Wednesday, 15 October 2014

History: Causes of the wars up to 1455



History


Causes of the Wars up to 1455


A) Henry VI


·         Ruled by regent as a child

·         Schizophrenic

·         Easily manipulated

o   Dominated by court favourites

§  Somerset and Suffolk

·         Pious and focused on education

o   Built Eton and Kings College Cambridge

o   Built churches and cathedrals

·         No desire for war

o   left military leadership to other

o   Naval and coastal security neglected

o   Never led his country into war

·         Margaret of Anjou

o   Made King look weaker

o   Increases tension between Yorkists and Lancastrians



B) Economics


·         Poor economy

o   Agricultural slump in 1430s

§  Land revenues fell by 10%

o   Commercial slump in 1450s

o   Wool trade collapsed due to breakdown of Anglo-Burgundy alliance

o   100 Year War drained finances

§  Land lost

§  Wool trade restricted

§  Wine trade restricted

·         Noble economics

o   Economic slump meant nobles needed the kings financial support

o   Some nobles (Nevilles, Buckingham) were rich, whilst others (Beauforts) were poor

o   Often rewarded with French land that was lost après. 100 year war

·         Henrys economics

o   Crown owed £370,000 in debt

§  Calais Garrison cost £17,000 per annum

§  Previous pursuits in French war

§  Cost of court remained high

o   Annual income fallen from £120,000 to £45,000 per annum

o   Inefficient taxation of land

o   Continued to give land and titles to favourites

§  Rendered himself powerless


D) Divided Nobility


·         Less money going around = more competition for Kings support

·         Increase in power due to bastard feudalism and retinues

·         Traditional family feuds

o   Nevilles vs Percys

·         Ambition



E) French Wars


·         Fuelled pre existing noble rivalries

o   York withdrawn from war – less responsibility

o   Somerset responsible for surrender of Caen 1450

·         Diminished status and prestige of England

o   Ended dual monarchy

·         Exposed Henrys weaknesses

·         Economic problems

o   Merchants lost wine and wool trade

§  Looked to York for stronger Government

·         Nobles now have more time and a need to regain what they may have lost in France

Monday, 13 October 2014

Chemistry: Atomic Structure



Chemistry

Atomic Structure

A)     Describe protons, neutrons and electrons in terms of relative charge and mass

Particle
Relative Mass
Charge
Position in atom
Proton
1
1+
Nucleus
Neutron
1
0
Nucleus
Electron
1 / 2000
1-
Orbiting nucleus








B)      Define ions and isotopes

Isotope: different atomic forms of the same element, which will thus have the same number of protons and electrons but different numbers of neutrons.

Ion: Atoms that have gained or lost electrons and are represented with their charge.


C)      Define and calculate relative atomic and isotopic masses

Relative Isotopic Mass: Mass of one isotope of an element relative to 1/12 the mass of carbon-12

Relative Atomic Mass: The average mass if an atom of an element relative to 1/12 the mass of carbon-12.  Is calculated by:

         [(% x isotopic mass) + (% x isotopic mass) + (% x isotopic mass)] / 100

Chemistry: Moles and equations



Chemistry

Moles and equations


A)     Define and use Mole calculations


One Mole: the amount of substance that contains as many elementary entities as there are carbon atoms in 12g of carbon-12


-Moles can be calculated using these equations:
- Amount in moles = mass in grams / molar mass                                      =     n = m / M

-Amount in moles = volume of gas / 24                                                     =     n = Vgas / 24


 -Amount in moles = concentration x volume                                              =     n = c x V

-Amount in moles = number of particles / Avogadro's number                           =     n = # / 6.02 x 10²³




B)      Define Avogadro’s number


Avogadro’s number: the number of carbon-12 atoms in 12g of carbon-12. Is equal to 6.022045 x 10²³




C)      Explain and calculate Empirical and Molecular formula


Empirical formula: gives the simplest whole number ratio of the atoms of each element present in the compound.


Molecular formula: gives the actual number of atoms of the different elements in one molecule of a compound


-e.g

A compound contains 73.47% carbon, 10.2% hydrogen, and 16.33% oxygen. Its relative Molecular Mass in 98. Calculate the empirical and molecular formulas of the compound.

Method:


1) Divide percentage of each element by the elements Atomic mass

2) Divide all figures by the smallest calculated

3) Use this ratio to generate the empirical formula

4) Use the formula show to calculate how many times the empirical formula goes into the molecular formal mass

5) Multiply the empirical formula by this number to calculate the molecular formula



                      C                                             H                                             O


                       73.47/12                                   10.2/1                                 16.33/16


                       =6.1225                                    =10.2                                     =1.02


                    6.1225/1.02                             10.2/1.02                              1.02/1.02


                           =6                                        =10                                           =1



Empirical Formula :              C₆H₁₀O


Molecular Formula Mass      =      Empirical Formula Mass x n

     98         =       [(6 x 12) + (10 x 1) + (16)] x n

     98         =       98 x 1

                                1 x Empirical Formula         =       Molecular formula

                                                   1 x C₆H₁₀O          =        C₆H₁₀O




D)     Be able to calculate percentage errors


Percentage error = (maximum error / measurement taken) x 100