Introduction to Biochemistry A - Topic 1, Week 2, 2024-25 PDF

Introduction to Biochemistry A Topic 1 Week 2 Atomic theory continued – electrons Electron shells and valency – forming bonds and ions Electronegativity and valency The mole continued – the unit of definition for chemical amount Molarity - mass, moles...

Introduction to Biochemistry A Topic 1 Week 2 Atomic theory continued – electrons Electron shells and valency – forming bonds and ions Electronegativity and valency The mole continued – the unit of definition for chemical amount Molarity - mass, moles and concentration - practice for chemistry calculations - Standard (SI) units required for chemistry and biology Introducing Practical 1 (in week 3) Proper use of the Lab Notebook Crowe & Bradshaw Prep work assessed in practical class Chemistry for the Biosciences (4th Ed) Completion work peer assessment Tuesday Week 4 Chapters 2 & 5 Quick recap - key terms from Week 1 Atomic number states the number of protons in the nucleus of an atom - also tells us the number of electrons in an electrically neutral atom - in a neutral atom, the number of protons (+ve charge) and electrons (-ve charge) are equal The Mass number (atomic mass) is the total number of protons and neutrons in a single atom Atomic mass is a measure of the mass of a single atom relative to the mass of a single proton (1 amu) (A single neutron also weighs 1 amu) The mass of any molecule is the sum of all atomic mass values of all the atoms in the molecule  Quick recap - key terms from Week 1 Isotopes of an element have different numbers of neutrons in their nucleus Relative atomic mass = the ratio of the average mass of all atoms of an element - relative to 1/12th of the mass of a carbon atom - also referred to as atomic weight - calculated from the atomic mass and abundance of each isotope Ar = [(amu isotope 1) x (% abundance isotope 1)] + [(amu isotope 2) x (% abundance isotope 2)] etc. (amu Carbon-12) / 12 or Ar = [(amu isotope 1) x (% abundance isotope 1)] + [(amu isotope 2) x (% abundance isotope 2)] etc. 1 Quick recap - key terms from Week 1 Avogadro’s constant states the number of atoms in 1 mole of an element (6.023 x 1023 ) or the number of molecules in 1 mole of a compound (6.023 x 1023 ) 1 mole is the amount of an element that has the same numerical value in weight (g) as its atomic mass or the amount of a molecule that has the same numerical value in weight (g) as its molecular mass 1 mole of the element magnesium weighs 24.3 g and contains 6.023 x 1023 atoms 1 mole of the molecule carbon dioxide weighs 44.0 g and contains 6.023 x 1023 molecules Core (inner) and Valence (outer) electrons 1. Core electrons (the inner electrons) These electrons are not used in forming chemical bonds with other atoms Crowe & Bradshaw Chemistry for the Biosciences (4th Ed) 2. Valence electrons (electrons) Chapter 2 Valence electrons are always – and only – in the outermost electron shell Some or all valence electrons can be involved in forming chemical bonds with other atoms We need to think a little bit more about electron shells (and orbitals) Erwin Schrödinger proposed that Niels Bohr proposed that electrons electrons move continuously each orbit the nucleus at a particular around a nucleus in specific distance, implying that they occupy volumes of space - orbitals - but fixed shells. without following a fixed path. This is reflected in the Molecular Orbital theory This is reflected in the Valence Bonding theory  Each of the electron shells contains 1 or more electron orbitals Schrödinger’s Molecular Orbital The electron shells of Bohr’s theory proposes that electrons can Valence Bonding theory contain move in four different types of orbital different combinations of these orbitals They are called the s, p, d & f orbitals – each has a distinctive shape in space The combination of orbitals defines around the nucleus the maximum number of electrons each shell can hold s orbitals – 1 per shell The number of VALENCE electrons in the outermost shell defines how an atom will react p orbitals – 3 per shell d orbitals – 5 per shell orbitals – 7 per shell  How many orbitals are found in each shell ? Orbitals are named s, p, d, f – as atoms get larger, they fill up in this order p-orbitals only accept electrons when the s-orbital is already full d-orbitals only accept electrons when the p-orbitals are already full Each orbital can hold up to a maximum of 2 electrons Each electron shell contains specific orbitals Each shell therefore has a defined maximum electron capacity = this differs because shells have different numbers of orbitals Each electron shell is most stable when full = when all its orbitals have 2 electrons Shell 2 only accepts electrons when shell 1 is already full Shell 3 only accepts electrons when shells 1 & 2 are already full Shell 4 and beyond get a bit more complicated ….. Electron shell 1 contains 1 orbital s-orbitals are spherical in shape They are named according to the electron shell they occupy Shell 1 = has 1 orbital 1-s; 2-s; 3-s; etc. = it is an s-orbital In shell 1, the s-orbital is referred to as the 1-s orbital orbital = maximum 2 electrons Shell 1 = maximum 2 electrons All elements have electrons in Shell 1 The two smallest elements ONLY have electrons in Shell 1  We can link the electron shells with the Periodic Table…. Elements in Period 1 have electrons that occupy Shell 1 only Shell 1 holds the valence electrons of these elements Hydrogen has 1 valence electron in the 1-s orbital Helium has 2 electrons in the 1-s orbital - the 1-s orbital of Helium is full - Shell 1 is therefore full 2 Helium is more stable than Hydrogen, because its valence electron Shell is full Helium VALENCY - atoms can gain or lose valence electrons to become more stable Atoms tend to react with each other to achieve a full outer electron shell The VALENCY of an atom indicates the maximum number of electrons an atom needs to lose or gain, to reach a full outer electron shell Hydrogen cation H+ (0 electrons = no orbital) Hydrogen atom H Hydride anion H- (2 electrons = full 1-s orbital) Hydrogen has a valency of 1 and can either gain or lose 1 electron to have a full outer shell When Hydrogen gains or loses 1 electron in a reaction:  this leaves a stable charged ION = unbalanced numbers of protons and electrons  VALENCY - atoms can also share valence electrons to become more stable We can also define the VALENCY of an atom as the maximum number of hydrogen atoms the atom could combine with to form a stable molecule Hydrogen atom H Hydrogen molecule H2 Hydrogen has a valency of 1 and combine with 1 hydrogen atom to form a stable molecule When 2 Hydrogen atoms combine, they share 2 electrons:  this produces a stable uncharged molecule with balanced numbers of protons and electrons  The outer shell of both atoms (shell 1) has a share of 2 electrons = full  These are not ions – the number of protons and electrons for each atom in the molecule What about helium? Noble (inert) gases have a VALENCY of zero Atoms with a full outer electron shell are not reactive The VALENCY of an atom with a full outer electron shell = 0 Helium atom He All elements in the last Group of the Periodic table (Group 18) have full outer shells and valency = 0 Shell 1 is full because the 1-s orbital can hold a maximum of 2 electrons Electron shell 2 contains 4 orbitals Higher level Shells also contain a spherical s-orbital PLUS other orbitals 2-s orbital Shell 2 = has 4 orbitals 2-p orbitals = ONE s-orbital + THREE p-orbitals = maximum 8 electrons Shell 2 = maximum 8 electrons p-orbitals are lobed in shape Remember all elements with electrons in shell 2 already have a full shell 1 = ONE 1-s orbital (2 electrons) – now behave as core electrons + ONE 2-s orbital (2 electrons) these are now the + THREE 2-p orbitals (3 x 2 electrons valence electrons Elements in Period 2 have electrons that occupy Shell 1 and Shell 2 Shell 1 holds 2 core electrons for all these elements (1-s orbital) Shell 2 holds the valence electrons of these elements Lithium has 2 core electrons (in the 1-s orbital) and 1 valence electron (in the 2-s orbital) Beryllium has 2 core electrons (in the 1-s orbital) and 2 valence electrons (in the 2-s orbital) The 2-s orbital is full, but Shell 2 is not full Electrons occupy the lowest available orbital and shell As the number of electrons increases (with atomic number), they occupy the shells in order Shell 1 is occupied first >> The electrons of Hydrogen & Helium are in the 1-s orbital Shell 2 is occupied next, starting with the 2-s orbital >> Two of the electrons of Lithium are in the 1-s orbital and one is in the 2-s orbital The p-orbitals of Shell 2 (2-p) are occupied next, but only when both the 1-s and 2-s orbitals are full >> Carbon has atomic number 6 (6 proton, 6 electrons) Elements in Period 2 have electrons that occupy Shell 1 and Shell 2 Shell 1 holds 2 core electrons for all these elements (1-s orbital) Shell 2 holds the valence electrons of these elements Boron has 2 core electrons (in the 1-s orbital) and 2 valence electron (in the 2-s orbital) PLUS 1 valence electron (in a 2-p orbital) Carbon has 2 core electrons (in the 1-s orbital) and 2 valence electrons (in the 2-s orbital) PLUS 2 valence electrons (in a 2-p orbital) VALENCY rises then falls across Period 2 Atoms of elements with fewer shell 2 electrons are more likely to lose these electrons and shed the outermost shell: Valency 1 2 3 4 3 2 1 0  this leaves a positively charged ION with more protons than electrons (cation)  the loss leaves a full inner core shell as the outermost electron shell – these still behave as core electrons  the ion is more stable that the atom, because it now has a full outer electron shell Lithium can shed 1 valence electron (from the 2-s orbital) to become a Lithium ion Li+ Beryllium can shed 2 valence electrons (from the 2-s orbital) to become a beryllium ion Be2+ Boron can shed 3 valence electrons (from the 2-s and 2-p orbitals) to become a boron ion B3+ Lithium atom Li Beryllium atom Boron atom Valency = 1 Valency = 2 Valency = 3 Lithium ion Li+ Beryllium ion Be2+ Boron ion B3+ VALENCY rises then falls across Period 2 Atoms of elements with nearly full valence shells are more likely to gain electrons to fill their outermost shell: Valency 1 2 3 4 3 2 1 0  this leaves a negatively charged ION with fewer protons than electrons (anion)  the gain of electrons allows the ion to have a full outermost electron shell  the ion is more stable that the atom, because it now has a full outer electron shell  Fluorine can gain 1 valence electron to become a Fluoride ion F- Oxygen can gain 2 valence electrons to become an Oxide ion O2- Nitrogen can gain 3 valence electrons to become a Nitride ion N3- Fluorine atom F Oxygen atom Nitrogen atom Valency = 1 Valency = 2 Valency = 3 F N O O Fluoride ion F- Oxide ion O2- Nitride ion N3-  The Valency of atoms suggests how they bond together Valency = how many electrons need to be gained or lost to become stable charged ions with full valence electron shells Both elements have a Valency of 1 – their atoms would most likely react in a 1-to-1 ratio Lithium gives up its valence electron to fluorine Lithium atom Li Fluorine atom F Valency = 1 Valency = 1 Li+ F– Lithium ion Li+ Fluoride ion F F- F Stable ions, full valence shells, attracted by resulting ionic charge (ionic bond)  The Valency of atoms suggests how they bond together Valency = how many electrons need to be gained or lost to become stable charged ions with full valence electron shells The Valency of these elements suggests their atoms would react in a 2-to-1 ratio 2 Lithium atoms give up their valence electrons to 1 Lithium atom Li oxygen atom Oxygen atom (Li+)2 O2- Valency = 1 Valency = 2 Lithium ion Li+ Oxide ion O O2- O Stable ions, full valence shells, attracted by resulting ionic charge (ionic bond) Try predicting how these pairs of elements might combine …. Valency 1 2 3 4 3 2 1 0 Elements Compound name Molecular formula Beryllium and oxygen Lithium and nitrogen Beryllium and fluorine Boron and nitrogen Beryllium and nitrogen Why do some atoms lose electrons more easily than others? Across each Period …… Atomic number rises, increasing the nuclear charge Valence electrons are all in the same shell The increased nuclear attraction pulls the valence electron shell closer = atomic radius (size) DECREASES across a Period Down each Group…. Atomic number rises, increasing nuclear charge Valence electron shells are further from the nucleus More core electron shells shield the nuclear charge Decreased nuclear attraction = less pull on the valence electron shell = atomic radius INCREASES down a GROUP Why do some atoms lose electrons more easily than others? The combination of atomic number (nuclear charge) and atomic radius (distance of valence electrons) impacts ELECTRONEGTIVITY The tendency of an atom to attract the electrons, when it reacts with other elements to form a chemical bond Increasing electronegativity means atoms are more likely to capture extra electrons to fill their outer shell – they are more likely to become anions Decreasing electronegativity means atoms are more like to release electrons from their valence shell – they are more likely to become cations. Electronegativity increases across Periods & decreases down Groups Across a Period …. Rising atomic number increases the nuclear charge = greater attraction for electrons = more electronegative Down a Group… Rising atomic number increases the nuclear charge BUT Rising nuclear shielding = weaker attraction for electrons = less electronegative Electronegativity INCREASES across a Period, but DECREASES down a group  Fluorine is the most electronegati ve element Caesium is the least electronegative element (of those with confirmed values) What about Carbon? Does carbon tend to lose or gain electrons….? Remember - we can also define the VALENCY of an atom as the maximum number of hydrogen atoms the atom could combine with to form a stable molecule 1 Carbon atom C 4 Hydrogen atoms H Methane molecule CH4 Carbon has a valency of 4 and combines with 4 hydrogen atom to form a stable molecule When Carbon combines with 4 hydrogen atoms, they share 8 electrons:  this produces a stable uncharged molecule with balanced numbers of protons and electrons  The outer shell of both atoms (shell 2) has a share of 8 electrons = full  These are not ions – the number of protons and electrons is balanced for each atom Electron shell 3 contains 9 orbitals For elements that have electrons in Shell 3: Shell 1 and Shell 2 are already full, and hold 10 core electrons for these elements 3-s orbital Shell 3 = One s-orbital = maximum 2 electrons 3-p orbitals PLUS Three p-orbitals = maximum 6 electrons PLUS Five d-orbitals = maximum 10 electrons Shell 3 = maximum 18 electrons 3-d orbitals Elements in Period 3 have electrons that occupy Shell 1 and Shell 2 and Shell 3 Valency 1 2 3 4 3 2 1 0 Shell 1 holds 2 core electrons for all these elements Shell 2 holds 8 core electrons for all these elements Shell 3 holds the valence electrons of these elements Sodium and Magnesium have valence electrons in the 3-s orbital From Aluminium to Argon, the elements also have valence electrons in the 3-p orbitals Valency values mirror those seen in elements of Period 2 Sodium can shed 1 valence electron (from the 3-s orbital) to become a sodium ion Na+ Magnesium can shed 2 valence electrons (from the 3-s orbital) to become a magnesium ion Mg2+ Aluminium can shed 3 valence electrons (from the 3-s and 3-p orbitals) to become an aluminium ion Al3+ Sodium atom Na Magnesium atom Aluminium atom Valency = 1 Valency = 2 Valency = 3 Na O Mg O Al O Sodium ion Na+ Magnesium ion Mg2+ Aluminium ion Al3+  Chlorine can gain 1 valence electron to become a chloride ion Cl - Sulfur can gain 2 valence electrons to become a sulfide ion S 2- Phosphorus can gain 3 valence electrons to become a phosphide ion P 3- Chlorine atom F Sulfur atom Phosphorus atom Valency = 1 Valency = 2 Valency = 3 Cl S P chloride phosphide sulfide ion P 3- ion Cl - ion S 2- The d-orbitals of electron Shell 3 are only filled by elements in Period 4 Elements in Period 3 have valence electrons that occupy ONLY the 3-s and 3-p orbitals under most physiological conditions Elements in Period 4 have valence electrons that occupy the 3-d orbital Elements in Period 4 also have electrons in Shell 4 (in one 4-s and three 4-p orbitals) Summary? Learn these valency shell & electron numbers Valency 1 2 3 4 3 2 Outer Maximum 1 0 Electron valence shell electrons Shell 1 2 Shell 2 8 Shell 3 8 Shell 4 18 Elements in Groups 1 to 13 lose valence electrons more easily than gaining them Elements in Groups 15 to 17 gain electrons more easily than losing them Elements in Group 14 are Elements in Group 18 more likely to share electrons have full electron shells & than gain or lose them are unreactive Linking stoichiometry with chemical amount and concentration Quantitative mass relationships – how elements react together Magnesium and oxygen Calcium and chlorine Valency Valency Valency =2 =2 Valency =1 =2 Crowe & Bradshaw Chemistry for the Mg O Biosciences (4th Ed) O Cl Ca Cl Chapter 5 Magnesium oxide Mg2+O2- Calcium chloride Cl2+(Cl –)2 Stoichiometry – how do we indicate the ratio of atoms within molecules? Mg O O Cl Ca Cl Magnesium oxide Mg2+O2- Calcium chloride Ca2+(Cl–)2 Brackets allow the formula to show the individual ion charges No number needed Subscript number used when when only 1 atom in the 1 or more atoms are present in formula = (1-to-1 ratio) a greater than 1-to-1 ratio Mg2+O2- + Ca2+(Cl–)2 Ca2+O2- + Mg2+(Cl–)2 Superscript number used to indicate ion charge Stoichiometry – brackets are also used for multi-atom ions or groups Mg O O Ca Magnesium oxide Mg2+O2- Calcium bicarbonate Ca2+(HCO3–)2 Supercript number used to indicate the ion charge Brackets also allow the formula to show whole ions or functional groups present in the molecule Subscript number AND brackets used when whole functional groups or ions are present in a greater than 1-to-1 ratio Mg O 2+ 2- + Ca (HCO ) 2+ – 3 2 Ca2+O2- + Mg2+(HCO3–)2 Stoichiometry – is also needed to indicate the ratio of molecules in reactions + H2 + O2 H2O Reactions must be balanced on both sides, because of the law of conservation of mass = the total mass of the reactants equals the total mass of the products = the total number of atoms must be the same before and after the reaction = the amount of reactants and of products is expressed as a ratio of whole numbers (moles) Does the reaction above balance all atoms in all reactants and products….? A COEFFICIENT value must be added to indicate the ratio of molecules in the reaction. Can you determine what components need balancing, and what number to apply? Coefficient values indicate the molar amounts of each molecule in the reaction We introduced the mole – last week to quantify the amount of an atom or molecule = link numbers of molecules to a measurable mass unit, the gram. 1 mole = 6.023 x 1023 atoms or molecules or particles Add the right coefficients to the reactions below, to conform to the law of conservation of mass Methane oxidation CH4 + O2 CO2 + H2O Glucose oxidation (Cellular respiration) C6H12O6 + O2 CO2 + H2O Urea hydrolysis NH2-CO-NH2 + H2O NH3 + CO2 Last week you calculated relative molecular mass, molar mass and numbers of atoms Glucose C6H12O6 Relative molecular mass = sum (Ar for all atoms in molecule) = (12 x 6) + (1 x 12) + (16 x 6) = (72) + (12) + (96) = 180 Molar mass (g mol-1) has a value equal to the Relative atomic (or molecular) mass = (12 x 6) + (1 x 12) + (16 x 6) = (72) + (12) + (96) = 180 g mol-1 1 mol of glucose = 6.023 x 1023 molecules Each molecule of glucose has 24 atoms 1 mole of glucose = 24 x (6.023 x 1023) atoms = 1.44 x 1025 atoms The number of atoms and molecules inevitably changes if the amount (of moles) changes HOW MANY ATOMS IN 0.5 mol H2O 1 mol H2O = 6.023 x 1023 molecules 0.5 mol H2O = (0.5) x (6.023 x 1023 molecules) = 3.0115 x 1023 molecules 1 molecule = H2O = 3 atoms 0.5 mol H2O = (3) x (3.0115 x 1023) atoms = 9.0345 x 1023 atoms By rearranging the formula, we can calculate the reverse = the amount (moles) of a molecule, from a known number of either atoms or molecules 3.01 x 1023 molecules H2O 6.023 x 1023 molecules = 1 mol H2O 3.01 x 1023 molecules = (3.01 x 1023 ) / (6.023 x 1023 ) = 0.4997 mol H2O 3.01 x 1023 atoms H2O 1 molecule H2O = 3 atoms 3.01 x 1023 atoms H2O = (3.01 x 1023) / 3 molecules H2O = 1.003 x 1023 molecules H2O 1.003 x 1023 molecules = (1.003 x 1023) / (6.023 x 1023) = 0.1665 mol H2O It’s important to link chemical amount (moles) with mass (grams) and molar mass X ÷ Molar mass (g = mass mol-1) Amount (g) (mol Amount mass We can always (mol) = Molar (g) mass (g calculate molar mol-1) mass, using the Periodic Table mass =Amount (mol)Molar mass (g (g) x mol-1) By understanding the relationship, we can calculate the amount (mol) from a known HOW MANY MOLES IN 700 g nitrous oxide N2O mass (g) of atoms or molecules Molar mass (g mol-1) N2O == [ (2 x 14 ) + (16) ] = 44 g mol-1 Amount (mol) = 700 g N2O = 15.91 mol 44 g mol-1 By understanding the relationship, we can calculate the mass (g) of a given amount (moles) of an element or WHAT IS THE MASS OF 1.5 MOLES of water H2O molecule Molar mass (g mol-1) H2O == (2 + 16) mass (g) = Amount (mol) x Molar mass (g mol ) -1 = 18 g mol-1 mass H2O (g) = amount H2O (mol) x Molar mass H2O (g mol-1) = 1.5 mol x 18 g mol-1 = 27 g Finally today, learn and use these rearrangements of the mathematical formula linking amount, volume and concentration amount of solute (mol) concentration (mol L ) = -1 volume of solution (L) amount of solute (mol) volume of solution (L) = concentration (mol L-1) amount of solute (mol) = concentration (mol L-1) x volume of solution (L) Molecules in cells are in solution at carefully controlled concentrations CONCENTRATION = The amount of a substance in a particular volume​ Groceries 6 eggs per box or 6 eggs / box or 6 eggs box-1 Smoke pollution 1970 parts per million or 1970 ppm or 0.0197 % Sugar syrup 35 g in 100 mL or 350 g L-1 For chemistry and biology, we express concentration as moles per litre = mol / L or = mol L-1 or = M (Molar) This is also known as the molarity of a solution Example: A healthy blood sugar concentration is below 0.006 mol L-1 or below 0.006 M Note – some diagnostic measurements are still expressed as amount per dm3 (= 1L) For this module we will use the Litre as the standard measurement of volume A word about numbers and their units ….. We must use standardised units when expressing measurable quantities in experimental chemistry and biology International System of Units SI units (Système International) Concentration molarity mol L -1 Most SI scales have the same prefix for values at the same magnitude For example: Weight = milligrams (mg) Length = millimetres (mm) Volume = millilitres (mL) Amount = millimoles (mmol) Concentration = millimolar (mM) 0.9 g of lithium chloride is dissolved in 0.5 L water - calculate the concentration (mol L-1) of the solution. Step 1 – calculate the amount (mol) within the known mass (g) Amount (mol) = mass (g) 0.9 g LiCl = = 0.021 mol Molar mass (g mol-1) 42.4 g mol-1 Step 2 – calculate the relationship between amount (mol) and volume (L) amount of solute (mol) concentration (mol L-1) = volume of solution (L) 0.021 mol = = 0.042 mol L-1 0.5 L You have 0.5 kg of sodium chloride and want a solution of 15 mol L-1. Calculate the volume you need to dissolve the salt in. Step 1 – calculate the amount (mol) within the known mass (g) mass (g) 500 g NaCl Amount (mol) = = = 8.547 mol Molar mass (g mol-1) 58.5 g mol-1 Step 2 – calculate the relationship between amount (mol) and volume (L) amount of solute (mol) volume of solution (L) = concentration (mol L-1) 8.547 mol = = 0.5698 L = 569.8 mL 15 mol L -1 What mass of glucose (C6H12O6) must be dissolved in 2 L of water, to produce a concentration of 0.25 M? The answer should be in grams (g) Step 1 – calculate the amount (mol) needed in the given volume, for the required concentration amount of solute (mol) = concentration (mol L-1) x volume of solution (L) = (0.25 mol L-1) x ( 2 L ) = 0.5 mol Step 2 – calculate the mass (g) needed to give this amount mass (g) = Amount (mol) x Molar mass (g mol-1) = (0.5 mol) x (180 g mol-1) = 90 g What mass of hydrogen cyanide (HCN) must be dissolved in 2 mL of water, to produce a concentration of 0.75 M? The answer should be in grams (g) Maths support, workshops and tutorials are run by the Centre for Academic Development  Introduction to Biochemistry A Summary of Week 2 We have covered several concepts : You can now :  Electron shells, valency and  Explain how electrons are positioned around the atomic nucleus. electronegativity  Describe the difference between core and valence electrons.  Explain why atoms tend to give up or gain valence electrons to form ions.  Predict how some atoms might interact with each other  Indicate how carbon is different, and shares electrons within molecules.  Relating moles and mass  Explain the relationship between moles, atom mass and number  Calculate these concepts using simple formula and the Periodic Table  Learn the SI units relevant for the various concepts  Molarity (concentration)  Interrelate amount, mass, volume and concentration  Calculate these concepts using simple formula and the Periodic Table Use the recommended module textbook to add to your weekly lecture notes The bulk of this week’s lecture content is referred to in Chapters 2 and 5 BUT Use the index to find key terms from the lecture content and expand your understanding. Make revision-ready notes, so that your test performance doesn’t just rely on these slides Do it THIS WEEK – in good time to prepare for your lectures next Monday Crowe & Bradshaw Chemistry for the Biosciences (4th Ed) Introduction to Biochemistry A Topic 1 Week 2 You can now also answer the second Numerical Quiz questions on Moodle - - - THIS QUIZ CONTRIBUTES TO YOUR MODULE MARK - - - There will be 9 questions this week, based on concepts we have explored last week and today The questions will be posted (in a MS-Word document) on Moodle, at 2pm TODAY Calculate your answers, and use the Quiz link to submit them The quiz link will open on Moodle, at 2pm TODAY - it will close on FRIDAY at 11.59pm Remember: 1. You can have 2 attempts at the quiz, and you will get a hint after the first attempt 2. You must answer ALL QUESTIONS again on your 2nd attempt Let’s look ahead to your Practical …