Thursday 20 April 2017

Hi again!!

Hi all,

It's been so long! Long story short, sorta locked myself out of this account and thus...well...haven't been able to post hahahaha. All good now:))

It's been like 8 months so let's recap...

For those asking, I did manage to grab A*s in all three of my sciences:)). Despite this though I only carried on with physics and biology...well...and chemistry too...for a bit (long story)...

Basically with the new A level system I started out with 4 (Chemistry, Maths, Physics and Biology) but our school quickly made us cut down to 3 (much to my sadness) so I decided to drop Chemistry because it's simply the one I didn't need as much (regarding my career path).

I know that sounds a bit strange because I am hoping to read Law at university, but still, Chemistry had to be dropped):

So here we are now, i'm currently mid-way through 1st year A levels (equivalent to AS year) and it's a bit hell at times but mostly all good. What i'm really getting at in this post is whether anyone would like me to do revision blogs such as this one, and my others, for Biology and Physics A level? (I could also do math, but it may be a bit trickier but if you want, I will:D). For biology I do the AQA spec and for physics I do OCR Physics A (maths is also OCR covering the topics Core 1, Core 2, Statistics 1, Core 3, Core 4, and Mechanics 1).

Even if we take different exam boards, it may be useful since a lot of stuff crosses over all boards.

Have a mull over:) tbh I love to do this for the enjoyment of helping people, so let me know if these blogs have/will help you, and i'll continue them:))

Thank you for all your support,

Millie xx

Wednesday 18 May 2016

2.38 describe tests for the anions...

Halide ions
To test for halide ions, add dilute nitric acid (HNO3) followed by silver nitrate solution (AgNO3). A precipitate will be produces, the colour of this precipitate will determine what ions are present.
Cl -  > white precipitate
Br- > cream precipitate
I-   > yellow precipitate

NOTE: The silver nitrate solution determines which halide ions are present. The dilute nitric acid is added to 'get rid' of an carbonate or sulphite ions (as these would react with the silver nitrate).


Sulphate ions
To test for sulphate ions (SO42- ), add dilute HCl and then barium chloride solution (BaCl2). If sulphate ions are present, a note precipitate would form (this precipitate is barium sulphate)

NOTE: The HCl is used to 'get rid' of any carbonate ions, and these may impede results if present as they would also produce a precipitate).


Carbonate ions
To test for carbonate ions (CO32- ), add dilute HCl to the sample you are testing. If a gas is produced, collect it and bubble it through limewater. If carbonate ions are present, the limewater will go cloudy as carbon dioxide will be released.

1.55 write ionic half equations representing the reactions at the electrodes during electrolysis

Half equations just show what happened at each electrode. Like any equation, they need to be balanced. here are some examples...

2Br- > Br2 + 2e-


Above is an equation of a reaction at an anode. We can tell this because the Br2 loses an electron. This can be inferred as on the left it is 2Br- and on the right it is Br2 (meaning it has lost an electrode as it no longer is a negative ion, it is an atom). As it is 2Br-, 2 electrons must be lost.


2H+ + 2e- > H2



Above is an equation of a reaction at a cathode as 2H+ gains 2 electrons to become H2. The charges cancel out and it is neutral on both sides. As there are 2 'H's, there must be 2 electrons gained.

1.46 understand that a metal can be described as a giant structuree of positive ions surrounded by a sea of delocalised electrons

In a metal, the atoms are ionically bonded in a giant 3-D structure. The outer shell electrons become detached (making positive ions, cations). The cations are 'floating' in a sea of delocalised electrons. The metal structure stays together because the cations (+) are attracted to the electrons (-). This is known as a metallic bond.

1.31 deduce the charge of an ion from the electronic configuration of the atom from which the ion is formed

An atom with less then four electrons on its outer shell will want to lose electrons because that is the quickest way for it to have a full outer shell. So we know the atom will lose electrons (this makes positive ions).

Atoms with more than four electrons will gain electrons to fill their outer shell (as it is easier than losing electrons). This will result in negative ions.

Here are some examples...

Na has the electronic configuration of  2.8.1 (it has 1 electron on it's outer shell, so it is easiest to lose 1 electron to make a full outer shell). This results in the positive ion (a cation), Na+, which has the electronic configuration of 2.8

Cl has the electronic configuration of 2.8.7 (it has 7 electrons on it's outer shell, so it is easiest to gain 1 electron to make a full outer shell). This results in the negative ion (an anion), Cl-, which has the electronic configuration of 2.8.8

Tuesday 17 May 2016

5.22 understand that nitrogen from air, and ydrogen from natural gas or the cracking of hydrocarbons, are used in the manufacture of ammonia

Ammonia is manufactured using hydrogen and nitrogen (as ammonia is NH3). The equation for the reaction is...

N2 (g)  + 3H2 (g)   + 2NH3 (g) (+heat)

Hydrogen is sourced from natural gas. Nitrogen is sourced from the air (as the air is 78% nitrogen).

The manufacture of ammonia with hydrogen and nitrogen is reversible, therefore, as soon as ammonia is made, it starts to turn back into nitrogen and hydrogen. This mean that not all of the nitrogen and hydrogen is converted to ammonia and not all ammonia is converted into nitrogen and hydrogen (because, as soon as nitrogen and hydrogen are produced, it starts to turn into ammonia). This means the reaction reached a dynamic equilibrium.