THE USE OF HYDROCARBON COMPOUNDS IN EVERYDAY LIFE

Before that, what is hydrocarbon intake?

            Hydrocarbons are carbon compounds that are composed only of carbon and hydrogen elements. Grouped into two groups, namely:

1. Aliphatic hydrocarbons include alkanes, alkenes and alkynes.

2. The aromatic hydrocarbons include benzene and its derivative compounds.

            The main sources of hydrocarbons are all fossil fuels (coal, oil and gas). Hydrocarbons have a very large compound derivatives, and it is arguable that all carbon compounds or organic compounds are hydrocarbon derived compounds because the major constituents are hydrogen and carbon.

Here are some useful hydrocarbons in everyday life:

A. In the field of Clothing

            From hydrocarbon materials that can be used for clothing is PTA (purified terephthalic acid) made from para-xylene where the basic ingredient is kerosene (kerosene). From this Kerosene all the ingredients are formed into aromatics, namely para-xylene. It is a benzene compound (C6H6), but there are two methyl groups on C1 and C3 atoms of the benzene molecule.

B. In the Board field
1. Wood               

           Our house is built using wood that comes from trees. Wood is made up of various molecules, such as cellulose, lignin, tannins, and resins. Cellulose is the main constituent of wood consisting of tightly bound glucose molecules. Lignin is a form of hard structure in wood. The structure is very complex and large include the incorporation of protein and sugar molecules. Tannins are widely present in wood and can be used for the skin equation industry. Resins are often found in certain wooden bases and the types are different. For example balm, turpentine, and oil. The oil produced by pine trees can be used as a disinfectant. Resins are easily obtained from tree sap and some of them have a sweet smell.
2. Plastic pipe          

     

           To drain the excess water, an efficient and strong drainage channel is required. These plastic pipes are now widely used in homes for waterways. The type used is, PVC (polyvinylchloride) and PE (polyethene) both are produced in industrial synthesis. Both are resistant to heat and not easily cracked because the constituent molecules are strongly bonded. The molecular formula of PVC (C2H3Cl) n 'whereas PE is (C2H4) n'. 

C. Trade field
1. Paper

               Paper industry is one of the industries that have an important role in the economy in Indonesia. Indonesia is a country that supplies paper raw materials. Because Indonesia has a vast forest. The supply of paper for the country has been sufficient and currently attempted to export paper abroad.
2. Coal and natural gas

               Indonesia's coal production in 2005 reached 150 million tons. Most of the production is used to meet the needs of exports to other countries. The rest is used domestically, for example for power plants, cement plants, and pulp industries. Coal contains a lot of hydrocarbons and some minerals, such as pyrite and headache.

D. The Use of Hydrocarbon Compounds in the Field of Food

            Some chemicals consist only of carbon and hydrogen (hydrocarbons). Hydrocarbons are used in industry, especially in the petroleum and asphalt industries. Chemical energy stored in hydrocarbons, the constituent elements are carbon and hydrogen. Hydrocarbons derive energy from the sun when plants use sunlight during photosynthesis to produce glucose (food) Glucose, the simplest carbohydrates flow in the bloodstream making it available to all body cells. U8

               Please watch my video below.

https://www.youtube.com/watch?v=Yu5ebm2sKSg

Lesson Plan of Classroom Learning (RPP) of Chemistry Class XII

(Implementation of Curriculum 2013)

A. Identity

School Identity:  SMA N 1 DOLOK PANRIBUAN

Subject: Chemistry

Class / Semester: XII / 1

Basic Material: Chemical Elements

Time Allocation: 1 meetings (1 x 45 Minutes)

B. Core Competence

KI  1 : Living and practicing the religious teachings it embraces.
KI 2: Living and practicing honest, disciplined, responsible, caring (polite, cooperative, olerant, peaceful) behavior, courteous, responsive and proactive and showing attitude as part of the solution to problems in interacting effectively with the social environment and Nature and in placing ourselves as a reflection of the nation in the association of the world.
KI 3: Understand, apply, analyze and evaluate factual, conceptual, procedural, and metacognitive knowledge based on their curiosity about science, related causes of phenomena and events, and apply procedural knowledge to specific areas of study according to their talents and interests to solve problems .
KI 4: Cultivate, reason, present, and create in the realm of concrete and abstract realms related to the development of the self-study in schools and act effectively and creatively, and able to use methods according to scientific.

Indicators:1. Identify the existence of elements that exist in Indonesia, especially Indonesia.2. Identify products that contain these elements.Identify the physical properties of the main elements and transition elements (boiling point, melting point, hardness, color, solubility and specific properties of each element).3. Identify the chemical properties of the main elements and transition elements (reactivity to other elements).Indicators:
1. Identify the existence of elements that exist in Indonesia, especially Indonesia.2. Identify products that contain these elements.Identify the physical properties of the main elements and transition elements (boiling point, melting point, hardness, color, solubility and specific properties of each element).3. Identify the chemical properties of the main elements and transition elements (reactivity to other elements).

C. Learning Objectives1.By observing the demonstration or experiment activities, students can independently identify the abundance of elements in nature and the properties of the elements.
2. Given experimental data, students can independently make experimental reports about the physical properties and chemical properties of alkaline element elements, earth-algae, noble gases, halogens, third period elements and elements of the fourth period.
3. Through exercises and discussions, students can analyze the abundance of elements in Nature, especially in Indonesia, as well as the physical and chemical properties of alkaline element elements, earth-algae, noble gases, halogens, third period elements as well as elements of the fourth period.
4. Developing curiosity, thoroughness, diligence / tenacity, and mutual respect for opinions through group discussion, question and answer, and individual assignments.
5.Growing self-awareness of God's majesty and awareness of God's ordinance YME is the best determination for human life through demonstration activities, watching video or animation, and imaginative group / individual exercises.
 D. Learning MaterialsAbundance of Elements in NaturePhysical properties and chemical properties of the elements E. Learning Approach and MethodApproach: scientificModel: Cooperative Learning TPS type (Think Pair and Share), Discovery LearningMethod: Discussion, guided discovery, question and answer, practice questions  F. Media and Learning Resources· LCD projector· Media Power Point· Student Discussion Sheet· High School Chemistry Book Class X

G. Learning Steps 

1. Meeting 1

Activities


Event Descrition




TimeAllocation





preliminary





The teacher creates a religious class atmosphere by greeting the opening, praying with the students prior to the lesson, and checking the attendance of the students.1. Apperception  The teacher explains the purpose of learning / basic competencies that must be achieved by students communicatively.2. Motivation  The teacher asks motivational questions to the students.





10 meinutes


core



                                 













                           











                                                    
















·        
Observe  

Observe
the literature on the abundance of elements in nature, especially in Indonesia.

·        
Questioning
 

The
teacher asks the students questions, "What is the meaning of the element?
How is the abundance in nature? 

·        
Data
collection

T Teachers divide their students into groups to discuss the
abundance of elements / compounds in nature, especially in Indonesia. 

Teachers
provide some practice questions to be done by students through discussion 

·        
Associating

Recognize
the existence of order in the properties of the elements obtained thanks to the
creative discrateful for the abundance of elements in nature as a gift of God
YME.ommunicate Some students come forward to do the exercises on the board.

   Teachers
provide feedback and review on student work on the board.

      The
teacher discusses the exercises together with the students and answers the
students' questions.


70 minutes


Cover




· The teacher guides the students in concluding what has been discussed.· Teachers provide homework in the form of exercise questions.· Teacher informs the student of material to be discussed at next meeting.· The teacher closes the learning activities by saying hello.




10 minutes

H. Assessment (instrument attached)
1. Cognitive2. Presentations and discussions3. PsychomotorExercises :1. The flame test for Cl2 salt gives a green color. From the analysis data specify the name of Cl2's salt?2. The calcium metal can be prepared by electrolysis of fused calcium chloride with graphite electrode. Write down the reaction that occurred?3. How to separate Mg2 + and Ca 2+ ions from a solution containing Mg2 +, Ba2 + and Ca2 + ions?Procedure for flame retardation of alkaline earth elements1. Tools and materialsReaction glass test tubeWire nikrom burnerCrystal CaCl2 Crystal SrCl2Hydrochloric acid chloride BaCl22. Ways of working1. sample the sample crystals on the watch glass2. pour concentrated HCl solution in two test tubes3. Dip the nichrome wire on the HCl and heat it in a bunsen flame4. dip the nikrom wire back into the second reaction tube then into the next sample crystals heated over the flame bunsen and observe the flame color that occurs.               The experimental procedure of precipitation reaction of calcium, strontium, and barium compounds
1. Tools and materialsTest tube measuring glassPipette drops test tubeCaCl2 solution solution SrCl2BaCl2 solution of NaOH solutionNa2CO3 solution Na2SO4 solutionNa2SO4 solution Na2C2O4 solutionK2CrO4 solution

2. Ways of working

a. input the sample solution into 5 test tubesb. add into it each ion OH-, CO32-, SO42-, C2O42-, CrO42-c. observe the precipitate that occurred

I. Assessment of Learning Outcomes
1. Test the learning outcomes (mastery of the concept) using a chemical peskoran (each question is given a score of 1 if the answer is correct, and the score is zero if wrong).2. Assessment Attitude (behavior) using the rubric of behavior assessment3. Assessment of data processing skills of experimental results using performance rubric.

Behavioral Observations

No.	Rated aspect	SCOR			Information
		1	2	3
1.	Curiosity
2.	Accuracy in using experimental data and perform data processing
3.	Perseverance / tenacity in learning both in groups and individuals in solving problems that exist in the LKS.
4.	Honesty in processing data to identify carbon compounds and in solving problems in LKS

PREDICT RENDEMENT OF PRODUCT A REACTION

    In chemistry, the chemical yield, the yield of the reaction, or only the rendement refers to the amount of reaction product produced in the chemical reaction. Absolute rendement can be written as weight in grams or in moles (molar yield). The relative yield used as a calculation of the effectiveness of the procedure is calculated by dividing the amount of product obtained in moles by the theoretical yield in moles

 I. Estimated Formation of Deposition Based on Ksp Price

            The solubility product is generally denoted by Qc and the method of calculating Ksp is the product of the concentration of dissolved ions in a soluble salt. The difference is that the Qc value shows the solubility product in the unconfirmed state whether the solution is unsaturated, precisely saturated or saturated (precipitated) while Ksp is the product of the concentration of ions decomposed in a soluble salt at the maximum saturated). To determine the state of the solution, we must know the value of Ksp first.

If the value of Qc <Ksp then the solution is not saturated.

If the value of Qc = Ksp then the solution is exactly saturated.

If the value of Qc> Ksp then the solution is saturated (characterized by the formation of the precipitate because the solvent is no longer able to dissolve the solute which in this case is the soluble salt).

Problems example:

500 mL of Pb (NO 3) 2 10-3 M solution mixed with 1 liter of 10-2 M NaI solution. If Ksp PbI2 = 6 is known. 10-9, Determine whether sediment is formed or not?  

Answer:

Mol Pb2 + = V. M

                 = 0.5 liters × 10-3 M

= 5. 10-4 mol   Mol I-

= V. M = 1.0 liter × 10-2 M

= 1. 10-2 mol  

Concentration after mixing:

[Pb2 +] = mol Pb2 + / total volume

= 5. 10-4mol / 1.5L

= 3.33. 10-4 M [I-]

= total mole / total volume

= 1. 10-2 mol / 1.5 L

= 6.67. 10-3M  

Qc = [Pb2 +] [I-]

= (3.33.10-4) (6.67.10-3)

= 1.5.10-8 M

Price Qc> Ksp then PbI2 deposition occurs.

II. Predicting Chemical Reaction Balance Products

            Some rules that apply in determining the value of chemical equilibrium constant when the equilibrium reaction is manipulated (modified) include:

1. If the reaction can be expressed in the sum of two or more reactions, the equilibrium constant value of the whole reaction is the product of the equilibrium constant equilibrium of each reaction.

A + B <===> C + D  Kc '

C + D <===> E + F  Kc ''

A + B <===> E + F  Kc = Kc 'x Kc' '

2. If the reaction is written in reverse of the original reaction, the equilibrium constant value becomes the inverse of the original equilibrium constant value.

A + B <===> C + D  Kc '= [C] [D] / [A] [B]

C + D <===> A + B  Kc = [A] [B] / [C] [D] = 1 / Kc ' 3.

If an equilibrium reaction is multiplied by a factor n, the equilibrium constants value becomes the original equilibrium constant value raised by a factor of n.

A + B <===> C + D Kc '= [C] [D] / [A] [B]

2 A + 2 B <===> 2 C + 2 D Kc = [C] 2 [D] 2 / [A] 2 [B] 2 = {[C] [D] / [A] [B] 2 = (Kc ') 2

               One useful use of chemical equilibrium constant is to predict the direction of the reaction. To study the tendency of the reaction direction, the quantity Qc is used, ie the product of the initial concentration of the product is divided by the multiplication of the initial concentration of the reactant, each of which is removed by the reaction coefficient. If the value of Qc is compared to the value of Kc, there are three possible relationships that occur, among others:

A. Qc <Kc, Reversible reaction system excess reactants and product deficiencies. To achieve equilibrium, a number of reactants are converted into products. As a result, the reaction tends toward the product (to the right).

B. Qc = Kc, The system is in an equilibrium state. The reaction rate, both toward reactants and products, is the same.

C. Qc> Kc, Reversible reaction system of excess product and lack of reactants. To achieve equilibrium, some products are converted into reactants. As a result, the reaction tends toward the reactant (to the left).

Problems example:

Consider the following reaction: H2 (g) + I2 (g) <===> 2HI (g) Kc = 49.5 (4400C). For example in a closed container there is H2 0.1 mol; I2 0.2 mol; And HI 0.1 mol. Has the equilibrium been achieved?

Solution:

Calculate quota of reaction Q.

Q = [HI] 2 / [H2] [I2] = [0.1 mol] 2 / [0.1 mol] [0.2 mol] = 0.5

Since Q <Kc then the reaction is not balanced. The reaction will proceed to the right until equilibrium is obtained.

 III. Calculating Percentage of Results
When you do Stoichiometric Calculations to determine how many products are produced by the reaction, this is called "Theoretical Results". The expected number will be generated if everything goes right in the reaction but "Science" is like "alive" and not always right and something can happen is not as it should be And when it does, the results can be reduced, this is called "The Real Results ". So what could be wrong in chemical reactions to reduce the results?               
The reaction is not always complete and the dynamic balance is achieved which makes the reaction stop before all reactants are used.
Some products or reactants may be lost during filtering or moving from one container to another.The reactants may also be impure. Quantity can also be wrong measured. Sometimes the actual results can be greater than the theoretical results. This is usually due to the weight of an improper reactor or a mixed product,Percentage of Results = (Actual Results / Results in theory) x 100 
To calculate Percentage of Results from a reaction the following steps should be followed:
Step 1: Calculate the results theoretically. Theoretical results are what is expected from stoichiometric calculations using chemical equations.Step 2: Determine the actual result. This will usually come from doing experiments and measuring the amount of each product. But here you will get is quantity, like mass, volume or mole.Step 3: Percentage of Results = (Actual Results / Results in theory) x 100
Example 1:
Two other students did a burning reaction by burning 1.0kg of charcoal according to C (s) + O2 (g)> CO2 (g) reaction. They get 1050 liters of carbon dioxide on LTP.What Percentage of Results from CO2 (g)(RAM C = 12.0 and 1 mol of gas is 24.5 liters in LTP)
Step 1: The Results TheoryReacted mol C = mass / RMM C = 1000 / 12.0 = 83.3 mol.Mol Theory of CO2 (g) produced = Mol C x CO2 / C ratio = 83.3 x 1/1 = 83.3 mol.Volume Theory of CO2 generated on LTP = Mol CO2 x 24.5 = 83.3 x 24.5 = 2.041 liter
Step 2: Actual yield 1050 liters CO2 (g)
Step 3: Percentage of Results = (Actual Results / Theory Results) x 100 = (1050/2041) x 100 = 51.4%

REPORT

  I. PRACTICAL TITLE

    TERMOKIMIA AND LEGAL HESS

 II. DATE AND TIME
    SATURDAY, 9 NOVEMBER 2016

III. OBJECTIVES OF EXPERIMENT
1. Measuring the heat of reaction with a simple tool.2. Collecting and analyzing thermochemical data.3. Apply the Hess law. 
Answer:A. Enthalpy is the sum of the energy of all forms of energy possessed by the substance whose quantity can not be measured.B. Isolated system is a system with its environment can not exchange both energy and material.C. Open systems are systems and environments can exchange energy and material.D. Closed system is a system which enables the transfer of energy (heat) to the environment, but can not transfer mass.E. The environment is anything outside the system that affects and limits the system.F. The calorimeter is the process of measuring the heat of the reaction / measuring the temperature change of a certain amount of water over the solution. As a result of a chemical reaction in an isolated container. 2. What is the difference in enthalpy with energy in (ΔE)?Answer:• Enthalpy is the sum total of all forms of energy possessed by the substance whose quantity can not be measured, whereas• Deep energy is the total amount of total potential energy and kinetic energy of substances contained in a system. 

V. THEORETICAL BASIS               
           Thermodynamics explains the relationship between heat and other forms of energy. Its development, which was an important scientific achievement in the 19th century, was due to the efforts of physicists and engineers who wanted to achieve high efficiency in a heat engine. The interest in improving the engine once again becomes important because of the necessity of using fossil fuels effectively. However, in the last 75 years, the important application of thermodynamics is in the field of chemistry. The law of thermodynamics is an important tool for studying chemical reactions. Thermochemistry is the influence of the heat that accompanies the chemical reaction. The second law of thermodynamics is primarily the basis for deriving the equilibrium constant of the properties of thermodynamic properties, in the third law of thermodynamics will be unveiled the starting point for looking at the properties of experimental thermodynamic properties (Petrucci.1987: 225).
                Each system has energy because the material particles (solid, liquid, or gas) always move randomly and diverse. There is translational motion, rotation, and ubrasi (vibrate). In addition, there may be a shift in the energy levels of electrons in atoms or molecules. Each movement, influenced by many factors and can change shape when colliding with each other. As a result, the energy of a particle's ganga will be different from the others. The total energy of all the particles will be different from the others. The total energy of all the particles in the system is called the inner energy (U). Therefore, the absolute value of U can not be calculated.               The first law of thermodynamics discusses the energy changes that accompany the event, and is useful for calculating the incoming or outgoing heat of the system. By equation: q = aU - w. The second law, to be discussed about spontaneous and non-spontaneous changes. The second law of thermodynamics sounds the natural process of adding natural entropy or entropy as it increases, and the third thermodynamic law sounds a pure element or compound in the form of perfect crystals having zero entropy at 0 ° C (Syukri.1999: 74).
               The application of the first law of thermodynamics to chemical events is called thermodynamics, which deals with the heat that accompanies chemical reactions. Chemical reactions include isothermal processes, and when done in the open air then the reaction calories qp = ΔH. Consequently, the heat can be calculated from the enthalpy change of the reaction q = ΔH = Result H-reaction. So that there should be a uniformity of standard conditions, ie temperature 25 ° C and pressure 1 Atm. Thus, thermodynamic calculations are based on standard conditions (Shukri.1999: 84).
               Thermochemistry is a part of thermodynamics that studies heat changes that follow chemical reactions. The amount of heat that arises or is required in a chemical reaction is called reaction heat. The heat of the reaction at P remains the same as the change in the ental strength, and the heat of reaction in U remains the same as the change in power.
The magnitude of the reaction pans depends on the type of reaction, the phase state of the substances in the reaction, the amount of the reacting agent, and the reaction temperature. In thermodynamic equations, the amount of substances in the reaction is expressed in moles while the heat is expressed in Kilocalories (Sukardjo 1990: 192).• Change of enthalpy (ΔH) and internal energy changes (ΔU) in chemical reactions.The reaction at constant pressure ΔH and heat of reaction at constant volume, ΔU is connected through the equation:ΔU = ΔH-PΔUIf the heat of reaction is carved under constant pressure conditions at a constant temperature of 298 K, it is -566.0 KJ, indicating that the 566.0 KJ energy has left the system as heat ΔH = -566.0 KJ. To evaluate the pressure-volume workPΔU = P (Vt -Vi)Then we can use the ideal gas equation. The kinetic equationPΔV = RT (nf-ni)Nf is the mole of gas in this product (2 moles CO2) is the number of moles of reactant solid gas (2 moles CO + 1 mole O2), soPΔV = 0,0083145 KJ / mol K-1. 298 K × [2 (2 + 1)] mol        = -2.5 KJInternal energy change isΔU = ΔH-PΔV     = -566,0KJ - (- 2,5KJ)     = -563.5KJ• Change of enthalpy (ΔH) accompanying material changes.When the liquid comes into contact with the atmosphere, energized molecules on the liquid surface can overcome the attraction with each other and enter the form of gas or steam.• Indirect determination ΔH: Hess's law of equilibrium of enthalpy changesΔH is an extensive natureHess's law concerning the sum of the constant heat (Petrucci.1992: 239-244).


VI. TOOLS AND MATERIALS
A. Tool
1. Measuring cup
2. Calorimeter
3. Mixer
4.Cup of trophies
B. material
1. 40 mL distilled water
2. 40 mL HCl 1 M
3. 40 mL NaOH 1M
4. 1M acetic acid
5. Sodium Hydroxide 1M
6. Sodium Acetate 1M
7. 1M Nitrate Acid
8. Ammonia 1M
A. Determination of calorimeter constants


VIII. OBSERVATION DATA
A. Determination of calorimeter constants
 
B. Determination of ΔH of neutralization for acid-base

IV. PRAKTEK QUESTIONS
1. Provide an understanding of: A) enthalpy; B) isolated system; C) open system; D) closed system; E) the environment; F) calorimeter; G) exothermic

  VII. WORK PROCEDURES



B. Determination of ΔH of neutralization for acid-base














IX. DISCUSSION  


         In this experiment the calorimeter is cleaned and dried. Enter 20 mL of distilled water into the calorimeter, record the weight and measure the temperature, then take 40 mL of distilled water with a measuring cup, heat and record the hot water weight. Combine hot water and cold water into the calorimeter, note the temperature. Result of experiment of determination of calorimeter constant, by formula:  

C.Mp (Tp-Tm)                             =   C.Md(Tm-Td)+ W(Tm-Td)

4,184 J/g°C.20(60°C-39°C)         =   4,184J/g°C.20(39°C-28°C) + W(39°C-28°C)

4,184 J/g°C (21°C)                      =   4,148J/g°C (11°C) + W (11°C)

87,864 J                                       =   46,024 J + 11 W



46,024 J                                       =   3,803 J°C



11 W                                            =   87,864 J –11 W                

    W                                              =   41,8   

B. Determination of ΔH of neutralization for acid-base


      In this experiment we made observations of different mixtures of acid-base solutions. After conducting an experiment in accordance with work procedures, the data obtained are:

1. The temperature of the basic solution (NaOH) = 30 ° C

Mixed temperature = 39 ° C

And get it:

Qreaksi = C. M. (Tf-Ti) + W (Tf-Ti)

            = 4.184.80 (30 -29.75) + (39 -29.75) .3,80

              = 334.72. (9,25) + 35,17

              = 3131.33 J

Qreaksi = -Range

-Round about = 3131.33 J

The equation of the reaction :




HCl + NaOH → NaCl + H2O

                                   

                               = 3131.33 J
         0.02 mol
                               = 156566,5J / mol



Then ΔH = reaction -Range around / The reacting mole
              = 3131.33 J /0.02 mol
              = 156566,5J / mol


2. The temperature of 1M CH3COOH acid solution: 26 ° C
The temperature of 1M NaOH base solution is: 30 ° C
Mixed temperature: 27 ° C
To get Qreaksi used formula:
Qreaksi = C. M. (Tf-Ti) + W (Tf-Ti)
              = 4.184J / g ° C.80gr (27 ° C-28 ° C) + (27 ° C-28 ° C)
                   . 3,803 J / ° C
              = -334,72 J / ° C. (-1 ° C) + (- 3,803 J)
              = -338,523 J
-Qsecured = Qreaksi
                   = -338,523 J
The reacting mole
CH3COOH + NaOH → NaCH3COO + H2O
Then ΔH = reaction -Range around / The reacting mole
              = 338,523  J /0.02 mol
              = 116926.15 J  / mol





X. DISCUSSION

A. DETERMINATION OF CALORYMETER STIPULATION     A calorimeter is a tool used to measure the heat of a reaction. While the heat absorbed by the calorimeter is called the calorimeter constant. The calorimeter constant can be determined by experimental temperature measurements on cold water, hot water, and a mixture of cold and hot water. Then we put the water into the calorimeter alternately to measure the temperature. After getting the temperature respectively, then we can determine the calorimeter constant that has been discussed in the discussion. From the discussion obtained the calorimeter constant (W) 3.803 J / ° C.Our group has determined from our experimental results that the calorimeter constant is 3.803J / ° C. This calorimeter constant is also used to determine ΔH neutralization. According to the theory of the determination of the calorimeter is 0 or the smaller the value the better the calorimeter used.So our experiments got results greater than zero. This experiment has been performed in accordance with work procedures, it's just possible tool error at the time of lab work, other things can also happen to praktikan 
itself.


XI. CONCLUSION

1. In measuring the reaction can use a simple


calorimeter.calorimeter constant, it takes data in the form

of weight,  Cheat type, and temperature of the substance.
2. To determine the calorimeter constant based 

on the black principle, the equation is:
C=Waterheattype4,14J/g°C
M=Weightofhotwater
Md=Coldwaterweight
Tp=Hotwatertemperature
Td=Coldwatertemperature
Tm=mixedtemperature
W = The calorimeter constant J / g
3. The calculation of heat of reaction can be done by using:
A. Hess's Law
B. Standard enthalpy entries
C. Average reactant energy

3. The calculation of heat of reaction can be done by using:

A. Hess's Law

B. Standard enthalpy entries

C. Average reactant energy



XII. 








































2. To determine the


C.MP. (Tp-Tm) =c.Md (Tm-Td) + W (Tm-Td)

information







XII. BIBLIOGRAPHY 

Agus. 2009.Kimia Dasar University.Jakarta:Erland. 

Petrucci, Raip H. 1992. Basic Chemistry.Jakarta: Erland 

Shukri. 1999. Basic Chemistry I. Bandung: ITB