Sunday, October 18, 2009

Elementary Law And Calculation

Elementary Law And Calculation [of] Kimia(Stoikiometri) : In chemistry, stoikiometri ( [is] sometime referred [as] [by] stoikiometri react to differentiating it from composition stoikiometri) [is] science studying and [counting/calculating] quantitative [relation/link] from product and reaktan in reaction of chemistry ( chemical equation). This word come from stoikheion Greek ( elemen) and three? ( size measure).

chemical laws in fact represent applied physics law in chemical system. most elementary concept in chemistry [is] Conservation laws of mass expressing that [there] no change of[is amount of measured Iihat vitamin at the (time) of reaction of ordinary chemistry. Modern physics indicate that in fact everlasting energilah, and that mass and energi each other is interconnected. This energy conservation instruct to is important [of] balance concept nya, termodinamika, and kinetics

Chemistry [is] often conceived of " center science" because connecting various other science, like physics, materials science, nanoteknologi, biological, pharmacy, doctor, bioinformatika, and geology [ 1]. This Koneksi arise to [pass/through] various subdisiplin exploiting concepts from various science discipline. For example, physical chemistry entangle principal applying [of] physics to items [at] atom storey;level and molecule.

Chemical relate to items interaction which can entangle two Iihat vitamin or [among/between] energi and items, especially in its [relation/link] with first law of thermodynamics. traditional Chemical entangle interaction [among/between] chemical Iihat vitamin in reaction of chemistry, altering one or more Iihat vitamin become one or more other Iihat vitamin. Sometime react this moved by enthalpy consideration, like when two Iihat vitamin have high enthalpy [to] like elemental oxygen and hydrogen react to form water, Iihat vitamin with lower enthalpy. Reaction of chemistry earn facility with a[n katalis, what generally represent other chemical Iihat vitamin in media react but [is] not consumed ( the example [is] sulphate acid which [is] water electrolysis mengkatalisasi) or immaterial phenomenon ( like electromagnet radiasi in reaction of fotokimia). traditional Chemical also handle chemical Iihat vitamin analysis, either in in and also outside a[n reaction, like in spectroscopy.

All normal items consist of subatom components or atom which forming atom; proton, electron, and neutron. Atom can be combined to yield more complex items form like ion, molecule, or crystal. Structure a world of us experience everyday and nature of items which [is] have interaction [to] with we [is] determined by nature of chemical Iihat vitamin and interaction between them. Harder steel from iron because its atoms tied in stiffer crystal structure. Wood burnt or experience of oxidation quickly because he earn to react spontaneously with oxygen at one particular reaction of chemistry if residing in to the a[n certain temperature.

Iihat vitamin tend to to be classified pursuant to energi, phase, or chemical composition [his/its]. Items can be classified in 4 phase, sequence from owning lowest energi [is] is solid, liquid, gas, and plasma. From is fourth [of] this phase type, plasma phase can only be met [by] outside space which is in the form of star, because its requirement [of] very big nya. Solid Iihat vitamin have structure remain to [at] room;chamber temperature which can fight against other weak style or gravitation which try change [his/its]. Hydrogen have limited tying, without structure, and will emit a stream of with gravitation. Gas [do] not have tying and act as free particle. Meanwhile, plasma only consist of free peripatetic ion; excessive energi pasokan prevent this ion coalesce to become element particle. One way of to differentiate third first phase [is] with volume and its for is: harsh [of] nya, queer solid Iihat vitamin [of] form and volume which remain to, hydrogen have volume remain to but without form which remain to, while gas [do] not have volume goodness and or form which remain to.

Water ( H2O) in form of dilution in room;chamber temperature because its molecules tied by antarmolekul style of[is so-called hydrogen bond. On the other side, sulfide hydrogen ( Gasiform H2S) [at] room;chamber temperature and standard pressure, because its molecules tied with dipolar interaction ( feebler dipole). Hydrogen bond [at] water have enough energi to maintain molecule irrigate [in order] not to [be] separate one another, but [do] not to emit a stream of, making extant [it] dilution in temperature [among/between] 0 ° C until 100 ° C [at] sea level. Degrading furthermore energi or temperature permit tighter form organization, yielding a[n solid Iihat vitamin, and discharge energi. Make-Up of energi will liquefy temperature although ice will not change until all liquid ice. Make-Up of temperature irrigate in turn will causing it boil ( see evaporation heat) time there are enough energi to overcome antarmolekul attractive force and hereinafter enable molecule to make a move to to avoid one another.

Man of science studying chemistry [is] often referred [as] [by] chemistry. Most chemistry [do/conduct] specialization in one or more subdisiplin. taught [by] chemistry [at] high school [is] often referred [as] " [common/ public] chemistry" and addressed as deliverer to many elementary concepts and to give appliance student to continue to its continuation subjek. Many presented [by] concept [at] this storey;level [is] often assumed do not be inaccurate and complete technically. Even though, [the] mentioned represent remarkable appliance. Kimiawan by reguler use simple clarification and appliance and this elegan in their masterpiece, because proven can in accurate figure make chemical reaktivitas model very is varying.

chemistry historyly represent new development, but this science take root [at] alchemy which have dipraktikkan during for centuries in all the world.

A.Hukum - Elementary Law [of] Chemistry
A. Conservation Laws Of Mass ( Punish Lavoisier)
Have You pay attention let iron rasher open air, and at one time we will find, that that iron have turned into iron rust. If us deliberate iron mass before rusting with yielded iron rust, in the reality bigger iron rust mass . Is real correct that way?

You who often see paper or wood burnt, obtained [by] result [is] a number of rest of combustion in the form of ash. If You consider the the ash, hence dusty mass will be more [is] light the than wood mass or paper before burned. Is real correct that way?

From the occurence, we get picture that impressing in a[n reaction of chemistry, there [is] difference [of] Iihat vitamin mass, before and after reaction.

A. Conservation Laws Of Mass ( Punish Lavoisier)
Have You pay attention let iron rasher open air, and at one time we will find, that that iron have turned into iron rust. If us deliberate iron mass before rusting with yielded iron rust, in the reality bigger iron rust mass . Is real correct that way?

You who often see paper or wood burnt, obtained [by] result [is] a number of rest of combustion in the form of ash. If You consider the the ash, hence dusty mass will be more [is] light the than wood mass or paper before burned. Is real correct that way? From the occurence, we get picture that impressing in a[n reaction of chemistry, there [is] difference [of] Iihat vitamin mass, before and after reaction

B. Punish Comparison Remain to ( Punish Proust)
[At] previous module, You have studied chemical formula [of] compound. And You have recognized various formed compound by two element or more for example, irrigate ( H2O). Water formed by two element that is Hydrogen element and Oxygen. Like You know that items have mass, including oxygen and hydrogen. How us know hydrogen element mass and oxygen which [is] terda, chemical expert [of] French, so called [of] Joseph Louis Proust ( 1754-1826), trying to join oxygen and hydrogen to form water.

From above tables seen, that each;every 1 hydrogen gas gram react by 8 oxygen gram, yielding 9 water gram. This matter prove that hydrogen mass and oxygen mass which implied in water have comparison which remain to that is 1 : 8, any to the number of formed water. From [done/conducted] attempt [it], Proust tell its famous theory with the title, Law Comparison Remain to, sounding:

" Mass compiler elements comparison a[n compound always remain to
Your Understanding? You pay attention example [of] hereunder!
Follow the example of 
If us react 4 hydrogen gram by 40 oxygen gram, how much/many formed water gram 
[Reply/ answer]:
Comparison Mass hydrogen with oxygen = 1 : 8.
Mass hydrogen comparison with mixed oxygen = 4 : 40.
Because hydrogen comparison and oxygen = 1 : 8, hence 4 needed [by] hydrogen gram 4 x 8 oxygen gram that is 32 gram.

For the case of this [of] mixed oxygen [do] not react altogether, oxygen still remain over counted ( 40 - 32 ) gram = 8 gram. So, now we will [count/calculate] how much/many formed of [by] water gram 4 hydrogen gram and 32 oxygen gram? Of course 36 gram

[Is] written as H2 + O2==> H2O
Comparison Mass 1 gram : 8 gram 9 gram
If early reaction 4 gram 40 gram ….. gram?
reacting 4 gram 32 gram 36 gram
Oxygen remain over = 8 gram.
How Your understanding? [So that/ to be] You more understanding, try to do practice following 

If/When magnesium metal burned with oxygen gas will be obtained [by] Magnesia compound. Result of attempt as described [at] tables following.

Tables 06.2 Reaction of Magnesium with Oxygen 
  1. above data show going into effect comparison law it remain to ( Proust)? If going into effect, how much/many magnesium mass comparison and oxygen in Magnesia compound
  2. In AB compound known [by] A mass comparison : B mass = 2 1. If there are 60 AB compound gram, determining mass of[is each element in the compound
  3. Perbandingnan, Fe mass : S mass = 7 : 4, to form sulfide iron compound. If/When 30 iron gram ( Fe) and 4 brimstone gram ( S) formed to become sulfide iron compound, how much/many sulfide iron mass gram ( Fes) which can happened
If You have doing it, cocokanlah with answer key following.

1. legalistic Above data [of] comparison remain to because from data 1, 2, 3, 4, Magnesium mass comparison : Oxygen mass in Magnesia compound always 12 : 8 or 3 : 2 2. Mass A comparison : B mass = 2 : 1 amount of comparison 3. To form AB compound

Amount of AB compound = 60 gram

Hence, A mass in the compound = x 60 = 40 gram

B mass in the compound = x 60 = 20 gram 3. Mass Fe comparison : S mass = 7 : 4

If/When all Fe element [used up/finished], hence S needed = 4 x 30 = 17,1 gram

This matter not possible (to), because available only 4 gram S. Become which [used up/finished] form compound [is] S element, as heavy as 4 gram.

Hence, needed Fe = x 4 gram = 7 gram

............ Mass Fes that happened = 7 gram + 4 gram = 11 gram

................ Iron ( Fe) which remain ( 30 - 7 ) gram = 23 gram

C. Punish Doubled Comparison ( Dalton law 
Chemical composition shown by its chemistry formula. In compound, uliginous, two element join each rendering a number of certain atom to form a[n compound. From two element can be formed [by] some compound with comparison different each other. FOR EXAMPLE, brimstone with oxygen can form SO2 compound and SO3. From hydrogen element and oxygen can be formed [by] H2O compound and H2O2.

Dalton investigate the elements comparison in each compound and got [by] a[n regularity pattern. The pattern expressed as Doubled Comparison law which [is] its sound.

: " If/When two element can form more than one compound, where mass one of [the] the element remain to be ( [is] same), hence other element mass comparison in the compound represent integer and modestly"

Example [of]:

Nitrogen And oxygen can form N2O compound, NO 

N2O3, and N2O4 with mass composition seen [at] tables following.

Tables 06.3 Comparison Nitrogen and oxygen in compound

From the tables, seen that if/when N mass made remain to be ( [is] same), counted 7 gram, hence oxygen mass comparison in 

N2O : NO : N2O3 : N2O4 = 4 : 8 : 12 : 16 or

..................................... 1 : 2 : 3 ..: 4 

Your Understanding? [So that/ to be] You more understanding, try to do practice following


Composition two A sample and B after analysed in the reality only contain carbon atom and oxygen. Result of analysis can be seen [by] [at] tables following::

Tables 06.4. Mass carbon comparison : oxygen.

a. Do both sampel represent [is] same compound? Or both .... differing b. Do the data support comparison law remain to or law .... doubled comparison 
KEY PRACTICE a. Unegual b. Yes, supporting doubled comparison law because mass comparison :
Carbon : Oxygen , [at] I compound = 4 : 11
Carbon : Oxygen , [at] II compound = 4 : 6

D. Punish Comparison Volume ( Gay Lusssac 
Initially [all] man of science find that, Hydrogen gas can react with Oxygen gas form water. Volume Hydrogen gas comparison and Oxygen in the reaction [is] remain to, namely 2 : 1.

Later;Then [in] year 1808, French man of science, Joseph Louis Gay Lussac, making a success of attempt about gas volume in concerned [at] various reaction by using is assorted [of] gas

Following [is] example [of] from [done/conducted] attempt

Picture 06.1 Attempt [of] Gay Lussac

According to Gay Lussac 2 Hydrogen gas volume react by 1 Oxygen gas volume form 2 aqueous vapour volume. [At] reaction of forming [of] aqueous vapour, [so that/ to be] reaction of perfection, to each;every 2 Hydrogen gas volume needed [by] 1 Oxygen gas volume, yielding 2 aqueous vapour volume " All reacted gas with result of reaction, measured [by] [at] [is] same client and temperature or ( [is] same T.P) 

To be more comprehend Law volume comparison, Your pay attention, data result of attempt with reference to gas volume reacting [at] [is] same pressure and temperature 

Data result of attempt shall be as follows

Tables 06.5 Data Attempt reaction of gas.

Pursuant to attempt data [at] above tables, gas volume comparison reacting and result of reaction, in the reality compare as integer. the Attempt data legalistic volume comparison or recognized with Law Gay Lussac that " [At] [is] same pressure and temperature [of] gas volume comparison reacting and result of reaction compare as integer "

So… now You have studied Elementary Laws [of] Chemistry covering Law kekalan [of] mass, Punish comparison remain to, Punish comparison fold and Law volume comparison. Elementary Law [of] This Chemistry will be applied [by] [at] chemical calculation, therefore comprehend better, this items to facilitate You in studying next topic

B. Conception Mole 
CONGENIALITY CONCEPTION MOLE Each;Every Iihat vitamin exist in nature lapped over to the particle in the form of atom, molecule, and ion. the Particle Iihat vitamin size measure very small so that we difficult to [counting/calculating] it. Amount of particle in a[n Iihat vitamin also plenty (of) and that thing make we difficult to [counting/calculating] it. [All] chemist succeed to find the way of counting particle, Iihat vitamin mass, and gas volume, so-called: STOIKIOMETRI ( Quantitative [Relation/Link] [among/between] Iihat vitamin in concerned in reaction 

: Is mass 1 dozen of marble [is] equal to mass 1 dozen of ping-pony ball 

: In chemistry, set of used to express the amount of particle in Iihat vitamin named [by] MOLE. Amount of atom particle, ion molekul,atau in 1 Iihat vitamin mole will [is] equal to amount of particle in 1 other Iihat vitamin mole. But, mass each;every Iihat vitamin in 1 unegual mole. How much is amount of particle in 1mol Iihat vitamin.

: used standard to express 1 mole [is] the amount of particle in C-12 atom isotope have mass [to] 12 gram. Amount of carbon atom particle which is there are in 12 C-12 atom gram represent a[n very big number and referred [as] [by] Avogadro constant ( L ). Level of Avogadro constant: 1 sma = 1,661 x 10-24 Mass gram 1 C-12 atom = 12 sma ( 12 x 1,661.10-24gram = 1,9932.10-23gram) Amount of C-12 atom particle in 12 C-12 element gram 12 gram = 6,02 x 10 23 C-12 atom 1,9932.10-23gram / atom

1 mole a[n Iihat vitamin ( compound or element) [is] to the number of that Iihat vitamin contain 6,02 x 10 23 particle ( atom, molecule, ion). : 1 mole a[n Iihat vitamin ( compound or element) [is] to the number of that Iihat vitamin contain 6,02 x 10 23 particle ( atom, molecule, ion). 1 gold element mole ( Au) contain 6,02 x 10 23 Au atom 1 NH3 compound mole contain 6,02 x 10 23 NH3 molecule 1 Nacl mole contain 6,02 x 10 23 Nacl molecule 1 Na mole contain 6,02 x 10 23 Na atom 1 Na mole+ or Cl- containing 6,02 x 10 23 Na ion+ or Cl- Conclusion of[is Amount of particle = Amount of x mole 6,02.1023 Amount of particle = L x n of[is Amount of particle = atom for the element of = molecule for the compound of kovalen = ion for the compound of ion 

: Problem example [of] : If known [by] Avogadro constant = 6,02. 1023, determining : Amount of H2S molecule in 0,4 H2S mole b. Amount of H atom in 0,4 H2S mole c. Amount of S atom in 0,4 H2S mole d. Amount of element atom in 0,4 H2S mole.

: Amount of H2S molecule = L x n = 0,4 x 6,02.1023 = 2,408.1023 H2S molecule 

: b. Amount of H atom = L x n = 0,4 x 2 x 6,02.1023 = 4,816.1023 H atom c. Amount of S atom = L x n = 0,4 x 1 x 6,02.1023 = 2,408.1023 S atom 

: d. Amount of element atom in H2S = L x n = 0,4 x 3 x 6,02.1023 = 7,224.1023 atom 

: How much is atomic mass and molecule which is there are in 1 Iihat vitamin mole ? To knowing it you have to comprehend beforehand atomic mass relative ( And molecule mass Ar) relative ( Mr). Ar Element = atomic mass comparison a[n element to one per twelve times mass one carbon atom which [is] have mass [to] 12. Compound Mr = molecule mass comparison a[n compound to one per twelve times mass one carbon atom which [is] have mass [to] 12. Mr = S Ar 

: [Count/Calculate] molecule mass relative ( Mr) Iihat vitamin following: H2S Cl2 HCL C6H12O6 Mgso4.7H20 Known : Ar H=1; S=32; Cl=35,5; C=12; O=16; Mg=24. How [relation/link] [among/between] Ar and Mr 

H2S lapped over to the 2 H atom and 1 S Mr H2S atom = ( 2 Ar x + ( 1 Ar x = ( 2 x ( 1 x 32 = 34 . Cl2 lapped over to the 2 Cl Mr Cl2 atom = ( 2 Ar Cl x = ( 2 x 35,5 = 71

HCL lapped over to the 1 H atom and 1 Cl Mr HCL atom = ( 1 Ar x + ( 1 Ar Cl x = ( 1 x ( 1 x 35,5 = 36,5 

: d. C6H12O6 lapped over to the 6 C atom , 16 H atom and 6 O Mr C6H12O6 atom = ( 6 Ar x + ( 12 Ar x + ( 6 Ar x = ( 6 x 12 + ( 12 x + ( 6 x 16 = 180 

: Mgso4.7H2O lapped over to the : 1atom Mg, 1atom S, 11 O atom and 14 H Mr Mgso4.7H2O atom = ( 1 Ar Mg x + ( 1 Ar x + ( 11 Ar x + ( 14 Ar x = ( 1 x 24 + ( 1 x 32 + ( 11 x 16 + ( 14 x = 246 

: How much is mass 1 element mole and compound ? Element Monoatomik Atomic Mass Element = Amount of Ar Molecule Diatomik Mass Molecule Compound x mole= Amount of Mr Compound Compound Mass Compound ion ion x mole = Amount of Mr Compound Mass Compound kovalen kovalen x mole = Amount of Mr x mole

Mass Iihat vitamin = Amount of Ar x mole / G mr = Ar x n / mr 
: Example [of] [Count/Calculate] Iihat vitamin mass following ! 1 Ba atom mole 2 O2 gas mole 0,5 H2O mole 0,02 Nacl O,3 C6H12O6 mole mole Known : Ar H=1; C=12; O=16; Na=23; Cl=35,5; Ba=137 a. 1 Ba atom mole : a. 1 Ba Atomic mass Ba atom mole = Amount of Ar Ba x mole = 1 x 137 = 137 gram b. 2 O2 gas mole : b. 2 O2 Mass O2 gas gas mole = Amount of Mr O2 x mole = 2 x 32 = 64 gram c. 0,5 H2O Mass H2O mole = Amount of Mr H2O x mole = 0,5 x 18 = 9 gram d. 0,02 Nacl mole : d. 0,02 Nacl Mass Nacl mole = Amount of Mr Nacl x mole = 0,02 x 58,5 = 1,17 gram e. 0,3 C6H12O6 mole : e. 0,3 C6H12O6 Mass C6H12O6 mole = Amount of Mr C6H12O6 x mole = 0,3 x 180 = 54 gram 

How much is volume 1mol gas ? : How much is volume 1mol gas ? Volume 1 mole a[n gas in the situation standard ( Atm O0C,1) referred [as] [by] gas molar volume, what is the level of = 22,4 litre. According to Avogadro hypothesis, [at] [is] same volume, different gas ( [at] [is] same pressure and temperature) containing the amount of particle which [is] its amount [is] [is] same. Become can be concluded that Volume gas compare diametrical with amount of mole ( n)

Gas volume [at] STP : Volume Gas [at] STP Volume = Amount of V molar volume x mole = Vm V x n = x mole 22,4 L / mole 

Gas volume in the situation pressure and temperature known. : Gas volume in the situation pressure and temperature known. P V = R T V n = R T P n 

: P = Pressure ( Atm) V = Volume ( Liter) N = R mole = Ideal Gas Constant ( 0,082 L atm / mole K) T = Temperature in Kelvin ( O0C = 273 

Gas volume in the situation other gas known. : [At] [is] same pressure and temperature, volume compare diametrical with amount of its mole. Gas volume in the situation known other gas 

: Determining gas volume following : 0,5 Cl2 gas mole 1 N2 gas mole 2 CO gas mole 0,5 Cl2 V gas mole = 0,5 x 22,4 = 11,2 litre 0,5 Cl2 V gas mole = 0,5 x 22,4 = 11,2 litre b. 1 N2 V gas mole = 1 x 22,4 = 22,4 litre c. 2 CO V gas mole = 2 x 22,4 = 44,8 litre 

Example [of] : 2 How much is volume 6,4 measured [by] SO2 gas gram [at] temperature 27oC pressure 38 cmHg? ( Ar S=32, O=16) P V = R T V n = R T P V n = 0,1 x 0,082 x 300 0,5 V = 4,92 n litre = G SO2 / n mrso2 = 6,4 / 64 n = 0,1 P mole = 38 cmHg / 76 cmHg = 0,5 T atm = ( 27 + 273)K T = 300 K

Example [of] : 3 [At] certain pressure and temperature, volume from 8 SO3 gas gram [is] 2,5 litre. [is] same in the situation determine volume from 18 NO.(AR S=32 gas gram; N=14; O=16) n n = V V ( gasSO3) ( gasNO) 0,1 mole 0,6 mole = 2,5 V NO V NO litre = ( 2,5 x 0,6) / 0,1 = 15 n litre = G / SO3 n mr = 8 / 80 = 0,1 NO n mole = 18 / 30 = 0,6 mo

Ditulis Oleh : Efendi Pakpahan // 6:56 AM

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