akan diberikan bagi pendaftar yang terseleksi. Program beasiswa dibagi dalam dua periode. Tidak ada sistem kuota berdasarkan daerah dan atau sekolah/perguruan tinggi. Hal ini bertujuan agar beasiswa dapat diterima secara merata bagi seluruh pengguna DataPrint

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    Di tahun 2018 ini ada hadiah khusus juga loh untuk 5 orang peserta yang paling banyak merekomendasikan program ini ke teman-teman. Jika nama lengkapmu dan asal sekolah atau kampus kamu tertera di formulir pendaftaran temanmu (nama pemberi rekomendasi), siap-siap deh kamu akan jadi pemenangnya di akhir periode!

    Beasiswa yang dibagikan diharapkan dapat meringankan biaya pendidikan sekaligus mendorong penerima beasiswa untuk lebih berprestasi. Jadi, segera daftarkan diri kamu, dikolom PENDAFTARAN pada www.beasiswadataprint.com

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  • What is your body made of?
     Your first thought might be that it is made up of different organs—such as your heart, lungs, and stomach—that work together to keep your body going. Or you might zoom in a level and say that your body is made up of many different types of cells. However, at the most basic level, your body—and, in fact, all of life, as well as the nonliving world—is made up of atoms, often organized into larger structures called molecules.

          Atoms and molecules follow the rules of chemistry and physics, even when they're part of a complex, living, breathing being. If you learned in chemistry that some atoms tend to gain or lose electrons or form bonds with each other, those facts remain true even when the atoms or molecules are part of a living thing. In fact, simple interactions between atoms—played out many times and in many different combinations, in a single cell or a larger organism—are what make life possible. One could argue that everything you are, including your consciousness, is the byproduct of chemical and electrical interactions between a very, very large number of nonliving atoms!
    So as an incredibly complex being made up of roughly 7,000,000,000,000,000,000,000,000,000 atoms, you'll probably want to know some basic chemistry as you begin to explore the world of biology, and the world in general.
    Matter and elements

            The term matter refers to anything that occupies space and has mass—in other words, the “stuff” that the universe is made of. All matter is made up of substances called elements, which have specific chemical and physical properties and cannot be broken down into other substances through ordinary chemical reactions. Gold, for instance, is an element, and so is carbon. There are 118 elements, but only 92 occur naturally. The remaining elements have only been made in laboratories and are unstable.
    Each element is designated by its chemical symbol, which is a single capital letter or, when the first letter is already “taken” by another element, a combination of two letters. Some elements follow the English term for the element, such as C for carbon and Ca for calcium. Other elements’ chemical symbols come from their Latin names; for example, the symbol for sodium is Na, which is a short form of natrium, the Latin word for sodium.
    The four elements common to all living organisms are oxygen (O), carbon (C), hydrogen (H), and nitrogen (N), which together make up about 96% of the human body. In the nonliving world, elements are found in different proportions, and some elements common to living organisms are relatively rare on the earth as a whole. All elements and the chemical reactions between them obey the same chemical and physical laws, regardless of whether they are a part of the living or nonliving world.
    The structure of the atom

                  An atom is the smallest unit of matter that retains all of the chemical properties of an element. For example, a gold coin is simply a very large number of gold atoms molded into the shape of a coin, with small amounts of other, contaminating elements. Gold atoms cannot be broken down into anything smaller while still retaining the properties of gold. A gold atom gets its properties from the tiny subatomic particles it's made up of.
    An atom consists of two regions. The first is the tiny atomic nucleus, which is in the center of the atom and contains positively charged particles called protonsand neutral, uncharged, particles called neutrons. The second, much larger, region of the atom is a “cloud” of electrons, negatively charged particles that orbit around the nucleus. The attraction between the positively charged protons and negatively charged electrons holds the atom together. Most atoms contain all three of these types of subatomic particles—protons, electrons, and neutrons. Hydrogen (H) is an exception because it typically has one proton and one electron, but no neutrons. The number of protons in the nucleus determines which element an atom is, while the number of electrons surrounding the nucleus determines which kind of reactions the atom will undergo. The three types of subatomic particles are illustrated below for an atom of helium—which, by definition, contains two protons.Structure of an atom. The protons (positive charge) and neutrons (neutral charge) are found together in the tiny nucleus at the center of the atom. The electrons (negative charge) occupy a large, spherical cloud surrounding the nucleus. The atom shown in this particular image is helium, with two protons, two neutrons, and two electrons.

            start superscript, minus, 24, end superscript grams. Since grams are not a very convenient unit for measuring masses that tiny, scientists chose to define an alternative measure, the dalton or atomic mass unit (amu). A single neutron or proton has a weight very close to 1 amu. Electrons are much smaller in mass than protons, only about 1/1800 of an atomic mass unit, so they do not contribute much to an element’s overall atomic mass. On the other hand, electrons do greatly affect an atom’s charge, as each electron has a negative charge equal to the positive charge of a proton. In uncharged, neutral atoms, the number of electrons orbiting the nucleus is equal to the number of protons inside the nucleus. The positive and negative charges cancel out, leading to an atom with no net charge.
    Protons, neutrons, and electrons are very small, and most of the volume of an atom—greater than 99 percent—is actually empty space. With all this empty space, you might ask why so-called solid objects don’t just pass through one another. The answer is that the negatively charged electron clouds of the atoms will repel each other if they get too close together, resulting in our perception of solidity.

    Make Video about : Matter

    8



  • Limiting reagent and theoretical yield

    It’s a classic conundrum: We have five hot dogs and four hot dog buns. How many complete hot dogs can we make?

         A reaction with five hot dogs and four hot dog buns reacting to give four complete hot dogs and one leftover hot dog. The hot dog buns are the limiting reagent, and the leftover single hot dog is the excess reagent. The four complete hot dogs are the theoretical yield.

    Assuming the hot dogs and buns combine in a one-to-one ratio, we will be limited by the number of hot dog buns we have since we will run out of buns first. In this less than ideal situation, we would call the hot dog buns the limiting reagent or limiting reactant.

    In a chemical reaction, the limiting reagent is the reactant that determines how much of the products are made. The other reactants are sometimes referred to as being in excess, since there will be some leftover after the limiting reagent is completely used up. The maximum amount of product that can be produced is called the theoretical yield. In the case of the hot dogs and hot dog buns, our theoretical yield is four complete hot dogs, since we have four hot dog buns. Enough about hot dogs, though! In the following example we will identify the limiting reagent and calculate the theoretical yield for an actual chemical reaction.
    Problem solving tip: The first and most important step for any stoichiometric calculation—such as finding the limiting reagent or theoretical yield—is to start with a balanced reaction! Since our calculations use ratios based on the stoichiometric coefficients, our answers will be incorrect if the stoichiometric coefficients are not right.

    How to Calculate Percent Yield in Chemistry ?

    1.Start with a balanced chemical equation. A chemical equation describes the reactants (on the left side) reacting to form products (on the right side). Some problems will give you this equation, while others ask you to write it out yourself. Since atoms are not created or destroyed during a chemical reaction, each element should have the same number of atoms on the left and right side.[1]
    For example, oxygen and glucose can react to form carbon dioxide and oxygen:  → 
    Each side has exactly 6 carbon (C) atoms, 12 hydrogen (H) atoms, and 18 oxygen (O) atoms. The equation is balanced.
    Read this guide if you are asked to balance an equation yourself

    2.Calculate the molar mass of each reactant. Look up the molar mass of each atom in the compound, then add them together to find the molar mass of that compound. Do this for a single molecule of the compound.
    For example, one molecule of oxygen () contains two oxygen atoms.
    Oxygn's molar mass is about 16 g/mol. (You can find a more precise value on a periodic table.)
    2 oxygen atoms x 16 g/mol per atom = 32 g/mol of .
    The other reactant, glucose () has a molar mass of (6 atoms C x 12 g C/mol) + (12 atoms H x 1 g H/mol) + (6 atoms O x 16 g O/mol) = 180 g/mol.

    3.Convert the amount of each reactant from grams to moles. Now it's time to look at the specific experiment you are studying. Write down the amounts of each reactant in grams. Divide this value by that compound's molar mass to convert the amount to moles.[2]

    For example, say you started with 40 grams of oxygen and 25 grams of glucose.
    40 g  / (32 g/mol) = 1.25 moles of oxygen.
    25g  / (180 g/mol) = about 0.139 moles of glucose.
    Image titled Calculate Percent Yield in Chemistry Step 4
    4.Find the ratio of your reactions. Remember, a mole is just a large number that chemists use to "count" molecules. You now know how many molecules of each reactant you started with. Divide the moles of one reactant with the moles of the other to find the ratio of the two molecules.
    You started with 1.25 moles of oxygen and 0.139 moles of glucose. The ratio of oxygen to glucose molecules is 1.25 / 0.139 = 9.0. This means you started with 9 molecules of oxygen for every 1 molecule of glucose.

    5.Find the ideal ratio for the reaction. Go back to the balanced equation you wrote down earlier. This balanced equation tells you the ideal ratio of molecules: if you use this ratio, both reactants will be used up at the same time.
    The left side of the equation is . The coefficients tell you there are 6 oxygen molecules and 1 glucose molecule. The ideal ratio for this reaction is 6 oxygen / 1 glucose = 6.0.
    Make sure you list the reactants in the same order you did for the other ratio. If you use oxygen/glucose for one and glucose/oxygen for the other, your next result will be wrong.

    6.Compare the ratios to find the limiting reactant. In a chemical reaction, one of the reactants gets used up before the others. This limiting reactant determines how long the chemical reaction can take place. Compare the two ratios you calculated to identify the limiting reactant:[3]

    If the actual ratio is greater than the ideal ratio, then you have more of the top reactant than you need. The bottom reactant in the ratio is the limiting reactant.
    If the actual ratio is smaller than the ideal ratio, you don't have enough of the top reactant, so it is the limiting reactant.
    In the example above, the actual ratio of oxygen/glucose (9.0) is greater than the ideal ratio (6.0). The bottom reactant, glucose, must be the limiting reactant.
    Summary

    The limiting reagent is the reactant that gets used up first during the reaction and also determines how much product can be made. We can find the limiting reagent using the stoichiometric ratios from the balanced chemical reaction along with one of the many nifty methods in Example 1.
    Once we know the limiting reagent, we can calculate the maximum amount of product possible, which is called the theoretical yield. Since the actual amount of product is often less than the theoretical yield, chemists also calculate the percent yield using the ratio between the experimental and theoretical yield.

    USING ENGLISH TO : PREDICT RENDEMENT OF PRODUCT A REACTION

    8



  • 1.TITLE: Practical report on making alcohol from glutinous tape

    2. PRACTICAL OBJECTIVES:

    O To know how to make black sticky tape.
    O To know the alcohol content contained in black sticky tape.
    O To know the benefits of black sticky tape as bioethanol
    O To obtain alcohol from the glutinous tape fermentation product
    3. DAY / PRIVATE DATE : Tuesday, August 24, 2015

    4. BRIEF THEORY :
        The process of making black glutinous tape is a fermentation process of glutinous tape by Saccharomyces Cerivisiae mushrooms that convert fructose and glucose carbohydrates into alcohol and carbon dioxide. In addition there are also fungi microorganisms that turn starch into glucose, namely Mucor chlamidosporus and Endomycopsis fibuligera. Both these microorganisms assist in converting starch into simple sugars (glucose). Benefits of black sticky tape is beneficial for the health of the body because it contains lactic acid bacteria. These foods are beneficial for body immunity, lowering cholesterol and suppressing cancer cells. In order for lactic acid bacteria to remain on glutinous tape it must be stored in cold temperatures.

    5. TOOLS :
    Ø Pans
    Ø Stove
    Ø Jars
    Ø Tray
    Ø Bakul Spoon

    6. MATERIALS :
    Ø Centong
    S 1 liter of black sticky rice
    S Water
    S Banana Leaf
    S 5 tablespoons tape yeast

    7. STEPS OF WORK:
    1) The black glutinous rice is washed
    2) Soaked Black Glutinous Rice for 24 Hours
    3) Rinse again Black Sticky Rice with water several times until clean
    4) Then steamed black sticky rice until cooked
    6) Yeast pounded until smooth
    5) Once cooked is placed on top of the winnow then cooled using a fan
    7) The sticky rice has been sprinkled with yeast, the top surface is covered up tightly with several layers so as not to be contaminated with air.
    8) After the cold tape and then mixed yeast that has been smooth and stirred until evenly distributed. A) The first layer is covered by using banana leaves.
    9) Then the tape is silenced for 1 week at room temperature for the fermentation process.
    B) The second layer is closed using a jar. C) The third layer is closed by using a solatip in the lid of the jar.
    12) To know the existence of alcohol content, quantitative analysis is done by burning the alcohol obtained to see the fire.
    10) After 1 week, the glutinous tape and water are separated by way of filtering or squeezing.
    11) Then the glutinous water obtained is distilled to produce steam to become alcohol.


    8. DATA RESULTS EXPERIMENTS
    A. Observation Table

         Tape
        Before Fermented
        After Fermented
       Volume (ml)


       from black sticky rice
       Flavor
        Taste
     Teksture
       Flavor
      Taste
     Teksture
       Water Of Sticky Rice
         alcohol


       -


        -


        Hard

        
    Alcohol
          Sweet,sour


         Soft


       200


       100


    9. DATA ANALYSIS
       Based on experimental results of glutinous tape making by fermentation obtained 200 ml black sticky water. This is because the results of fermentation done within 7 days will produce enough glutinous water while a little alcohol because within 7 days the glutinous tape made is too ripe.
    From the results of the experiment there is a change of taste, smell, and texture. That is sweet and sour taste has an alcoholic taste, strong smell of alcohol, aqueous texture, Because in the process of permentation changes the hydrolysis of starch into glucose and maltose that will give sweetness and sugar changes to alcohol and organic acids caused by the fungus Saccharomyces cerevisiae .
    In the process of distillation of glutinous water used from each glutinous tape is 200 ml so that the alcohol obtained as much as 100 ml for black sticky tape. This is because in the presence of black sticky rice is used in the manufacture of tape as much as 0.5 kg. When the steam distillation process is generated long, so the resulting alcohol content is small.
    Reaction

    The reaction in the fermentation of black sticky rice into tape is glucose (C6H12O6) which is the simplest sugar, through fermentation will produce ethanol (2C2H5OH). This fermentation reaction is carried out by yeast, and is used in food production.
    Chemical Reaction Equation: C6H12O6 + 2C2H5OH + 2CO2 + 2 ATP Translation: Sugar (glucose, fructose, or sucrose) + Alcohol (ethanol) + Carbon dioxide + Energy

    10. CONCLUSION
    1. Preparation of alcohol or ethanol can be produced from black sticky tape.
    2. Alcohol produced by 100 ml black sticky tape.
    3. From the results of the experiment there are changes in taste, odor, and texture after fermentation

    11.DOCUMENTATION
        

        

         


    USING ENGLISH TO : REPORT

    12

  • "Stoichiometry is the study of measuring or predicting the amount of reactants or products in a chemical reaction based on the variables such as the mass of reactants or products, the limiting reactant and the balanced chemical equation."

    Early Stoichiometry
    The masses of the starting materials and of the products of chemical reactions were of obvious interest to early chemists. The earliest measurements may have been made by prehistoric metal workers who weighed a metal ore with a primitive balance and compared the weight with that of the extracted metal. Weighing was the most common and most accurate measurement that chemists could make for many centuries. An early example is the work of Belgian chemist Johann van Helmont in the early seventeenth century. Van Helmont weighed a large pot containing a growing plant at intervals and tried to show that the gain in weight was fully accounted for by the water added. He did not measure the carbon dioxide gas taken up or the oxygen released by the plant and so his conclusion was not valid, although the measurements were roughly correct.

    Whether or not pure substances have the same proportion by mass of their constituents was by no means initially obvious. Around 1800 two French chemists, Claude Berthollet and Joseph Proust, supported opposite views on this topic. If a metal such as lead is heated in air, there is a gradual color change as lead oxide is slowly formed. Berthollet argued for a combination of "indefinite proportions" as this transformation occurs: the reactant is lead, the product is lead oxide, and there is an indefinite number of intermediates . Proust argued for "definite proportions" in that the system would at all times consist only of lead mixed with lead oxide (for simplicity we can ignore that more than one oxide of lead exists). The ratio of lead to lead oxide would change as the reaction proceeded but the system would have only two components. 

    The term "stoichiometry" was devised by German chemist Jeremias Richter in 1792 to describe the measurement of the combining ratios of chemical elements by mass. The term has since been expanded to include the combining ratios of substances in any chemical reaction. Richter studied mathematics with philosopher Immanuel Kant and wrote a thesis on the use of mathematics in chemistry. He was convinced that all chemical changes could be described in terms of simple whole-number ratios. He put forward the Law of Reciprocal Proportions, stating that if two chemical elements unite separately with a third element, the proportion in which they unite with the third element will be the same or a multiple of the proportion in which they unite with each other. This law has disappeared from most chemistry textbooks, but a companion law, the Law of Multiple Proportions, has survived.

    The Law of Multiple Proportions states that when two elements combine to form two or more different compounds, the weights of one compound that can combine with a given weight of the second compound form small whole number ratios. For example, consider one experiment in which 10.0 grams of sulfur is combined with 10.0 grams of oxygen to form an oxide of sulfur, and another experiment under different conditions in which 3.21 grams of sulfur is combined with 4.82 grams of oxygen to form a different oxide. For each 10.0 grams of sulfur used in the second experiment, 15.0 grams (4.82 × 10.0/3.21) of oxygen is used. The ratios of the masses of oxygen that combine with a fixed mass of sulfur are 10.0:15.0, which is equal to the whole number ratio 2:3. This conforms to the Law of Multiple Proportions. 

    The Laws of Reciprocal and Multiple Proportions have ceased to have predictive scientific value. Their importance lies in the fact that they provided evidence that Dalton needed in 1807 to postulate his atomic theory. The reason for Richter's whole number ratios has since become obvious: the simple ratios occur because atoms, although having different masses, react in simple ratios. Dalton's insistence that atoms cannot be split in chemical reactions holds true in modern chemistry.

    Balancing Chemical Equations
    Chemical equations are an indispensable way of representing reactions. They are routinely used to calculate masses of reactants and products. In the case of the examples used above for the Law of Multiple Proportions, the equations are:

    S + O 2 = SO 2          (1)

    2S + 3O 2 = 2SO 3          (2)

    Note that we do not write the second equation as:

    S + 3O = SO 3          (3)

    because O (an oxygen atom) means something very different from O 2 (an oxygen molecule). Chemical equations also introduce the concept of a limiting reagent , or the reactant that is used up first in a reaction, when one or more components are in excess of the stoichiometric amount.

    The balancing of chemical equations is a common exercise in elementary stoichiometry. It is not always appreciated, however, that some chemical equations are ambiguous in that they can be balanced in more than one way. Consider, for example, the following equation:

    H + + ClO 3 − + Cl − → Cl 2 + ClO 2 + H 2 O          (4)

    where the dashed arrow signifies an unbalanced equation. It may be balanced as follows:

    4H + + 2ClO 3 − + 2Cl − = Cl 2 + 2ClO 2 + 2H 2 O          (5)

    Both sides of this equation have four H-atoms, six O-atoms, four Clatoms, and a total charge of zero. Equation 5 can also be balanced as:

    16H + + 4ClO 3 − + 12Cl − = 7Cl 2 + 2ClO 2 + 8H 2 O          (6)

    Here both sides have 16 H-atoms, 12 O-atoms, 16 Cl-atoms, and a total charge of zero. How can both equations balance, and which is correct? To answer the first question, many equations can be written as the sum of two or more component reactions. In this case the following related reaction can be used:

    8H + + 2ClO 2 + 8Cl − = 5Cl 2 + 4H 2 O          (7)

    If equation (5) is doubled and added to equation (7), the result is equation (6). Alternatively, equation (5) could be tripled and added to equation(7) to obtain yet another balanced equation with the same reactants and products in different stoichiometric amounts. There is therefore no limit to the number of balanced equations.

    Deciding which equation is "correct" is often difficult because one of many competing pathways may take precedence in a reaction, depending on the energy requirements of the system (the thermodynamic limitations) and the speed of the reactions (the kinetics of the system). In the example above, analysis shows that equation (5) is thermodynamically unfavorable at room temperature while equation (6) is favorable.

    Non-Stoichiometric Compounds
    Most of chemistry is governed by simple whole-number ratios of molecules and atoms. Simple stoichiometry, although valid for the vast majority of mole ratios, is not universal: there are compounds with non-integral mole ratios. Substances such as alloys and glasses created problems for the initial acceptance of Dalton's atomic theory. There are, in addition, simple nonstoichiometric compounds that have varying ratios of constituent atoms. Such compounds are generally crystalline solids with defects in their crystal lattices; the lack of simple stoichiometry may give them important properties. Wustite, an oxide of iron, is an example of a non-stoichiometric compound. Its formula can be written Fe n O 1.000, where n may have values varying from 0.88 to 1.00 and its physical and chemical properties will vary somewhat depending on the value of n.

    Current Applications of Stoichiometry
    Most chemical reactions are complex, occurring via many steps. In such cases, can an overall reaction be written that describes the stoichiometry of a system under consideration? Consider an example in which sulfur is burned in oxygen to simultaneously form sulfur dioxide (mostly) and some sulfur trioxide:

    S + O 2 → SO 2          (8)

    S + 1.5 O 2 → SO 3          (9)

    (Note that the "1.5" in reaction (9) means 1.5 moles, not 1.5 molecules.) If the two reactions are added, the resulting equation is: 2S + 2.5 O 2 → SO 2 + SO 3 . This representation of the reaction is plainly wrong because it states that one mole of SO 2 is obtained for every mole of SO 3 , whereas most of the products consist of SO 2. The reason for this inconsistency is that the arrows in reactions (8) and (9) mean "becomes"; they are not equivalent to equals signs because they involve time dependence. In order to obtain an overall stoichiometric description of the reaction, both equations (8) and (9) are necessary, as is knowledge about their relative importance in the over-all reaction.

    Stoichiometry also has biochemical applications. In this case, the systems are biological networks. A typical biological network might be the central metabolism of a bacterium living in the gut under anaerobic conditions. This system consists of multiple processes that occur simultaneously involving reactions catalyzed by many enzymes. At the same time that reactants such as glucose are being consumed, many different metabolic products are being formed, and the combined reactions provide energy for the overall process. By doing experiments in which some genes in the bacterium have been deactivated, and then analyzing the "metabolic balance sheets," it becomes possible to identify which genes are essential for the overall process and which have no effect. It then becomes possible to predict the properties of mutants of the bacterium.



    Source : http://www.chemistryexplained.com

    Using English To : Calculate

    10




  • RENCANA  PELAKSANAAN  PEMBELAJARAN / LESSON PLANNING
    (RPP)

    School name            :  MAN 1 KUALA TUNGKAL
    Study                         :  Chemistry
    Class                         :  X / 2
    Topic                          : STOICHIOMETRI
    Time Allocation         : 2 x 45 Minute

               CORE COMPETENCY
             Living and practicing the teachings it embraces.
             Realizing and practicing honest, disciplined, responsible, caring (polite, cooperative, tolerant, peaceful) behavior, polite, responsive and pro-active and showing attitude as part of the top issues in interacting effectively with the social and natural environment and in placing oneself As a reflection of the nation in the association of the world.
             Understanding, applying, analyzing factual, conceptual, procedural knowledge based on his knowledge of science, technology, art, culture and humanities with the insights of humanity, nationality, state and civilization on the causes of phenomena and events, and applying procedural knowledge to specific areas of study With his talents and interests to solve problems.
             Processing, reasoning and chanting in the realm of concrete and abstract realms are linked to the development of what they studied in schools independently, and able to use methods according to scientific rules.
           BASIC COMPETENCIES
        Recognizing the orderliness of the structure of material particles as a manifestation of the greatness of God YME and the knowledge of the structure of matter particles as the result of human creative thinking whose truth is tentative.
              Demonstrate scientific behavior (having curiosity, discipline, honest, objective, open, able to distinguish facts and opinions, resilient, meticulous, responsible, critical, creative, innovative, democratic, communicative) in designing and conducting experiments and discussions embodied in Everyday attitude.
              Demonstrate cooperative, courteous, tolerant, peace-loving and environmentally-friendly behavior and thrifty in utilizing natural resources.
              Demonstrate responsive, and proactive and wise behavior as a form of problem-solving ability and decision making
              Applying the concept of relative atomic mass and relative molecular mass, reaction equations, fundamental laws of chemistry, and the concept of moles to solve chemical calculations
              Indicators:
              Write down the correct reaction equation
              Write equation of equivalent reaction
              Apply a reaction equation in chemical calculations

              Process and analyze data related relative atomic mass and relative molecular mass, reaction equations, basic chemical laws, and mole concepts to complete chemical calculations.
    Indicators:
    1. Skillfully write down the equation of the correct and equal reactions and apply in chemical calculations

    A. LEARNING OBJECTIVES
              Write down the correct reaction equation
              Write down the equivalent reaction
              Applying a reaction equation in chemical calculations

    B. LEARNING MATERIAL
              Reaction Equation
              Fact
              Chemical reaction
              Concept
              The equation of the reaction, nomenclature of the compound
              Principle
              Equivalence of equivalent chemical reactions
              Procedure
               
                       C. MODELS AND LEARNING METHODS
                      •          Learning Model: STAD through a scientific approach
                      •          Learning Methods: Discovery, demonstration, discussion

                       D. MEDIA, TOOLS AND LEARNING SOURCES
                       •          Media: LCD projector, Laptop
                       •          Tools / Materials: Group drawing cards and student sweepstakes
                       •          Learning Resources: Chemistry Package SMA X, Internet


    E. LEARNING ACTIVITY

    No


    Activity

    Description

    Allocation Time
    1
    Preliminary
              Opening an event that begins by greeting
              Ask about the presence and condition of students for readiness in learning
              Associate the previous material with the material to be discussed through questions relating to the nomenclature of the compound and the chemical formula
              Delivering basic competencies and learning objectives

    10 minute
    2
    Core
             Students are divided into a piece of paper that contains writing for the division of groups consisting of 4-5 students based on similarity in the content of the writing.
             Students pay attention to experiment demonstration / animation about chemical reactions.
             Students ask questions related to the demonstration / animation shown
             Each group discusses chemical reactions based on the demonstration / animation.
             Each group writes chemical reactions that take place in groups
             Students try to write equivalent equation of reaction
             Each group manganalisis some equation of reaction related to writing and composition
             Each student in the group should understand the review of the material by asking for help from another student friend if they still do not understand
             Students from each group are drawn to determine the students who appear to present the results of the discussion
             Present the results of the discussion in turns from each group
             The group and teachers discussed the study
             Students in the group were given a quiz on the equation of the reaction

    60 minute
    3
    Closing
              Student concludes material about equation of reaction
              The teacher gives some homework a matter related to the material being taught


    20 minute

          

    F.    ASSESSMENT
    1. Form and Assessment Techniques         
    a. Form of Instrument in the form: Test description, observation, performance appraisal

    No
    Rated aspect
    Assessment Techniques
    Assessment Time
    1.
    Attitude
              Engage actively in learning Cooperate in group activities.
              Tolerable to different and creative problem-solving processes.
            Meticulous in working
    Observation
    During learning and during discussion
    2.
    Knowledge
           Able to write down the correct chemical reaction equation
           Write equation of the equivalent reaction
          Able to apply simple reaction equations in chemical calculations

    Written test, Observation, performance appraisal

    Completion of individual and group tasks
    3.

    Skills
    Skillfully create equations of the correct and equal reactions
    Observation
    Completion of tasks (both individual and group) and during discussion


    F.    EVALUATION INSTRUMENTS
           Test: Description
          Problem: Description
              The equation of the reaction!
              Copper metal is reacted with a solution of sulfuric acid to give a solution of copper (II) sulfate and Hydrogen gas
              Sodium carbonate is reacted with a solution of hydrochloric acid resulting in a solution of sodium chloride, water and carbon dioxide gas

               Consider the following equation of the reaction!
              KOH(aq) + H2SO4(aq)  K2SO4(aq) + H2O(l)
              Fe2O3(s) + HCl(aq)  FeCl3(aq) + H2O(l)

    KEY ANSWER
    Equal reaction equation:
              2KOH(aq) + H2SO4(aq)  K2SO4(aq) + H2O(l)
              Fe2O3(s) + 6HCl(aq)  2FeCl3(aq) + 3H2O(l)
    Equal reaction equation:
              Cu(s) + H2SO4(aq)  CuSO4(aq) + H2(g)
              Na2CO3(s) + 2HCl(aq)  2NaCl(aq) + H2O(l) + CO2(g)
          


          

                       APPROVAL ASSESSMENT ASSEMBLY SHEET         
          
    Study                            : Chesmitry
    Class/Semester             : X/2
    School year                   : 20  / 20
    Observation time           :

    Put the √ mark on the columns according to the result of the observation.
    No
    Student's Name
    Attitude
    Total Score
    Active
    Toleran-ce
    Team work
    Thorough

    NG


    G

    VG

    NG

    G

    VG

    NG

    G

    VG

    NG

    G

    VG

    NG

    G

    VG

    1


















    2


















    3


















    4


















    5

































    etc
















          
    Information:
    NG       :   Not Good      G         : Good                          VG       : Very Good

    Active attitude indicator in Stoichiometric learning :
              It is not good to show absolutely no part in learning
              It is good to show that there is an effort to take part in learning but not yet consistent
              It is good to show that you have taken part in completing group tasks continuously and consistentl


    Behavior indicators work together in group activities.
              It is not good to be at all tolerant of different and creative problem solving processes.
              It is good to point out that there has been an attempt to be tolerant of different and creative problem-solving processes but still inconsistent.
              It is good to show that there is an effort to be tolerant of different and creative problem-solving processes continuously and consistently.

    Tolerant attitude indicator to different problem solving process and creative.
              It's not good at all to come up with ideas for different and creative problem-solving processes.
              It's good to show that there's been an attempt to come up with ideas for different and creative problem-solving processes but still not consistent.
              It is good to show that there is an effort to come up with ideas for different and creative problem-solving processes continuously and consistently.

    Indicators of meticulous attitude in completing individual or group work.
              It is not good to make mistakes in completing individual or group work
              Good if showing accuracy in completing individual or group work but still not consistent.
              It is good to show accuracy and meticulous completion of individual or group work continuously and consistently

    SAFETY ASSESSMENT SHEET SHEET

    Study                            : Chesmistry
    Class/Semester          : X/2
    School Years                : 20  / 20
    Observation Time        :
    Put a mark √ on the columns according to the results of observation.
    No
    Student's Name
    Skills
    Apply the concept / principles and problem solving strategies
    NS
    S
    VS
    1




    2




    3




    4




    5








    etc





    Information :

    NS       : not to skilled
    S          : skilled
    VS        : Very skilled
    Skilled indicators apply relevant concepts / principles and problem-solving strategies relating to making equivalent equation of reactions of some of the compounds involved
              Unskillful if at all unable to apply relevant concepts / principles and problem-solving strategies related to making equation of equivalent reactions of some of the compounds involved
              It is skilled to show that there is already an attempt to apply relevant concepts / principles and problem-solving strategies relating to making equation of equivalent reactions of some of the compounds involved but not entirely precise.
              Extremely tersampill, if indicating the existence of efforts to apply relevant concepts / principles and problem-solving strategies related to making equation of the equivalent reaction of some of the involved compounds is appropriate.








    RPP IN ENGLISH ABOUT : STOICHIOMETRI

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