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Summary Full Chemistry AS Level OCR Notes £10.49   Add to cart
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  7. AS Unit F321 - Atoms, Bonds and Groups

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Summary Full Chemistry AS Level OCR Notes
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These notes are straight to the point covering every single section of the specification of OCR AS level in only 57 slides. These notes include examples and every slide only takes about 5 minutes to read saving time and energy. These notes are great for all types of students including ones struggli...

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  • July 3, 2023
  • 58
  • 2022/2023
  • Summary

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Chem

YEAR 1

, xH20 calculations

Example

13.52g of Mgso4.Xh20 is heated to remove all water

6.6 of mgs04 remains calculate x

6.6 divided by mr of mgso4 = 0.0548

13.52-6.6=6.92 divide mr of h20= 0.3844 0.0548:0.3844

0.3844 divide 0.0548=7

x=7

, Limiting reagent
Example

H2so4 +2Naoh -----> na2so4 + 2H20

25cm of 1moldm-3 of h2s04

1.5g of naoh

Moles of h2so4= 25 times 1 divide by 1000= 0.025

Moles of naoh= 0.0375

0.025 needs 0.05 moles of naoh because its a 1:2

Hence naoh is limiting reagent as theres only 0.0375 moles

, Ideal gas equation
Rmrbr pv=nrt where p=pascal v=volume n=mole r=8.31 t=temp in Kelvin
Example
cac03-------> cao + co2
Calculate mass in kg of calcium carbonate that produce 998.9dm3 of co2 at 840c and 100kpa
Conversions 840c+273=1113k 100kpa=100000pa and 998.9dm3=0.9980
Moles of co2 = pv divide by rt = 100,000 times 0.9989 divide by 1113 times 8.31
= 10.8 moles 1:1 so mr of caco3=100 times by 10.8=1080 grams divide by 1000 to get kg =
1.08kg
Conversions for standard room temp
1kpa=1pa make sure pressure is always in pascal
1m3= 1,000,000 cm3 make sure volume always in metre cubed
0c=273k make sure temp is always in Kelvin for room temp its 298 which is the only diff

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