Unit 1 - Chemical Changes and Structure

Unit 1 - Chemical Changes and Structure
This unit studies how chemist can control the rate of chemical reactions, and the enthalpy changes that take place. It studies trends in the periodic table and investigates the relationship between the arrangement of elements in the periodic table and their bonding, structure and properties. The unit also considers polar covalent bonds in the context of the bonding continuum, followed before studying intermolecular forces.

Chemical changes and Structure
From previous work you should know and understand the following: Collision theory Atomic structure Electron orbital's or energy levels Valency Covalent and ionic bonding Physical properties of metals. 2

1.1 Controlling the Rate Learning Intentions
Learn how chemists control reaction rates by careful consideration of the influence of concentration, pressure, temperature, surface area and collision geometry. I can use Collision Theory to explain the effects of concentration and collision geometry on reaction rates. I can use Collision Theory to explain the effects of particle size and collision geometry on reaction rates. I can use Collision Theory to explain the effects of temperature, and collision geometry on reaction rates. I can use Collision Theory to explain the effects of a catalyst on reaction rates. I can calculate the average rate of reaction from graphs of changing quantities versus time I can state Rate = 1/ t (s-1) I can give the definition of an activated complex. I can give the definition of activation energy. I can use potential energy diagrams to:- say if a reaction is exothermic or endothermic. calculate the enthalpy change for a reaction with the correct units and sign. calculate the activation energy needed for forward, reverse and catalysed reactions. I can explain why it is essential that chemists can predict the quantity of heat taken in or given out in an industrial process. I can give the definition of temperature. I can use energy distribution diagrams to explain the effect of temperature on reaction rate. I can state how catalysts affect the activation energy for a chemical reaction. I can draw a potential energy diagrams to show the effect of a catalyst on the activation energy.

Section 1 - Controlling Rates of Reaction
Reactions happen at different rates. Industry needs to control reaction rates to increase production and get a good return for the investment Rates may need to be controlled for safety, or to keep the rate of production within the limit of the plant 4

Collision Theory For a chemical reaction to occur, reactant molecules
must collide. The collision must provide enough energy to break the bonds in the reactant molecules Then new chemical bonds form to make product molecules. 5 5

Factors Affecting Reaction Rate
Particle Size The smaller the particle size, the greater the surface area, and the faster the reaction due to more chance of successful collisions. Adding a Catalyst Speeds up a chemical reaction by lowering the activation energy. Concentration The higher the concentration of the reactants, the more particles there are to collide so more chance of successful collisions and a faster reaction. Temperature The higher the temperature, the more energy the particles have so more chance of successful collisions and a faster reaction. These observations are explained by collision theory.

Progress of a Reaction C. Why does the graph curve?
Reactions can be followed by measuring changes in concentration, mass and volume of reactants and products. A. Where is the reaction the quickest? B. Why does the graph level off? Rate No more products formed. C. Why does the graph curve? A The concentration of the reactants decrease with time. time 7 7

Rates of reaction The rate of reaction can be followed by measuring changes in Mass Volume of gas produced Concentration 8

Measuring Reaction Rates
Products Change in mass (g) Reactants time (s) change in mass of product or reactant Average rate of reaction = time interval Units g s-1 9 9

Measuring reaction rates
Products Change in volume (cm3) Reactants time (s) Average rate of reaction = change in volume of product or reactant in time for the change to occur Units cm3 s-1 10 10

Measuring reaction rates
Products Change in concentration (mol l-1) Reactants Time (s) Average rate of reaction = change in concentration of product or reactant time interval Units mol l-1 s-1 11 11

Effect of surface area Particle size, the smaller the particles, the greater the surface area, the greater the chance of successful collisions. 4X4= 16 cm2 16x6=96 cm2 2x2 = 4 cm2 24X8= 192 cm2 4X6= 24 cm2 12 12

Rate and Particle Size Only the particles on the surface of a solid can be involved in a collision Crushing a solid increases the surface area more particles are available for collision therefore increased rate of the reaction Higher Chemistry Eric Alan and John Harris 13

Hydrochloric acid reacts with marble chips (calcium carbonate)
Effect of surface area Hydrochloric acid reacts with marble chips (calcium carbonate) 2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O (l) 14

How can we follow the reaction?
A gas is produced. What will happen to the gas if there is no lid on the container? What will happen to the mass? How can we follow the rate? 15

What to do You are going to follow the rate of the reaction by
Measuring the loss of mass over time Measuring the volume of gases produced over time 16

Measuring rate of reaction
Two common ways: 1) Measure how fast the reactants are used up 2) Measure how fast the products are formed 17

How can we follow the reaction?
If we use a container fitted with a delivery tube we could measure the amount of gas produced. How? 18

Exp 1.1 - What to do – Mass Loss
Weigh out 1 g marble chips into a beaker Measure 20 cm3 2 mol l-1 HCl into a conical flask Place on balance Add 2g marble chips to the beaker. Now add it to the acid and take mass readings every 60 seconds Repeat using crushed chips 19

What to do – Mass Loss Record your results in a the table.
Plot a graph of Mass of reactants vs time rate = change in mass ( the unit is g s-1) time interval Calculate the rate for the 0-60s and s for each set of results 20

Exp What to do - Volume Measure 20 cm3 of 2 mol l-1 HCl into a conical flask fitted with a stopper and a delivery tube Set up an inverted measuring cylinder of water to collect the gas Add 2g marble chips to the acid Measure the volume of gas every 60 seconds Repeat with 2g crushed marble chips 21

Exp 1.1 - What to do - Volume Record your results in a the table.
Plot a graph of volume vs time rate = change in volume ( the unit is cm3 s-1) time interval Calculate the rate for the 0-60s and s for each set of results 22

Method A results Plot the results on a graph with time on the x axis and volume on the y. Use the same set of axes for both sets of results. time (s) 10 20 30 40 50 60 70 80 volume (cm3) C (cm3) G Sample results 23

rate = change in volume = _____________ cm3 s-1
Volume of gas cm3 Time (s) Work out the rate of reaction over the first 25 seconds and the second 25 seconds using the formula rate = change in volume = _____________ cm3 s-1 time interval Rate over 1st 25 seconds (cm3 s-1) rate over 2nd 25seconds Whole chips (C) 32-0 25-0 =1.3 50-32 50-25 =0.72 Ground chips (G) 45-0 =1.8 50-45 =0.2 24

Method B results Plot the results on a graph with time on the x axis and mass on the y. Use the same set of axes for both sets of results. time (s) 20 40 60 80 100 120 140 160 Mass (g) C mass(g) G Sample results 25

rate = change in mass The answer will have the units g s-1
Loss in mass (g) Time (s) Work out the rate of reaction over the first 25 seconds and the second 25 seconds using the formula rate = change in mass The answer will have the units g s-1 time interval Rate over 1st 25 seconds (g s-1) rate over 2nd 25seconds Whole chips (C) 0.8-2 25-0 =0.05 50-25 =0.018 Ground chips (G) 0.3-2 =0.068 =1x10-3 26

Rate and Concentration
for a reaction to take place the particles must collide Increasing the concentration of a solution increases the number of particles in the same volume. Therefore more chance of collision i.e. increased rate of the reaction 27

Effect of concentration
The higher the concentration, the more particles in a given space, the more chance there is of successful collisions. 28

Effect of concentration –the chemical clock challenge
The iodine clock reaction changes from colourless to blue/black 29

You will carry out the reaction using a series of dilutions of the iodide solution. This will be diluted by replacing some of the volume with water. 30

2I- (aq) + H2O2 (aq) + 2H+ (aq)  2H2O (l) + I2 (aq) I2 (aq) + 2S2O32- (aa)  2I- (aq) + S4O62- (ag) The reaction mixture stays colourless as the iodine molecules are converted back to iodide molecules by the thiosulphate ions. Once all the thiosulphate ions have been used, a blue black colour appears suddenly as iodine reacts with starch. Relative Rate = 1 t Units s-1 t being a measure of how long it takes for the blue/black colour to form. (when excess I2 forms) 31 31

1) Using syringes measure out 10cm3 sulphuric acid 0.1moll-1 10cm3 sodium thiosulphate 0.005moll-1 1cm3 starch solution 25cm3 potassium iodide solution 0.1mol l-1 Into a dry 100cm3 beaker 2) Measure out 5cm3 of hydrogen peroxide 0.1moll-1 into a syringe. Add it to the mixture as quickly as possible and start the timer. 3) Stop the clock when the mixture suddenly turns dark blue. 4) Repeat, using 20 cm3 of potassium iodide solution and 5cm3 of water with, then using repeated dilutions 32

Volume of water (cm3) Volume of 0.5 mol l-1 KI (aq) (cm3) Time (s) Rate (1/t) O.0 25.0 5.0 20.00 10.0 15.0 20.0 33

RESULTS - Plot a graph showing the volume of potassium iodide x axis and the rate of reaction on the y axis. 34

Effect of concentration - Cola challenge
Your task is to make up a solution containing the same concentration of sugar as a can of coke which contains 35g of sucrose C12 H22 O11 35

Work out the mass of sucrose required to make up 100cm3 of sucrose solution of the same concentration as cola, assuming there are 24g in 330cm3 Make up your solution. What is the concentration of this solution? 36

Concentration (c ) is measured in moles per litre ( mol l-1) no moles = C x V 1000 To calculate the concentration you need to work out the number of mol of sugar present 37

No moles = mass (g) GFM GFM = gram formula mass 38

Sucrose is a non-reducing sugar – it does not react with Bendicts unless it is first hydrolysed. Boil 10 cm3 sugar solution with 5cm3 1mol/l HCl. Then neutralise the solution with 5cm3 1mol/l NaOH. Allow this solution to cool to room temperature. Repeat with the lemonade solution. Add 2cm3 of Benedicts to each sample. Prepare a beaker of boiling water. 39

Add the test tubes to the boiling water If you have made up the solutions correctly, all your solutions should take the same time to change colour. 40

Effect of temperature -the vanishing cross
Sodium thiosulfate solution is reacted with acid. A precipitate of sulfur forms. The time taken for a certain amount of sulfur to form is used to indicate the rate of the reaction. 41

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Results Temperature (0C) 19 32 38 51 60 Reaction time in seconds 105 46 36 18 12 1/time 0.0095 0.0217 0.0278 0.0556 0.0833 A 10 oC rise in temperature will approximately double the rate of the reaction 44

Temperature and energy
The piston can be dragged however the gas particles WILL NOT move the piston if the temperature is increased.

The experiment can be viewed at 46

Rate measurement and temperature. Oxalic acid/permanganate reaction
5(COOH)2 (aq) + 6H + (aa) + 2MnO4 2- (aq)  2Mn 2+ (aq) + 10 CO2 (aq) 8H2O (l) What colour change takes place? State the; Aim of the experiment. Method, which variables to control and change. What to measure and how. How to record your results. What graph to draw. Make a conclusion. Evaluate. 47 47

b) Reaction Profiles Learning intention
Learn how a potential energy diagram can be used to describe a reaction pathway, and to display activation energy and reaction enthalpy. The activated complex is the unstable intermediate formed at the peak of the potential energy diagram. 48

Reaction profiles – Enthalpy
Thermochemistry is the study of heat energy taken in or given out in chemical reactions. This heat, absorbed or released, can be related to the internal energy of the substances involved. Such internal energy is called ENTHALPY, symbol H. As it is only possible to measure the change in enthalpy, the symbol  H, is used.  H = Hp - Hr Enthalpy (products) – Enthalpy (reactants) Units kJ, kilojoules 49 49

Reaction profiles -Exothermic and Endothermic Reactions
Step 1: Energy must be SUPPLIED to break bonds: Step 2: Energy is RELEASED when new bonds are made: A reaction is EXOTHERMIC if more energy is RELEASED then SUPPLIED. If more energy is SUPPLIED then is RELEASED then the reaction is ENDOTHERMIC 50

Enthalpy of products Enthalpy of reactants +H -H Enthalpy of reactants Enthalpy of products Exothermic reactions give out thermal energy so the enthalpy of the products is less than that of the reactants. Endothermic reactions take in thermal energy from their surroundings. The enthaply of the products is greater than that of the reactants 51 51

Exothermic reactions give out heat, causing a rise in the temperature Endothermic reactions take in heat The energy change in a reaction can be shown in a potential energy diagram or reaction profile 52

PE kJmol-1 PE kJmol-1 -H Path of reaction Path of reaction Endothermic reactions take in thermal energy from their surroundings.  H = +ve Exothermic reactions give out thermal energy  H = -ve 53 53

A. Combustion of methane CH4 (g) + 2O2 (g) → CO2 (g) + 2 H20 (l) CH4 (g) + 2O2 (g) H reactants  H negative, exothermic reaction CO2 (g) + 2 H20 (l) kJmol-1 products B. Cracking of ethane C2H6 (g) → C2H4 (g) + H2(g) C2H4 (g) + H2(g) products C2H6 (g) H  H positive, endothermic reaction reactants kJmol-1 54 54

Mix the following pairs of chemicals in a polystyrene cup to discover if the reactions are exothermic or endothermic 55

Reaction Temp before mixing/oC Temp after mixing/oC
Endothermicor exothermic 10cm3 NaOH + 10cm3 HCl 10cm3 NaHCO3 + 4 spatulas citric acid 10cm3 CuSO4 + spatula of Zn powder 10cm3 H2SO4 + Mg ribbon 56

Use of the thermite reaction
This reaction is used to weld railway lines together 57

An endothermic reaction
Higher Chemistry Eric Alan and John Harris 58

Enthalpy changes and industrial processes
For industrial processes it is essential that chemists can predict the quantity of heat taken in or given out. Exothermic reactions lower the temperature, slowing the reaction rate Heat must be supplied to maintain the rate of reaction – this is an expense 59

Enthalpy changes and industrial processes
Exothermic processes produce heat Heat may need to be removed to prevent the reactions proceeding beyond the capacity of the plant 60

c) Temperature and kinetic energy
Learning intention Learn to describe the relationship between temperature, kinetic energy and activation energy. 61

Activation energy and reaction pathway
Potential energy diagrams give useful information about the energy profile of a reaction. The activation energy is the minimum kinetic energy required by colliding molecules for a reaction to occur. In the diagrams shown above the activation energy appears like a ‘energy barrier’ which reactants must get over to become products. Higher Chemistry Eric Alan and John Harris 62

Breaking bonds Making bonds Activation Energy EA P.E. 1 2 3 1 2 Activated complex 3 Reaction Path Activation Energy is the additional P.E. which has to be attained by colliding molecules to form an activated complex. Activated complex is the unstable arrangement of atoms formed at the maximum of the potential energy barrier. 63 63

As a reaction proceeds from reactants to products, an intermediate stage is reached at the top of the activation barrier at which a highly energetic species called an activated complex is formed. A + B → X → C+D Higher Chemistry Eric Alan and John Harris 64

This unstable activated complex only exist for a short period of time. From the peak of the energy barrier it can lose energy in one of two ways i.e. to the stable products or to form the reactants again. The higher the Ea the higher the barrier and the slower the reaction. A + B → X → C+D Higher Chemistry Eric Alan and John Harris 65

Enthalpy Enthalpy -H Endothermic reactions take in thermal energy from their surroundings.  H = +ve Exothermic reactions give out thermal energy  H = -ve 66 66

1. Mark Ea and ∆H on the PE diagrams and then calculate the value of each for the forward reaction. Ea Ea ∆H Ea ∆H ∆H C Ea = 40 KJmol-1 A Ea = 50 KJmol-1 B Ea = 30 KJmol-1 ∆H = -10 kJmol-1 ∆H = -40 kJmol-1 ∆H = +20 kJmol-1 Higher Chemistry Eric Alan and John Harris 67

2. Mark Ea and ∆H on the PE diagrams and then calculate the value of each for the reverse reaction. Ea ∆H Ea Ea ∆H ∆H C Ea = 20 KJmol-1 A Ea = 60 KJmol-1 B Ea = 70 KJmol-1 ∆H = +10 kJmol-1 ∆H = +40 kJmol-1 ∆H = -20 kJmol-1 Higher Chemistry Eric Alan and John Harris 68

Calculate Ea for the forward reaction Ea = 210 – 20 = 190kJ Higher Chemistry Eric Alan and John Harris 69

Temperature and kinetic energy
What do we mean by temperature and heat (thermal energy) ? The thermal energy of a system is a measure of both the potential and kinetic energy within the system. The temperature is a measure of how ‘hot’ a system is. The temperature is a measure of the average kinetic energy in a system.. 70 70

Temperature and energy
The piston can be dragged however the gas particles WILL NOT move the piston if the temperature is increased.

Energy Distribution No of molecules Kinetic energy Ea Number of collisions which result in new products being formed. Total number of collisions with sufficient K.E. energy is the area under the graph to the right of the Ea . 72 72

Temperature and Activation energy
Activation energy EA Increasing the temperature means a greater number of molecules have energies in excess of EA. Even a small rise in temperature causes a large increase in the number of particles with energy above EA. Therefore a greater proportion of collisions will be successful. The blue area is larger than the grey so at temperature T2 more particles have enough energy to cause successful collisions. At the higher temperature T2 the number of particles with energy in excess of EA is greater. The blue area now represents particles with energy in excess of EA. T1 The grey area represents the number of particles with energy in excess of the EA. These particles have enough energy to cause successful collisions T2 No of molecules Energy associated with molecules 73

Energy distribution Ea Increased temperature Increased concentration
Ea does not change but the number of successful collisions increases significantly, so rate increases. Increased temperature No of molecules Kinetic energy Ea Increased concentration Ea No of molecules Ea does not change but the number of successful collisions increases. Kinetic energy 74 74

Activation energy and reaction pathway Partially broken reactant
Reactants Products 2. In full collision Activated complex Partially broken reactant bonds and partially formed product bonds Before collision Reactants 3. After collision Products Energy changes If the reactants have enough combined K.E. to overcome Ea , their K.E. is converted into the energy needed to form the activation complex. 75 75

d) Catalysts Learning intention
Learn how a catalyst speeds up reaction rate by lowering the activation energy, and how to represent this on a potential energy diagram. 76

Catalysts at Work Heterogeneous Homogeneous
When the catalyst and reactants are in different states you have ‘Heterogeneous Catalysis’. They work by the adsorption of reactant molecules. E.g. Ostwald Process (Pt) for making nitric acid and the Haber Process (Fe) for making ammonia and the Contact Process (Pt) for making Sulphuric Acid. Homogeneous When the catalyst and reactants are in the same state you have ‘Homogeneous Catalysis’. E.g. making ethanoic acid from methanol and CO using a soluble iridium complex. Enzymes are biological catalysts, and are protein molecules that work by homogeneous catalysis. E.g. invertase and lactase. Enzymes are used in many industrial processes 77 77

How a heterogenous catalyst works
Heterogenous Catalysis are thought to work in three stages... Adsorption Reaction Desorption Higher Chemistry Eric Alan and John Harris 78

For an explanation of what happens click on the numbers in turn, starting with  79

Adsorption (STEP 1) Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier. 80

Adsorption (STEP 1) Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier. Reaction (STEPS 2 and 3) Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur. This increases the chances of favourable collisions taking place. 81

Desorption (STEP 4) There is a re-arrangement of electrons and the products are then released from the active sites Adsorption (STEP 1) Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier. Reaction (STEPS 2 and 3) Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur. This increases the chances of favourable collisions taking place. 82

Examples of heterogenous catalysts
Metals Ni, Pt hydrogenation reactions Fe Haber Process Rh, Pd catalytic converters Oxides Al2O3 dehydration reactions V2O5 Contact Process Format FINELY DIVIDED increases the surface area provides more collision sites IN A SUPPORT MEDIUM maximises surface area and reduces costs 83

In some cases the choice of catalyst can influence the products Ethanol undergoes different reactions depending on the metal used as the catalyst. The distance between active sites and their similarity with the length of bonds determines the method of adsorption and affects which bonds are weakened. Copper Dehydrogenation (oxidation) Alumina Dehydration C2H5OH ——> CH3CHO + H2 C2H5OH ——> C2H H2O 84

Poisoning Impurities in a reaction mixture can also adsorb onto the surface of a catalyst thus removing potential sites for gas molecules and decreasing efficiency. expensive because... the catalyst has to replaced the process has to be shut down examples Sulphur Haber process Lead catalytic converters in cars 85

Investigating Catalysis
86

The problem Transition metals and their compounds are said to be effective catalysts You are asked to link the transition metal compounds to their ability to act as a catalyst for the decomposition of hydrogen peroxide 87

Materials available hydrogen peroxide solution (5-10 vol)
cobalt(II) chloride (aq) copper(II) sulphate (aq) manganese(II) chloride (aq) nickel(II) nitrate (aq) bench dilute sodium hydroxide (approximately 1 mol l-1). 88

Decomposition of hydrogen peroxide
Hydrogen → water + oxygen 2H2O → 2H2O + O2 What will you see during the reaction? What is the test for oxygen? 89

Testing the catalysts Collect 5 test tubes in a rack
To each add 5 cm3 of hydrogen peroxide Add 3 cm3 of a different catalyst to 4 test tubes, leaving the last one as a control Note the results 90

Which type of catalysis?
Did the experiment involve heterogeneous or homogeneous catalysis? Which catalyst was the most effective? 91

Changing the solubility of the catalysts
The transition metal ion in the compound has the catalytic effect. The ion can be found in solution (aqueous) or in a solid state. How could you use the chemicals provided to create transition metal ions in the solid state i.e. a heterogeneous catalyst? 92

Investigate the effect of changing the state of the catalysts
Which was the best catalyst overall? 93

An Example of a homogenous catalyst
Dissolve 4 spatulas full of potassium sodium tartrate in a small beaker with about 2.5 cm depth of water. Repeat with another beaker (the control) To one beaker, add enough 10% cobalt chloride solution (catalyst) for the solution to be pink. Add 10 ml hydrogen peroxide to the beakers. Heat the control to around oC and note observations Now heat the one with the catalyst and note observations 95

An Example of a homogenous catalyst
Higher Chemistry Eric Alan and John Harris 96

Catalysts and Potential energy diagrams
Catalysts work by providing… “AN ALTERNATIVE REACTION PATHWAY WHICH HAS A LOWER ACTIVATION ENERGY” 97

Potential energy graphs and catalysts
P.E. 50 - 25 - Reaction path 75 - 60 - Reactants Products Uncatalysed Catalysed 98 98

Activation energy Ea for the forward uncatalysed reaction 75 - Activation energy Eafor the forward catalysed reaction 60 - Reactants 50 - P.E. Products 25 - Reaction path Catalysts lower the activation energy needed for a successful collision. 99 99

Activation energy Ea for the reverse uncatalysed reaction reaction 75 - Activation energy Ea for the reverse catalysed reaction 60 - Reactants 50 - P.E. Products 25 - Reaction path Catalysts lower the activation energy needed for a successful collision. 100 100

75 - Catalysts lower the activation energy needed for a successful collision. 60 - Reactants 50 - P.E. Products 25 - Reaction path ∆H Activation energy Effect of catalyst – forward reaction No change Lowered Effect of catalyst – reverse reaction 101 101

Catalysts A catalyst speeds up the reaction by lowering the activation energy. A catalyst does not effect the enthalpy change for a reaction A catalyst speeds up the reaction in both directions and therefore does not alter the position of equilibrium or the yield of product, but does decrease the time taken to reach equilibrium. 102

Energy distribution and catalysts
No of Collisions with a given K.E. Kinetic energy Ea Un-catalysed reaction Total number of collisions (area under the graph) with sufficient K.E. energy to create new products. Catalysed reaction Ea is reduced 103 103

2H2O2(aq) → 2H2O(ℓ) + O2(g) ΔH = −196∙4 kJ mol–1
Concentrated solutions of hydrogen peroxide are used in the propulsion systems of torpedoes. Hydrogen peroxide decomposes naturally to form water and oxygen: 2H2O2(aq) → 2H2O(ℓ) + O2(g) ΔH = −196∙4 kJ mol–1 Transition metal oxides act as catalysts in the decomposition of the hydrogen peroxide. Unfortunately, there are hazards associated with the use of hydrogen peroxide as a fuel in torpedoes. It is possible that a leak of hydrogen peroxide solution from a rusty torpedo may trigger an explosion. Using your knowledge of chemistry, comment on why this could happen. 104

Unit 1 - Chemical Changes and Structure

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