Minggu, 25 Maret 2012

TEOREMA DASAR KETERBAGIAN 1

Definisi Keterbagian
Jika a dan b adalah bilangan bulat dengan  a dikatakan membagi b, jika terdapat sebuah bilangan bulat m sedemikian sehingga b = am dan ditulis a│b dan jika a tidak membagi b, maka ditulis a ł b.

Beberapa sifat-sifat / Teorema keterbagian

Teorema 1                                                            
Jika a, b, dan c adalah bilangan bulat dengan a│b dan  b│c maka a│c.

Bukti
a│b dan b│c maka menurut Definisi, terdapat bilangan bulat m dan n sedemikian sehingga   c = bn = (am)n = a(mn). Jadi, c = a(mn). Untuk suatu mn = p  anggota bilangan Bulat maka c = ap Akibatnya menurut Definisi, a│c.  
Untuk lebih jelasnya, diberikan Contoh  berikut.

Contoh
Jika 2│6 dan 6│90 maka menurut Teorema 2│90 karena terdapat bilangan bulat 45 sedemikian sehingga (45)(2) = 90

Teorema 2
Jika a, b, dan c adalah bilangan bulat dengan c│a dan c│b maka c│(am+bm). untuk suatu m,n anggota bilangan Bulat

Bukti
c│a dan c│b maka terdapat bilangan bulat xdan y sedemikian sehingga a =cx  dan b =cy
Sehingga, am = c(xm) dan  bn =c(yn). untuk suatu xm = p dan (yn)=q, Maka:
 am + bn = c(p+q). Akibatnya,c│(am+bn).

Teorema 3 (Buchmann, 2002: 3) 
a. Jika a│b dan b ≠ 0 maka |a| ≤ |b|.
b. Jika a│b dan b│a maka |a| = |b|.

Bukti
a. Jika  a│b dan b ≠ 0 maka menurut Definisi, terdapat  m ≠ 0 sedemikian sehingga b = am.
    Karena b = am maka |b| = |am| ≥ |a| sehingga, |a| ≤ |b|.
b. Andaikan  a│b dan b│a.  Jika a = 0 maka b = 0 dan jika a ≠ 0  maka b ≠ 0.
   Selanjutnya,
   Jika  a ≠ 0 dan  b ≠ 0 maka sesuai dengan Teorema 3a, |a| ≤ |b| dan |b| ≤ |a| sehingga |a| = |b|.

Senin, 12 Desember 2011

flightless bird, american mouth - iron & wine

I was a quick wet boy, diving too deep for coinsAll of your street light eyes wide on my plastic toysThen when the cops closed the fair, I cut my long baby hairStole me a dog-eared map and called for you everywhereHave I found youFlightless bird, jealous, weeping or lost you, american mouthBig pill looming[ From: http://www.metrolyrics.com/flightless-bird-american-mouth-lyrics-iron-wine.html ]
Now I'm a fat house catNursing my sore blunt tongueWatching the warm poison rats curl through the wide fence cracksPissing on magazine photosThose fishing lures thrown in the coldAnd clean blood of Christ mountain streamHave I found youFlightless bird, grounded, bleeding or lost you, american mouthBig pill stuck going down

Right Here Waiting - Richard Marx


Oceans apart day after day and I slowly go insane
I hear your voice on the line but it doesn't stop the pain
If I see you next to never how can we say forever

* Wherever you go whatever you do
I will be right here waiting for you
Whatever it takes or how my heart breaks
I will be right here waiting for you

I took for granted, all the times that I though would last somehow
I hear the laughter, I taste the tears but I can't get near you now
Oh, can't you see it baby you've got me goin' crazy

[Repeat *]

I wonder how we can survive this romance
But in the end if I'm with you I'll take the chance
Oh, can't you see it baby you've got me goin' crazy

[Repeat *]

Minggu, 11 Desember 2011

Perbedaan Mani laki2 dan Wanita

penulis Al Ustadzah
new1 Muslimah Berta 13 - Agustus - 2004 07:55:09

Muslimah Bertanya

Perbedaan Mani laki2 dan Wanita

Apakah wanita juga keluar mani sebagaimana hal laki-laki? Bila ya bagaimana ciri-cirinya? Dan apa yg harus dilakukan?

Jawab :

Wanita juga keluar mani sebagaimana laki-laki. Dengan mani itu muncul sifat identik sang anak apakah memiliki kemiripan dgn ayah ataupun dgn ibunya. Ketika ditanyakan hal ini kepada Rasulullah shallallahu ‘alaihi wasallam beliau berkata:

نَعَمْ، فَمِنْ أَيْنَ يَكُوْنُ الشَّبَهُ؟

“Iya darimana ada persamaan anak ?”

Namun mani wanita berbeda dgn laki2 seperti yg disabdakan Rasulullah shallallahu ‘alaihi wasallam:

مَاءَ الرَّجُلِ غَلِيْظٌ أَبْيَضُ وَمَاءُ الْمَرْأَةِ رَقِيْقٌ أَصْفَرُ

“Mani laki2 itu kental dan berwarna putih sedangkan mani wanita tipis/ halus dan berwarna kuning.”
Al-Imam An-Nawawi rahimahullah berkata: “Adapun mani wanitaberwarna kuning tipis/ halus. Namun terkadang warna bisa memutih krn kelebihan kekuatannya. Dan mani wanita ini bisa ditandai dgn dua hal: pertama aroma seperti aroma mani laki-laki. Kedua terasa ni’mat ketika keluar dan setelah keluar syahwatpun mereda.”

Sebagaimana hal laki2 bila seorang wanita keluar mani krn senggama maupun ihtilam mk ia wajib mandi. Hal ini pernah ditanyakan oleh Ummu Sulaim radhiallahu ‘anha kepada Rasulullah shallallahu ‘alaihi wasallam. Ketika datang menemui Rasulullah Ummu Sulaim berkata:

فَهَلْ عَلَى الْمَرْأَةِ مِنْ غُسْلٍ إِذَا احْتَلَمَتْ؟ فَقَالَ رَسُوْلُ اللهِ صَلَّى اللهُ عَلَيْهِ وَسَلَّمَ : نَعَمْ إِذَا رَأَتِ الْمَاءَ

“Apakah wanita harus mandi bila ia ihtilam?” Rasulullah menjawab: “Ya apabila ia melihat keluar mani.”

Dalam Al-Majmu’ Al-Imam An-Nawawi rahimahullah berkata: “Ulama sepakat wajib seseorang mandi bila keluar mani dan tdk ada perbedaan di sisi kami apakah keluar krn jima’ ihtilam onani melihat sesuatu yg membangkitkan syahwat ataupun keluar mani tanpa sebab. Dan samasaja apakah keluar dgn syahwat ataupun tdk dgn rasa ni’mat atau tdk banyak ataupun sedikit walaupun hanya setetes dan sama saja apakah keluar di waktu tidur ataupun di waktu jaga baik laki2 maupun wanita.”

Sumber:
- www.asysyariah.com
- blog.re.or.id

KEUTAMAAN SABAR MENGHADAPI COBAAN


“Artinya : Dari Abu Sa’id Al-Khudry Radhiyallahu anhu, dia berkata. ‘Aku memasuki tempat Rasulullah Shallallahu ‘alaihi wa sallam, dan beliau sedang demam. Lalu kuletakkan tanganku di badan beliau. Maka aku merasakan panas ditanganku di atas selimut. Lalu aku berkata. ‘Wahai Rasulullah, alangkah kerasnya sakit ini pada dirimu’. Beliau berkata: ‘Begitulah kami . Cobaan dilipatkan kepada kami dan pahala juga ditingkatkan bagi kami’. Aku bertanya. ‘Wahai Rasulullah, siapakah orang yang paling berat cobaannya ? Beliau menjawab: ‘Para nabi. Aku bertanya. ‘Wahai Rasulullah, kemudian siapa lagi? Beliau menjawab: ‘Kemudian orang-orang shalih. Apabila salah seorang di antara mereka diuji dengan kemiskinan, sampai-sampai salah seorang diantara mereka tidak mendapatkan kecuali mantel yang dia himpun. Dan, apabila salah seorang diantara mereka sungguh merasa senang karena cobaan, sebagaimana salah seorang diantara kamu yang senang karena kemewahan”.

Ukhti Muslimah !

Selanjutnya perhatikan perkataan Ibnu Abdi Rabbah Al-Andalusy : “Asy-Syaibany pernah berkata. ‘Temanku pernah memberitahukan kepadaku seraya berkata. ‘Syuraih mendengar tatkala aku mengeluhkan kesedihanku kepada seorang teman. Maka dia memegang tanganku seraya berkata. ‘Wahai anak saudaraku, janganlah engkau mengeluh kepada selain Allah. Karena orang yang engkau keluhi itu tidak lepas dari kedudukannya sebagai teman atau lawan. Kalau dia seorang teman, berarti dia berduka dan tidak bisa memberimu manfaat. Kalau dia seorang lawan, maka dia akan bergembira karena deritamu. Lihatlah salah satu mataku ini, ’sambil menunjuk ke arah matanya’, demi Allah, dengan mata ini aku tidak pernah bisa melihat seorangpun, tidak pula teman sejak lima tahun yang lalu. Namun aku tidak pernah memberitahukannya kepada seseorang hingga detik ini. Tidakkah engkau mendengar perkataan seorang hamba yang shalih : “Sesungguhnya hanya kepada Allah aku mengadukan kesusahan dan kesedihanku”. Maka jadikanlah Allah sebagai tempatmu mengadu tatkala ada musibah yang menimpamu. Sesungguhnya Dia adalah penanggung jawab yang paling mulia dan yang paling dekat untuk dimintai do’a”.


Perbaharuilah imanmu dengan lafazh la ilaha illallah dan carilah pahala di sisi Allah karena cobaan yang menimpamu. Janganlah sekali-kali engkau katakan : “Andaikan saja hal ini tidak terjadi”, tatkala menghadapi takdir Allah. Sesungguhnya tidak ada taufik kecuali dari sisi Allah.

reference: blog.re.or.id

Senin, 21 November 2011

CHEMISTRY REPORT - ACID BASE TITRATION

ACID BASE TITRATION
I. DAY/ DATE OF EXPERIMENT:
We did the experiment on Friday, 4th November 2011.
II. COMPLETED THE EXPERIMENT
We finished the experiment also on Friday, 4th November 2011
III. THE PURPOSE OF THE EXPERIMENT
· To determine the concentration of NaOH by Oxalic Acid solution
· To determine the concentration of HCl by NaOH
IV. BASIC THEORY
Acid-base titration is a technique that is widely used to determine the exact concentration of an acid or alkaline solution. Titration is a measurement of a solution of one reactant needed to react completely with a number of other specific reactants. Acid-base titration is a neutralization reaction. If the solution is called acidic raw acidimetry and if the solution is called alkalimetri raw base. Titration is a quantitative chemical analysis techniques used to determine the concentration of a particular solution, where the determination using a standard solution of known concentration is appropriately. Solution used for the determination of the solution of the unknown concentration is placed in a burette (see figure) and the solution is referred to as the standard solution or titrant or titrator, while the solution of unknown concentration is placed in Erlenmeyer (see picture) and is referred to as the analyte solution. Titrant is added little by little on the circumstances in which the analyte to obtain equivalent titrant reacts with the analyte, which means that all titrant reacts with the analyte out of this situation is referred to as a point equivalent. Equivalent point can be determined by a variety of ways, a common way is to use indicators. The indicator will change color with the addition of as little as possible titrant, in this way then we can immediately stop the process of titration. In other words, the acid-base titration of acid equivalent amount equal to the number of equivalent base.

Phenolphthalein indicator
Phenolphthalein is an organic compound having the formula C20H14O4, solid grain shaped, colorless and soluble in alcohol and organic solvents. Generally, using phenolphthalein as an indicator of acid-base (in acid solution is colorless and colored red). Phenolphthalein is an acid-base indicators are included in kelasftalein. Fenolftalein an indicator of one color. Route pH at 8.3 to 9.8 ie discoloration from colorless to red. This indicator is very suitable as an indicator for the titration of acids with strong bases. The acid can be titrated strong acid or weakacid with Ka price limits are eligible within perfectness. Fenolftalein reaction are compounds containing phenol groups, so that weakly acidic.

Anthocyanin
Anthocyanins are the pigments of red, purple, and blue found in all plants except Fungus. Most of the anthocyanin in the form of glycosides, usually bound one or two units of sugar such as glucose, galactose, ramnosa, and silosa. If monoglikosida, so the part of sugar only bound at position 3, and at position 3 and 5 when a diglikosida danbagian
aglikionnya called anthocyanidins. Most of anthocyanin reddish in acid solution, but the purple and blue with increasing pH which ultimately damaged in strong alkalis solution (Sinaga, 2010). The color resulting from anthocyanin is influenced by medium acidity, anthocyanin in acidic situation pH 1-3 showed a red color while pH caused increase resulting decrease in color intensity (Abbas, 2003). It shows that anthocyanin pigments are not stable against changes in pH and temperature.
There are 2 kind of standard solution. A standard solution is prepared by dissolving an accurately weighed quantity of a highly pure material and diluting to an accurately known volume of volumetric flask is called primary standard. A solution standardized by titrating a primary standard itself a secondary standard. It will be less accurate than a primary standard due to the error of titration.
A primary standard should fulfill these requirement;
1. It should be 100% pure, although 0.01 to 0.02% impurity is tolerable if it is accurately known.
2. it should be stable to drying temperature, and it should be stable indefinitely at room temperature. 
The primary standard is always dried before weighing.*
3. It should be readily available and fairly inexpensive.
4. Although not necessary, it should have a high formula weight. This is so that relatively large amount of it will have to be weighed to get enough to titrate. The relative error in weighing a greater amount of material will be smaller than that for a small amount.
5. If it is to be used in titration, it should possess the properties required for a titration listed above. in particular, the equilibrium of the reaction should be far to the right so that a very sharp end point will be obtained.


V. EXPERIMENTAL METHODS
A. Tools and Materials:
1. Stative and clamps
2. Burette
3. Erlenmeyer flask 250 ml
4. Funnel
5. Pipette mumps 25 ml
6. Pipette drops
7. Beaker 100 ml
8. Measuring cup
9. NaOH 0.1 m
10. C2H2O4 0.1 m
11. HCl 0.1 m
12. Phenolptalein
13. Aquades
14. Plant extracts
15. Pipette filler
16. Mortar and alu

VI. EXPERIMENT RESULT
1. Determining of NaOH solution concentration with 5 ml of C2H2O4 0,05 M using PP indicator.
Titration
NaOH needed (ml)
First
9,7
Second
9,4
Third
9,6
2. Determining of HCl concentration with NaOH using PP indicator.
Titration
NaOH needed (ml)
First
9,6
Second
9,8
Third
9,25
3. Determining of HCl concentration with NaOH using plant extract indicator.
Titration
NaOH needed (ml)
First
9,5
Second
9,3
Third
9,2

VII. DATA ANALYSIS
1. Determining of NaOH solution concentration with C2H2O4 using PP indicator.
Titration
NaOH needed (ml)
NaOH Molarity
First
9,7
0,025
Second
9,4
0,0265
Third
9,6
0,026
Reaction:
2NaOH + C2H2O4 → Na2C2O4 + 2H2O

2. Determining of HCl concentration with NaOH using PP indicator.
Titration
NaOH needed (ml)
HCl Molarity
First
9,6
0,049
Second
9,8
0,050
Third
9,25
0,048
Reaction :
HCl + NaOH → NaCl + H2O
3. Determining of HCl concentration with NaOH using plant extract indicator.
Titration
NaOH needed (ml)
HCl Molarity
First
9,5
0,049
Second
9,3
0,048
Third
9,2
0,047
Reaction :
HCl + NaOH → NaCl + H2O


VIII. EXPLANATION
1. Determining of NaOH solution concentration with C2H2O4 using PP indicator.
Reaction happened is
2NaOH + C2H2O4 → Na2C2O4 + 2H2O
Titration
NaOH needed (ml)
NaOH Molarity
First
9,7
0,025
Second
9,4
0,0265
Third
9,6
0,026
NaOH molarity can be gotten like in the above data is obtained from calculation with diluting formula, that is M1.V1 = M2.V2 which solution 1 is NaOH, solution 2 is C2H2O4. For more complete, can be seen in calculation attachment pages.

2. Determining of HCl concentration with NaOH using PP indicator.
Reaction happened is
HCl + NaOH → NaCl + H2O
Titration
NaOH needed (ml)
HCl Molarity
First
9,6
0,049
Second
9,8
0,050
Third
9,25
0,048
HCl Molarity can be gotten like in the above data is obtained from calculation with diluting formula, that is that is M1.V1 = M2.V2 which solution 1 is NaOH, solution 2 is HCl. Here used average of NaOH Molarity from the previous experiment (First Experiment). For more complete, can be seen in calculation attachment pages.

3. Determining of HCl concentration with NaOH using plant extract indicator.
Reaction happened is
HCl + NaOH → NaCl + H2O
Titration
NaOH needed (ml)
HCl Molarity
First
9,5
0,049
Second
9,3
0,048
Third
9,2
0,047
Here plant extract indicator used is “bunga sepatu”
HCl Molarity can be gotten like in the above data is obtained from calculation with diluting formula, that is that is M1.V1 = M2.V2 which solution 1 is NaOH, solution 2 is HCl. Here used average of NaOH Molarity from the previous experiment (First Experiment). For more complete, can be seen in calculation attachment pages.

IX. CONCLUSION
· The average of NaOH Molarity obtained from First Experiment data is 0.026 M.
· The average of HCl Molarity obtained from Second Experiment data is 0,049 M.
· The average of HCl Molarity obtained from Third Experiment data is 0,048 M.

QUESTION AND ANSWER
1. Why in titration of NaOH with oxalic acid using phenolptalein as indicator?
2. What is the differences between equivalent point and final point?
3. On solution above, which serves as primary standard solution, secondary standard solution, and tertiary standard solution?

1. In this titration we use the indicator phenolptalein(PP) because oxalic acid is classified as a very weak acid in unionized condition but if in alkaline environment, PP will be ionized more and he will give a bright color and the color changes are easier to be observed
2. The equivalence point is the point in a titration when the amount of added standard reagent is chemically equal to the amount of analyte
(the point where the number of moles of base equal the number of moles of acid).
The end point is the point in a titration when a physical change occurring immediate after the equivalence point
3. primary standard solution : H2C2O4 , because primary standard solution is a solution that has been known its concentration carefully
secondary standard solution : NaOH, because secondary standard solution ia a solution standardized by titrating a primary standard itself
tertiary standard solution : HCl,


REFERENCES
§ http://www.wikipedia.org
§ http://www.newton.dep.anl.gov/askasci/chem03/chem03996.htm
§ http://robbaniryo.com/ilmu-kimia/larutann-standar-primer-dan-sekunder/


ATTACHMENT
CALCULATION

* 1st Experiment
2NaOH + H2C2O4 ® Na2C204 + 2H2O
Finding Molarity of NaOH
a. First titration
M NaOH = ?
V NaOH = 9,7 ml
V H2C2O4 = 5 ml
M H2C2O4 = 0,05 ml
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M H2C2O4. V H2C2O4
1. x. 9,7 = 2. 0,05. 5
9,7 x = 10. 0,05
x = 0,05154 M
The normality of NaOH = 0,051 M
The molarity of NaOH = 0,025 N
b. Second titration
M NaOH = ?
V NaOH = 9,4 ml
V H2C2O4 = 5 ml
M H2C2O4 = 0,05 ml
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M H2C2O4. V H2C2O4
1. x. 9,4 = 2. 0,05. 5
9,4 x = 10. 0,05
x = 0,0531 M
The normality of NaOH = 0,053 M
The molarity of NaOH = 0,0265 N
c. Third titration
M NaOH = ?
V NaOH = 9,6 ml
V H2C2O4 = 5 ml
M H2C2O4 = 0,05 ml
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M H2C2O4. V H2C2O4
1. x. 9,6 = 2. 0,05. 5
9,6 x = 10. 0,05
x = 0,05208 M
The normality of NaOH = 0,052 M
The molarity of NaOH = 0,026 N

* 2nd Experiment
NaOH + HCl ® NaCl + H2O
Finding molarity of HCl using PP indicator
Data : VHCl = 5 ml
NNaOH = 0,052 N
MNaOH = 0,026 M
a. First titration
Got 9,6 ml of NaOH
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M HCl. V HCl
1. 0,026. 9,6 = 1. X. 5
5x = 0,249
x = 0,049
The molarity of HCl = 0,049 M
The normality of HCl = 0,049 N
b. Second titration
Got 9,8 ml of NaOH
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M HCl. V HCl
1. 0,026. 9,8 = 1. X. 5
5x = 0,25
x = 0,05 M
The molarity of HCl = 0,05 M
The normality of HCl = 0,05 N
c. Third titration
Got 9,25 ml of NaOH
n. M1. V1 = n. M2. V2
n. M NaOH. V NaOH = n. M HCl. V HCl
1. 0,026. 9,25 = 1. X. 5
5x = 0,24
x = 0,048
The molarity of HCl = 0,048 M
The normality of HCl = 0,048 N
The molarity of HCl is found : = 0,049 M

* 3rd Experiment
NaOH + HCl ® NaCl + H2O
Finding molarity of HCl using Plant extract indicator
Data : VHCl = 5 ml (V2)
M NaOH = 0,052 M(M1)
a. First titration
Got 9,5 ml of NaOH ( V1)
M1. V1 = M2. V2
M NaOH. V NaOH = M HCl. V HCl
0,026. 9,5 = x. 5
5x = 0,247
x = 0,049 M
The molarity of HCl = 0,049 M
The normality of HCl = 0,049 N
b. Second titration
Got 9,3 ml of NaOH ( V1)
M1. V1 = M2. V2
M NaOH. V NaOH = M HCl. V HCl
0,026. 9,3 = x. 5
5x = 0,2418
x = 0,048 M
The molarity of HCl = 0,048 M
The normality of HCl = 0,048 N
c. Third titration
Got 9,2 ml of NaOH ( V1)
M1. V1 = M2. V2
M NaOH. V NaOH = M HCl. V HCl
0,052. 9,2 = x. 5
5x = 0,239
x = 0,047 M
The molarity of HCl = 0,047 M
The normality of HCl = 0,047 N