 # wolfson

Senior Members

630

## Community Reputation

192 Excellent

• Rank
Molecule
• Birthday 07/23/1983

## Contact Methods

• Website URL
http://masterdragon1.proboards19.com/index.cgi

## Profile Information

• Location
England
• Interests
Forensic Science General Science General Maths, Organic Chemsitry and inorganic, Abstract Algebra, A.P, G.P, Series and poly, bi, trinomials (also log and e)
• College Major/Degree
Bsc (Hons.) Chemistry & Bsc/Applied science with forensic investigation and crime scene science
• Favorite Area of Science
Applied science, Forensic Science, Chemistry, Biology and Mathematics.
• Biography
completed my Bsc Chemistry now continuing my education in the field i should have originaly studied Forensics, don't get me wrong i really like chemsitry but forensic is hopefully me real "choice"!
• Occupation
University Student
1. ## Why is the pH of pure water i.e. neutral 7?

In pure water, the hydrogen ion (hydroxonium ion) concentration must be equal to the hydroxide ion concentration. [OH-] term in the Kw expression by another [H+]. [H+]2 = 1.00 x 10-14 Taking the square root of each side gives: [H+] = 1.00 x 10-7 mol dm-3 Converting that into pH: pH = - log10 [H+] pH = 7
2. Example 2: How much copper is deposited if a current of 0.2 Amps is passed for 2 hours through a copper(II) sulphate solution ? Electrode equation: (-) cathode Cu2+(aq) + 2e- ==> Cu(s) and Ar(Cu) = 64 the quantity of electricity passed in Coulombs = current in A x time in secs (Q = I x t) = 0.2 x 2 x 60 x 60 = 1440 Coulombs, and 1 mole electrons = 96500 Coulombs therefore moles of electrons passed through circuit = 1440 / 96500 = 0.01492 it takes two moles of electrons to form one mole of copper therefore moles copper = 0.01492 7 2 = 0.00746 mass of copper = moles of copper x atomic mass of copper = 0.00746 x 64 = 0.4775g of copper deposited. Taken from http://www.wpbschoolhouse.btinterne...lysis%20product
3. The mass of a substance produced at an electrode during electrolysis is proportional to the quantity of electricity that has passed. Look up, Faraday's formula to relate these variables. look at site below: http://www.wpbschoolhouse.btinternet.co.uk/page04/4_73calcs.htm#13.%20Electrolysis%20product
4. Melting point of diamond = 3815.56oC, thats about 2.5 times more than steel. This information is the correct. Ref: RSC.
5. Sulphuric acid at that temperature will rapidly corrode steel.
6. Pulkit that was so wrong it's not even funny: Na2SiO3 + 8 HF --> H2SiF6 + 2 NaF + 3 H2O
7. To a chemist, that seems to me as a spelling mistake for Sialic Acid.
8. infinity = The limit that a function is said to approach at x = a when (x) is larger than any preassigned number for all x sufficiently near a.
9. I wrote a bit on electronic config at: http://dragonslair.europe.webmatrixhosting.net/Science/Chemistry/Eleccon2.htm It should help.
10. If all you wish to do is measure the number of mols, the easiest way would be to just weight the gas. weight the gas = calculate the volume. The higher the temperature the faster molecules are moving. Faster the molecules = higher the pressure, as are moving, the harder they hit the wall of the container. It's all kinetic theory/motion.
11. petrol heat of combustion:5460 kj/mol LPG heat of combustion 2220 kj/mol 2 C8H18 + 25 O2 = 18 H2O + 16 CO2 (combustion of octane) CH3-CH2-CH3 + 5 O2 = 3 CO2 + 4 H2O (combustion of propane) 1 mole of propane combusts 2 mole of Octane combusts Octane uses 10920 Kj Propane uses 2220 Kj So if Octane is 80 cents per 10920Kj Octane = 80 / 10920 = 0.00733 cents per Kj You get 2 moles of propane for the equivalent of octane So 0.00733 / 2 = 0.0037 So 0.0037 cents per kj And 2220 kj = 8.2 cents I am not sure if this is correct, it’s early in the morning
12. A list of metals arranged in order of their electrode potentials. A metal will displace, from their salts, metals lower down in the series. Lithium Potassium Calcium Sodium Magnesium Aluminum Zinc Iron Cobalt Nickel Tin Lead ---------------- Hydrogen ---------------- Copper Mercury Silver Platinum Gold
13. The norm H =37.1 pm The norm He(1+) = 31 pm
14. The process for preparing stable high oxidation state ferrocene derivatives, namely Fe(III)/Fe(IV), by reacting ferrocene derivatives in which one or more electron-donating groups are covalently bonded to the cyclopentadienyl rings with electron-donating ligands having strong coordinative groups for ferric species. (APP 2001)
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