Model Answers
1. Atomic and electronic structure
Atoms consist of a central nucleus which contain positively charged protons and neutral neutrons (nucleons) and electrons which orbit the nucleus in quantised energy levels called shells. A maximum of two electrons can fit in the first shell, 8 in the second and at GCSE, 8 in the third. Electrons are filled sequentially from the inner to the outermost shell, and each shell must be full before moving to the next shell. For example, oxygen which contains 8 electrons had the electronic configuration 2, 6, (and not 1, 7 or 2, 4, 2). Furthermore, the period number the element is in indicates the number of shells and the group number indicates the number of electrons in the outer shell of an atom of that element. The number of neutrons may be calculated by subtracting the atomic number from the mass number for that element.
This model changes at AS level and the maximum number of electrons in each energy level becomes 2, 8, 18. Electrons are found in orbitals (probability density functions), with a maximum of two electrons (one spin up, one spin down) in each orbital. Each energy level (previously called 'shell' at GCSE) is split into sub-shells (s, p, d and f) which correspond with the different blocks of the periodic table.
Atoms consist of a central nucleus which contain positively charged protons and neutral neutrons (nucleons) and electrons which orbit the nucleus in quantised energy levels called shells. A maximum of two electrons can fit in the first shell, 8 in the second and at GCSE, 8 in the third. Electrons are filled sequentially from the inner to the outermost shell, and each shell must be full before moving to the next shell. For example, oxygen which contains 8 electrons had the electronic configuration 2, 6, (and not 1, 7 or 2, 4, 2). Furthermore, the period number the element is in indicates the number of shells and the group number indicates the number of electrons in the outer shell of an atom of that element. The number of neutrons may be calculated by subtracting the atomic number from the mass number for that element.
This model changes at AS level and the maximum number of electrons in each energy level becomes 2, 8, 18. Electrons are found in orbitals (probability density functions), with a maximum of two electrons (one spin up, one spin down) in each orbital. Each energy level (previously called 'shell' at GCSE) is split into sub-shells (s, p, d and f) which correspond with the different blocks of the periodic table.
2. Chemical calculations
Answers to .pdf file
Question 1: As the worked example given (38.6%)
Question 2:
a. 26.2% b. 35.0 c. 21.2%
Question 3:
1) NH3 2) 14 3) 18 4) 77.8%
Question 4:
1) HNO3 2) 14 3) 63 4) 22.2%
You might like to access the videos when checking your blog posts:
Empirical Formula: http://my-gcsescience.com/revision/additional/the-empirical-formula
1. Write down the elements involved
2. Underneath, write the % or mass (g) of each given in the question
3. Divide each one by its relative atomic mass
4. Now you have a ratio, so simplify by dividing by the smallest number
5. Convert the ratio to a formula e.g. a C : H : O ratio of 2 : 4 : 1 would produce C2H4O
Relative Formula Mass: http://my-gcsescience.com/revision/additional/calculate-relative-formula-mass
1. For each element in the formula, multiply the number of times that atom appears by its relative atomic mass.
2. Then sum all these numbers.
% Yield: http://my-gcsescience.com/revision/additional/percentage-yield-of-a-product
1. Calculate the theoretical yield first:
a) Write the full balanced chemical equation
b) If you are given the mass of one of the reactants, calculate the number of moles by doing (mass / relative formula mass of reactant)
c) Use the molar ratio to deduce the number of moles of the desired product
d) Calculate the theoretical yield of the product by doing (moles x relative formula mass of product)
2. The actual yield will be given in the question so, % yield = (actual yield / theoretical yield) x 100
Answers to .pdf file
Question 1: As the worked example given (38.6%)
Question 2:
a. 26.2% b. 35.0 c. 21.2%
Question 3:
1) NH3 2) 14 3) 18 4) 77.8%
Question 4:
1) HNO3 2) 14 3) 63 4) 22.2%
You might like to access the videos when checking your blog posts:
Empirical Formula: http://my-gcsescience.com/revision/additional/the-empirical-formula
1. Write down the elements involved
2. Underneath, write the % or mass (g) of each given in the question
3. Divide each one by its relative atomic mass
4. Now you have a ratio, so simplify by dividing by the smallest number
5. Convert the ratio to a formula e.g. a C : H : O ratio of 2 : 4 : 1 would produce C2H4O
Relative Formula Mass: http://my-gcsescience.com/revision/additional/calculate-relative-formula-mass
1. For each element in the formula, multiply the number of times that atom appears by its relative atomic mass.
2. Then sum all these numbers.
% Yield: http://my-gcsescience.com/revision/additional/percentage-yield-of-a-product
1. Calculate the theoretical yield first:
a) Write the full balanced chemical equation
b) If you are given the mass of one of the reactants, calculate the number of moles by doing (mass / relative formula mass of reactant)
c) Use the molar ratio to deduce the number of moles of the desired product
d) Calculate the theoretical yield of the product by doing (moles x relative formula mass of product)
2. The actual yield will be given in the question so, % yield = (actual yield / theoretical yield) x 100
3. Structure and bonding
Answers to Intermolecular Forces .pdf
1. Enthalpy change of vaporisation.
2. More electrons on one side of an atom/molecule than the other forming an instantaneous dipole. This causes an induced dipole in a neighbouring atom/molecule, which is hence attracted.
3. The jet of water will always be attracted to the nylon rod… it will not be repelled. This is because if the positively charged rod can attract the oxygen (delta negative side of dipole) a negatively charged rod will also attract the water as the molecules will just rearrange themselves and the hydrogen side (delta positive side of dipole) will be attracted.
4. Any of the following:
Water has a much higher enthalpy change of vaporisation than expected of group 6 hydrides.
The boiling point of water is much higher than predicted by the trend in boiling points for the hydrides of other group 6 elements.
Water has a very high surface tension and a high viscosity.
The density of ice is less than the density of water.
These are due to the strong hydrogen bonds between molecules.
Answers to Bonding Quiz .pdf
Answers to Intermolecular Forces .pdf
1. Enthalpy change of vaporisation.
2. More electrons on one side of an atom/molecule than the other forming an instantaneous dipole. This causes an induced dipole in a neighbouring atom/molecule, which is hence attracted.
3. The jet of water will always be attracted to the nylon rod… it will not be repelled. This is because if the positively charged rod can attract the oxygen (delta negative side of dipole) a negatively charged rod will also attract the water as the molecules will just rearrange themselves and the hydrogen side (delta positive side of dipole) will be attracted.
4. Any of the following:
Water has a much higher enthalpy change of vaporisation than expected of group 6 hydrides.
The boiling point of water is much higher than predicted by the trend in boiling points for the hydrides of other group 6 elements.
Water has a very high surface tension and a high viscosity.
The density of ice is less than the density of water.
These are due to the strong hydrogen bonds between molecules.
Answers to Bonding Quiz .pdf
- False - molecules
- Some (molecular ions are possible)
- False - ion
- False - stable
- False - charge
- False - bond
- d
- a
- a
- d
- d
- d
- c
- c
- c
- a
- SCl6 covalent
- CaO ionic
- MgI2 ionic
- LiCl ionic
- CO or CO2 covalent
- N2O5 covalent
- SiO2 covalent
- sodium sulfide ionic
- nitrogen oxide covalent
- disulfur decafluoride covalent
- magnesium fluoride ionic
- aluminium sulfide ionic
- carbon disulfide covalent
- phosphorus bromide covalent
4. Summary
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