Senin, 15 November 2010

THE LAW OF DALTON

THE LAW OF DALTON

If two elements can form more than one compound, so the mass comparison one of element which bonds at the same other element is not decimal number.

Example - 1

A and B elements can form 3 compounds.

The first compound contains 40% A mass..

The second compound contains 50% B mass.

The third compound contains 25% A mass.

So that fullfill the law of Dalton, determine the mass comparison of B element bonds at the three above compounds!

Example - 2

Three compounds formed by X and Y elements.

The first compound containing 1/4 part of X element.

The second compound containing 2/3 part of Y element.

The third compound containing 3/5 part of Y element.

So that fullfill the law of Dalton, how much is the mass comparison of X bonds at the three above compounds?


Example - 3

P and Q elements can form three compounds. For each of three above compounds found the data:

The first compound found 0,4 grams of P element.

The second compound found 0,2 grams of Q element.

The third compound found 0,5 grams of P element.

Tentukan perbandingan massa Q pada ketiga senyawa ! Apakah memenuhi Hukum Dalton ?

Example - 4

R and S elements can form 4 compounds. For each of ten grams these compounds found that : the first compound contains 8 grams of R element.Unsur, the second compound contains 5 grams of S element, the third compound contains 2,5 grams of R element and the fourth compound contains 6 grams of S ekement.

How much is the mass comparison of R bonds to the four above compounds?Is the law of Dalton used in this item?


Concept 3 : Implementation of Gay Lussac’ s law.

Basic: The coefficient comarison = the volume comparison = the mole comparison = the number of molecule.

Example - 1

To form 60 Cm3 of ammonia gas (25 °C ; 1 atm) how much are nitrogen gas and hygrogen gas required (25 °C ; 1 atm) ?

Example - 2

10 Lof hydrogen gas sulfide and 25 L of oxygen gas are react produce water vapour and sulfur dioxide gas120 °C ; 1 atm. How much is the gas volume after reacting (120 °C ; 1 atm) ?

Example - 3

At 25 °C ; 1 atm is occurred by Carbon combustion with 100 L of air. How much litter carbon dioxide gas volume formed if at 25 °C ; 1 atm the air containing 20% volum O2?


Example - 4

Asetilene gas (C2H2) completely burnt byoxygen gas according to the reaction equation below at (P, T) :

C2H2 (g) + O2 (g) ® CO2 (g) + H2O (g) (not balancing yet)

The amount of carbon dioxide gas and water vapour is equal to 12 x 1023 molekul, how much is the molecule number of asetilene gas and oxygen (P, T) ?

Example - 5

Calculate the volume of oxygen gas is that required to complete oxidation one litter of gas compound consists of 60% vol CH4 dan 40 % vol C26 at the same(P, T)?

Example - 6

6 Lof gas compound consists of methane gas and methane gas completely burnt by 18 L O2 at the same (P, T). Determine the volume of each gas in gas compound !


Example - 7

As much of 3 litters (P, T) certainly hydrocarbon gas burnt according to the equation below:

CxHy (g) + O2 (g) ® CO2 (g) + H2O (g)

If used 7,5 liters oxygen gas volume, and 3 litters water vapour formed (P, T). Determine the molecule formula of this hydrocarbon !

Example - 8

Five litters of hydrocarbon gas completely burnt by 35 litters of oxygen gas excess(P, T). The volume gas after reacting = 25 l. After adding NaOH is found 5 litters gas. Determine the molecule formula of this above hydrocarbon gas !

Selasa, 21 September 2010

Equation Reaction

REACTION EQUATION

A reaction equation consists of 2 part, those are left part determine reactant and right part determine product.

Example:




The condition for a balance reaction equation must fulfill the law of Lavoisier; the number of every element on the left part is similar to the number of every element on the right part. We use coefficient ( not index , because when we change index we also change the agent )




Selasa, 20 Juli 2010

solution

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coligate properties solution

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persamaan reaksi

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Unsur Kimia

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persamaan reaksi

PERSAMAAN REAKSI

Suatu persamaan reaksi terdiri atas 2 ruas yaitu ruas kiri menyatakan zat yang bereaksi (reaktan) dan ruas kanan menyatakan hasil reaksi (produk).
Contoh :







Syarat persamaan reaksi yang telah setara, adalah memenuhi Hukum Lavoisier yaitu jumlah tiap unsur di ruas kiri harus sama dengan ruas kanan. Untuk menyetarakan jumlah tiap atom unsur di ruas kiri dan kanan digunakan koefisien bukan index (karena mengubah indeks berarti mengubah zat).
Ada 2 cara untuk menyetarakan reaksi, yaitu :
A. Langsung
Untuk reaksi-reaksi yang sederhana.

Contoh 1 :











































B. Cara Matematis

Untuk reaksi yang panjang dan rumit.

Contoh 1 :




Contoh - 2




Contoh - 3




Contoh - 4





Contoh - 5




Contoh - 6




Contoh - 7




Contoh - 8




Contoh - 9