# Chemistry Unit 3 Equations

All equations and 2 example questions, bond energy example question not included due to them being large and requiring alot of extra information.

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- Created on: 15-05-13 14:28

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- Chemistry Unit 3 Equations
- Specific Heat Capacity
- How much energy is needed to increase the temperature of 500 g of lead
from 20ºC to 45ºC? The specific heat capacity of lead is 128 J/kg ºC.
- mass of lead = 500/1000 = 0.5kg temperature change = 45-20 = 25c energy needed = 0.5 x 128 x 25 = 1600j or 1.6kj

- E =m x c x t
- m = mass
- c = specific heat capacity
- t = temperature change

- How much energy is needed to increase the temperature of 500 g of lead
from 20ºC to 45ºC? The specific heat capacity of lead is 128 J/kg ºC.
- Bond Energy Calculations
- You will be given bond energies for example h-h = 436kj per mole.
- Energychange = sum of bonds broken - sum of bonds made

- You will be given bond energies for example h-h = 436kj per mole.
- Titration Equation
- M1 x V1 divided by N1 = M2 x V2 divided by N2
- E.G. 25.00cm of an unknown strength HCL reacted with 38.90cm of 0.100 mole per cm solution of Na OH calculate the concentration of HCL?
- HCL + NaOH = NaCL +H2O
- So where m1 is unknown v1 = 25.00cm n1 = 1 m2 = 0.100 moles per cm v2 =38.9cm and n2 = 1
- M1 = M2 x V2 x N1 divided by N2 x V1
- M1 = 0.100 x 38.90 x 1 divided by 1 x 25.00

- M1 = M2 x V2 x N1 divided by N2 x V1

- So where m1 is unknown v1 = 25.00cm n1 = 1 m2 = 0.100 moles per cm v2 =38.9cm and n2 = 1

- HCL + NaOH = NaCL +H2O

- E.G. 25.00cm of an unknown strength HCL reacted with 38.90cm of 0.100 mole per cm solution of Na OH calculate the concentration of HCL?

- M1 x V1 divided by N1 = M2 x V2 divided by N2

- Specific Heat Capacity
- How much energy is needed to increase the temperature of 500 g of lead
from 20ºC to 45ºC? The specific heat capacity of lead is 128 J/kg ºC.
- mass of lead = 500/1000 = 0.5kg temperature change = 45-20 = 25c energy needed = 0.5 x 128 x 25 = 1600j or 1.6kj

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