TECHNOLOGY VIDEOS CREATED
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CHEMISTRY VIDEOS CREATED
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A Dedicated page to discuss Chemistry
What is the IUPAC name of H2O
Oxidane
Referenece: (http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/Chapter2-Sec20-24.pdf)
Oxidane
Referenece: (http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/Chapter2-Sec20-24.pdf)
Why coco cola is exploding when we add mentos to it?
(Question Ashwin C. S., through aroramesh.weebly.com, on 28-10-2014)
Coca cola contain compressed carbon dioxide. Other carbonated beverages will also readily respond to the addition of Mentos. The surface of the mint Mentos is covered with many small holes that increase the surface area available and act as "nucleation sites" (bubble growth sites) that encourage the rapid formation of bubbles. The buoyancy of the bubbles and their growth in size will eventually cause the bubbles to leave the nucleation site and rise to the surface of the soda. Bubbles will continue to form on the porous surface and the process will repeat, creating a nice foamy result.
The ingradients of the mentos such as the gum arabic / gelatin combined with the potassium benzoate, sugar or (potentially) aspartame, in Diet sodas, also help in this process. In these cases, the ingredients end up lowering the surface tension of the liquid, allowing for even more rapid bubble growth on the porous surface of the Mentos.
References: http://urbanlegends.about.com/library/bl_mentos_and_coke3.htm
http://en.wikipedia.org/wiki/Diet_Coke_and_Mentos_eruption#cite_note-nakedeggs-3
http://www.todayifoundout.com/index.php/2012/11/why-do-mentos-and-diet-coke-react/
(Question Ashwin C. S., through aroramesh.weebly.com, on 28-10-2014)
Coca cola contain compressed carbon dioxide. Other carbonated beverages will also readily respond to the addition of Mentos. The surface of the mint Mentos is covered with many small holes that increase the surface area available and act as "nucleation sites" (bubble growth sites) that encourage the rapid formation of bubbles. The buoyancy of the bubbles and their growth in size will eventually cause the bubbles to leave the nucleation site and rise to the surface of the soda. Bubbles will continue to form on the porous surface and the process will repeat, creating a nice foamy result.
The ingradients of the mentos such as the gum arabic / gelatin combined with the potassium benzoate, sugar or (potentially) aspartame, in Diet sodas, also help in this process. In these cases, the ingredients end up lowering the surface tension of the liquid, allowing for even more rapid bubble growth on the porous surface of the Mentos.
References: http://urbanlegends.about.com/library/bl_mentos_and_coke3.htm
http://en.wikipedia.org/wiki/Diet_Coke_and_Mentos_eruption#cite_note-nakedeggs-3
http://www.todayifoundout.com/index.php/2012/11/why-do-mentos-and-diet-coke-react/
What happens ammonium chloride added to aq. ammonia?
(Question from Aiswarya through aroramesh. weebly.com, 28/11/2015)
First look at th e chemicals in this process
- Aqueous ammonia = NH3 + H2O = NH4OH
- Ammonium Chloride = NH4Cl
- NH4OH is a weak electrolyte (base) - i.e., it will ionize (dissociate) in small amounts
- NH4Cl is strong electrolyte – i.e., it will ionize (dissociate) completely in water
We have the aqueous ammonia – NH4OH in water- So a particular number (not all) of NH4OH ionize (dissociate) in to NH4+ and OH- ions
On adding a strong electrolyte, NH4Cl will completely ionize in to NH4+ and Cl- ions. But in the solution, we have already NH4+ ions and could not cross a particular value. So the strongest chemical dominates the weak chemical. Hence the Strong electrolyte ionize into NH4+ and Cl- ions, which suppress the ionization of NH4OH ionization (dissociation) in to NH4+ and OH- ions (This termed as common ion effect). Actually the suppression of ionization of NH4OH is to maintain a constant equilibrium value. – see the explanation at the end
In the third group (intergroup separation of cations, B. Sc. Chemistry Lab), we are adding NH4Cl and NH4OH. However the third group reactant is NH4OH only, which react with third group cations Al3+ form Al(OH)3, precipitate. Here NH4Cl has NO role in the formation of Al(OH)3 precipitate. (If we are NOT adding NH4Cl in third group, 4th group cations also will precipitate).
But if we are not adding NH4Cl in 3rd group, the dissociation of NH4OH will be high, hence more NH4+ and OH- ions will be present in the solution. These OH- ions also will react with 4th group metal ions and form metalhydroxide. Since the concentration of OH- ions is high, the concentration of 4th group metal hydroxides also will be high, which result in the precipitation of 4th group cations in third group. (4th group hydroxides only precipitate at high concentration, because its solubility product is higher than 3rd group hydroxides).
Common ion effect: Why suppression of weak electrolyte?
Let me show you one simple example. (Here the values given are arbitrary values, not the absolute (exact) values).
Equilibrium constant is a constant (which do not change on addition of another electrolyte), which is taken as 1.1 for NH4OH
We have a solution of aqueous ammonia. 100 numbers of NH4OH added to water. Since it is a weak electrolyte only few of them dissociated to NH4+ and OH- ions. Let 10 NH4OH dissociated into 10 NH4+ and 10 OH- ions.
NH4OH = NH4+ + OH-
K = [NH4+][OH-] / [NH4OH]
[NH4OH] = 100-10 = 90 in water
[NH4+] = 10 in water
[OH-] = 10 in water
K = (10 x10) / 90 = 1.1
To this solution we are adding a strong electrolyte, NH4Cl – 5 numbers (or concentration). Since NH4Cl is a strong electrolyte all added NH4Cl dissociate in to 5 NH4+ and 5 OH- ions.
[NH4Cl] = 5-5 = 0 in water
[NH4+] = 5 in water
[Cl-] = 5 in water
Now check the equilibrium constant of NH4OH.
In water Now we have
[NH4OH] = 90
[NH4+] = 10+5 =15 in water (5 is from the dissociation of NH4Cl ions, equilibrium constant doesn’t have brain to think that these 5 ions came from NH4Cl !!!!!!!!!!)
[OH-] = 10 in water
K = [NH4+][OH-] / [NH4OH] = (15 x10)/90= 1.6, But equilibrium constant K is a constant K =1.1. It cannot change
So solution cleverly decides to adjust the value K to 1.1. Inorder to achieve the value 1.1, either numerator should decrease or denominator should increase.
Ie., [NH4+] = should decrease (or)
[OH-] = should decrease (or)
[NH4OH] = should increase.
This would possible if the backward reaction happens to NH4OH dissociation:
Ie, NH4+ + OH- = NH4OH suppression of dissociation of weak electrolyte.
Initially 10 NH4OH dissociated from 100 NH4OH molecules. Due to suppression of dissociation of NH4OH , consider only 8 NH4OH dissociated in to 8NH4+ and 8OH- ions.
Now we have
[NH4OH] = 100-8 = 92 in water
[NH4+] = 8+5 =13 in water (5 is from the dissociation of NH4Cl ions)
[OH-] = 8 in water
K = [NH4+][OH-] / [NH4OH] = (13 x8)/92= 1.1
Now equilibrium constant remains a constant by suppressing the dissociation of weak electrolyte (We cannot suppress the dissociation of a strong electrolyte because he is STRONG !!!!!!)
Note: NH4OH + NH4Cl can act as buffer solutions