India 2024 — INChO2024-Question-1.pdf

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Problema 1

Question No 1 2 3 4 5 Total Marks 15 23 23 21 14 96 Marks Obtained

Signature of Examiner Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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Problem 1

15 marks The Fifth Taste In 1908, Professor Kikunae Ikeda, a Japanese chemistry professor, was intrigued by the distinct yet indescribable taste of mushrooms, and kombu seaweed. He isolated a white substance X from these sources and proved that compound X was responsible for the unique taste of mushrooms and kombu seaweed. It is a taste, different from sweet, sour, bitter, and salty. He named it umami, now identified as the fifth taste. Compound X, molecular mass 169.11 g mol–1, is an odourless, crystalline compound soluble in water (solubility 740 g L–1) and practically insoluble in ethanol or ether. It melts at 232 ${}^{\circ }C$ and has specific rotation $[\alpha ]D$ = + $24^{\circ }$. X can be synthesized from acrylonitrile using the Oxo process, which is an industrial process for the hydroformylation of alkenes. This process involves the net addition of a formyl group $(-CHO$) and a hydrogen atom $(-H$) across a carbon-carbon double bond as shown:

A commonly used scheme for synthesis of compound X is outlined here.

B1 is treated with excess of NaOH (aq.) followed by adjusting the pH to 7 to obtain X. Treatment of X with excess HCl (aq) produces Z. If a solution of pure Z is titrated with NaOH (aq), the variation of pH obtained with the amount of NaOH added is shown graphically here. Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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1.1 Identify structures of A1, A2, B1, X and Y.

X obtained by above synthesis is a racemic mixture. The umami taste is due to the L-enantiomer and the D-form is tasteless.

1.2 Draw Fischer projection of the D-enantiomer of X.

1.3 Write the IUPAC name of A2.

A1

B1

X

A2

Y Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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1.4 Predict the predominant structure of X present in solutions of (i) pH = 3.2 and (ii) pH = 0.5.

1.5 When solid X was heated, another compound M was obtained having a molecular mass which is less than that of X by 18 g mol–1. Aqueous solution of M is slightly alkaline which on warming does not give back X. Draw the structure of M.

Later in 1950’s, the following two compounds (X1 and X2) were also found to lend umami flavour to food.

1.6 The class(es) of compound(s) to which X1 and X2 belong is/are? (Mark X in correct box(es))

Nucleotides

Peptides

Nucleosides Glycosides

Phospholipids The nitrogen heterocyclic part present in X1 is

M pH 3.2

pH 0.5 Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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R 1.7 Draw any one tautomeric structure for the heterocyclic structure in X1 shown above.

Ketones react with alcohols in the presence of an acid to form ketals. When X1 is heated with equimolar amount of acetone in the presence of an anhydrous acid, R is obtained. 1.8 Draw the structure of R.

X1 undergoes stepwise hydrolysis in acidic conditions. When X1 was mixed with vinegar (pH = 4) in a food preparation, the umami flavor was lost due to first hydrolysis step which produced two species N and O. 1.9 Draw the structures of N and O.

N

O Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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Topic: Thermodynamics, Nuclear & Particle Physics Metodi: First Law of Thermodynamics, Thermodynamic Cycle Analysis, Ideal Gas Law, Radioactive Decay Law Competenze: Mathematical Modeling Fonte: Testo (PDF) — p.1

Problema 2

Problem 2

23 marks A hand-made Freezer A student made an ice cream freezer as part of a school project. He kept the freezer in an open space where the temperature was $27^{\circ }$ C. Figure on the right shows the schematics of the freezer. Chamber A is the freezer chamber having an air volume of 100 L, which on five sides was enclosed by wooden panels having very low thermal conductivity. On the sixth side, the chamber is made of copper sheet which is in contact with chamber B, fully made of copper metal. Chamber B is a cuboidal cylinder having square cross section (25 cm $\times$ 25 cm) through which a piston (made of insulating material) can be moved up and down using a handle above. Compression and expansion of air in chamber B leads to alternate heating and cooling of this chamber, eventually cooling chamber A.

Cross sectional view (Expanded State)

Cross sectional view (Compressed State)

The chamber B was so designed that when the piston was at maximum height h1 = 39 cm (maximum air volume V1), 1 mol of air in it at 27 ${}^{\circ }C$ had pressure of 1 atm. Treat air to be an ideal gas throughout this problem. Heat Capacity Values Constants related to air Ice cream mix in chamber A 210 J K–1 Specific heat capacity 1.005 $\times$ 103 J kg–1K–1 The copper wall of Chamber A 1.5 kJ K–1 Density 1.16 g L–1 Each copper wall of Chamber B 1.5 kJ K–1 γ= Cp Cv 1.4 Base plate of Chamber B 1.0 kJ K–1

The student starts the cooling cycle with piston at the topmost position and all components in thermal equilibrium with the surrounding. Consider the cycle of 4 steps as shown in the following diagram: Chamber A h2 Chamber B h2 insulating wall Chamber A Chamber B H = 31 cm h1= 39 cm Conducting wall 25 cm Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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Step I. The piston was pushed down extremely fast to h2, such that the volume of chamber B reduced to V2 (V2 = V1/5). Assume that air compression in chamber B was completely reversible and adiabatic. 2.1 Determine the temperature T2 and pressure P2 of air inside chamber B at the end of Step I.

Step II. The piston was kept at the compressed height, h2 till chamber B again reached thermal equilibrium with the outside surrounding air. Assume that there was no heat exchange of chamber B with chamber A during this step. 2.2 Determine the pressure P3 of air in chamber B at the end of Step II.

(T1, P1, V1) (T2, P2, V2) (T3, P3, V3) (T4, P4, V4) (T5, P5, V5) Step I Step II Step III Step IV Step V (repeat of step I) Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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2.3 For air in chamber B to go from state (T1, P1, V1) to state (T3, P3, V3), number of possible thermodynamic paths are (Mark X for the correct option(s))

i) one only (given above)

ii) two (one given above and second involving isothermal compression) iii) infinite (speed of piston compression can be altered in many ways) Step III. The piston was pulled up extremely fast to reach volume V4 = V1. Assume that air in chamber B underwent completely reversible and adiabatic expansion in this step. 2.4 Determine the temperature T4 and pressure P4 of air inside chamber B at the end of Step III.

Step IV. The piston was kept in this position (V5 = V4 = V1) for sufficient time till chamber A attained thermal equilibrium with chamber B. Heat exchange of air in chamber B with surrounding air (outside the chamber) vs with chamber A depends on the respective contact surface area of heat exchange. Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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Assume that heat transfer between chamber A and B is proportional to the surface area of contact, and heat transfer with surrounding is proportional to the remaining surface area of chamber B. [This is approximately true because chamber A had same starting temperature as the surrounding air]. 2.5 Determine the fraction of heat exchange (x) of chamber B with chamber A in Step IV, where

x= Heat transferred from chamber A (including walls between the two chambers) to air in chamber B Total heat gained by air in chamber B

If you were unable to calculate T4 and P4 in 2.4, then take T4 = 220 K and P4 = 0.75 atm for further questions. 2.6 Determine the temperature T5 which chamber A and B attain at the end of Step IV. Assume that the chamber B gains x fraction heat from chamber A, and rest from the surrounding. Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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The student repeated this cycle multiple times to get the desired freezing temperature of chamber A. 2.7 The student discovered that the temperature drop of chamber A at the end of Step IV was much smaller than the temperature drop in chamber B after Step III. The factor(s) responsible for the observation is/are: (Mark X for the correct option(s))

i) heat gain by chamber B from surrounding air

ii) heat absorbed by copper walls iii) higher air volume of chamber A than B iv) heat produced during the compression Step I

2.8 As this cycle is repeated and if the assumptions remain the same, which of the following parameter values will remain the same over cycles, which will decrease with cycles and which will increase with cycles: P2, P3, P4, P5, T2, T3, T4, T5.

Parameters which will remain same:

Parameters which will decrease:

Parameters which will increase:

2.9 If the compression and expansions in a cycle are done slowly, what changes are expected from the above results (Mark the statements as T/F):

i) Work done in compression step would be lower. ii) Temperature drop of chamber A per cycle would be lower. iii) Work done in expansion step by the gas in chamber B would be lower. iv) Heat dissipated from chamber B to surrounding during Step II would remain the same. Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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Topic: Thermodynamics, Conservation of Energy Metodi: Ideal Gas Law, Thermodynamic Cycle Analysis, First Law of Thermodynamics, Physical Modeling Competenze: Physical Reasoning, Mathematical Modeling, Diagrammatic Reasoning Fonte: Testo (PDF) — p.6

Problema 3

Problem 3

23 marks Acetic acid Acetic acid is a colorless liquid with a pungent odour. The worldwide production of acetic acid, which has wide applications in food and other industries, is expected to reach 18 million tons in few years with an average growth of 5% per year. Traditionally, acetic acid is produced by fermentation of apple, grape, honey, cane, coconut, dates etc. The maximum concentration of acetic acid in vinegar that can be obtained by fermentation process is 10% v/v. This route is highly energy intensive and uneconomical for pure acetic acid production. Part-I: General Methods of Preparation 3.1 It is difficult to separate pure acetic acid from an aqueous solution of acetic acid by distillation alone. Four true statements about acetic acid and water are given below. The statement(s) consistent with difficulty in effective separation of acetic acid from water by distillation is/are (Mark X for the correct option(s))

i) Boiling points of water and acetic acid are close. ii) Acetic acid and water do not form an azeotrope. iii) Acetic acid molecule forms hydrogen bonding with water. iv) Acetic acid molecules form a cyclic dimer in the vapour phase. 3.2 One of the laboratory methods for preparation of acetic acid is by hydrolysis of acetonitrile (methyl cyanide). Identify the correct statement(s) about the method by marking X against it.

i) Acid or base catalysis is needed for this hydrolysis reaction. ii) Acid or base is not needed for this hydrolysis reaction. iii) Acetamide can be obtained if acetonitrile is heated with excess of alkali. iv) Ammonia or its salt is formed as a byproduct.

Part-II: Commercial method for acetic acid production A commercial process for acetic acid production is methanol carbonylation at high temperature (180 – 220 ${}^{\circ }C$) and high pressure (30 – 40 atm) in the presence of a transition metal coordinate complex catalyst. One such catalyst is [Rh(CO)2I2]–1. The net reaction is: CH3OH + CO $\to$ CH3CO2H

Reaction – 1 An interesting feature of this reaction is that it requires a small amount of CH3I to start the reaction. Also, under given reaction conditions, the rate of formation of acetic acid is essentially first order with respect to both catalyst and CH3I and zero order with respect to CO and methanol. Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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3.3 [Rh(CO)2I2] –1 shows stereoisomerism. Draw the structures of stereoisomers.

Transition metal ions in coordination complexes which act as catalysts often are able to increase or decrease their coordination number as well as can undergo oxidation and reduction. The overall carbonylation of methanol to acetic acid takes place in several steps. Three new Rh-containing complexes, C1, C2, C3 and one Rh-free compound V are identified to be formed in the reaction system sequentially once the reaction starts. The step C1 $\to$ C2 involves a rearrangement wherein two adjacent ligands combine to form a new ligand in the coordination sphere.

Two additional reactions happen in the system producing Y and Z (both are Rh-free compounds) as given below. V + Y Z + CH3COOH

Reaction – 2 Z + CH3OH CH3I + Y

Reaction – 3 3.4 Identify C1, C2, C3, V, Y and Z, along with stereochemical structures for the complexes.

C1

C2

C3

V

Y

Z Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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M

N

P

Q

R

S 3.5 Write the balanced chemical equation with stereochemical structure(s) for the rate determining step in this production method.

Part-III: Side reactions in commercial acetic acid production The percent conversion in the process presented in Part II is about 85% with respect to CO and 99% with respect to CH3OH. The loss of CO is due to several side reactions which happen in the system as given below. i) One of the reactions which happens in the presence of the [Rh(CO)2I2] –1 catalyst is - CO + Y M + N

Reaction – 4 (M has lower molar mass than that of N.)

ii) The rhodium catalyst system can generate acetaldehyde by reaction 5 given below. Acetaldehyde is then reduced by one of the products formed in the reaction process, to give an intermediate which subsequently yields propionic acid by the same carbonylation process. C2 + HI $\to [RhI4(CO)]-1$ + CH3CHO Reaction – 5 iii) Acetaldehyde undergoes aldol condensation in the presence of an acid and the product is converted to carboxylic acid S as shown below.

3.6 Draw structures of M, N, P, Q, R and S. Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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3.7 For the following group of compounds formed in the reaction with methanol as indicated above, identify for each group, which carbon chain lengths are possible and which are not possible (Indicate number of C atoms as n, 2n, 2n-1, 2n+1, 2n+2, 3n+1, 4n+1, etc. where n = 1, 2, 3, $\dots$).

Possible

Not Possible i) Aldehydes

ii) Carboxylic acids 3.8 Suppose the acetic acid process was performed with unlabelled methanol but with 13C-labeled 13CO. Draw the structures of the following products with 13C labelled carbon-

i) propionic acid (by-product)

ii) S (by-product)

3.9 Determine the H formation of acetic acid at 25 C.

H f values

CH3OH (l) – 239.2 kJ mol–1 CH3CO2H (l) – 386.1 kJ mol–1 CO (g) – 110.5 kJ mol–1 CH3I (l) – 13.6 kJ mol–1

3.10 The important advantages of Rh- catalyzed method over fermentation method of acetic acid production is (are) (Mark X against the correct option(s))-

i) Higher percent conversion of methanol to product. ii) Acetic acid is produced at much higher concentration than vinegar. iii) Uses reactants which are not toxic. Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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Topic: Conservation of Energy, Thermodynamics, Astrophysics Metodi: Energy Conservation Method, Conservation Laws, First Law of Thermodynamics, Thermodynamic Cycle Analysis Competenze: Physical Reasoning, Mathematical Modeling, Estimation & Approximation Fonte: Testo (PDF) — p.11

Problema 4

Problem 4

21 marks Inter-atomic Forces and Static Friction Frictional force between two surfaces in contact originates from inter-atomic forces between the contact atoms of the surfaces. This problem first explores variation in inter-atomic interaction with distance between two atoms (Part I) and uses this idea to understand atomic origins of friction (Part II). Part I: Electrostatic interaction between two atoms Philip Morse, an American scientist gave a simple functional form for inter-atomic (electrostatic) potential (energy) between two atoms in terms of inter-nuclear distance, r as: VMorse(r) = D[1 $-e-\alpha (r-re)]$ 2 $-D$ where D, re and α are constants independent of the mass of the nuclei. [Note that the Morse potential does not go to the expected limit of V= $\infty$ at r= 0. ] The corresponding force F(r) between two atoms along r direction can be defined as: F(r) = $-\partial VMorse$ ∂r

4.1 Derive an expression for F(r) in terms of D, re and α.

Potential is considered attractive when F(r) is negative, and repulsive when F(r) is positive. 4.2 Derive expressions for the following quantities in terms of D, re and α: (a) the inter-nuclear distance rmin where VMorse(r) is minimum, and (b) the energy difference ε= $VMorse(\infty$) $-VMorse(rmin$). Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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(d) (c)

If we ignore quantum mechanical effects, rmin represents the equilibrium bond length between the two atoms and ε, their bond dissociation energy. 4.3 A Vmorse vs. r plot is given below. Assuming αre= 1, indicate on the plot: (i) the potential values at r= 0, r= rmin and r= $\infty$. (ii) ranges of r where potential is repulsive and attractive respectively.

4.4 Among the four plots shown below, the plot of force F(r) as a function of r is likely to be (Mark X for the correct graph):

rmin V r

F(r) F(r) F(r) F(r) r r r r (a) (b) Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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Part II: The nature of static frictional force Static friction is the ‘resisting’ force experienced by an object A of mass M at rest on a surface B, when an external horizontal force is applied on this object. First proposed by Leonardo da Vinci and then formally theorized by French engineer Guillaume Amontons 200 years later in 1669, frictional force was largely found to be proportional to the normal force: F= μMg where μ is the static friction coefficient and g is the acceleration due to gravity.

A zoomed in picture of two surfaces in contact is shown above (right). We assume surface B is atomically flat as shown. Contact surface of A is rough (as most surfaces are rough). There will be multiple (n) atoms of A that will come in close contact with B. Consider an atom P on surface A, that is in contact with surface B. We assume A to be a rigid body and forces acting from B on n contact atoms (like P) to be equal, and for all other A atoms to be zero. 4.5 What is the net normal force felt by each contact atom like P?

If an external force Fext applied on A along x-direction is just more than the static frictional force, then A may microscopically move a small distance δx in the x-z plane. This movement would not be noticeable at bulk scale. A being a rigid body, Fext is experienced equally by all atoms of A. Consider the atom P with center at (0, 0, z) interacting with two atoms of surface B with centers at: (-a,0,0), (a,0,0) as shown here.

We consider interactions between P and B atoms using two models.

Fext A B A B Normal (N) Mg z x P r z -a a x 0 P (0, 0, z) z B surface atoms Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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i) Hard Sphere Model- Assume atoms of A and B as hard spheres which can come a closest distance rAB and do not have any inter-atomic potential (or bond).

In this model, as Fext moves atom P from x= 0 to x= a, object A (being a rigid body) will get lifted by the same height as atom P. The net work (W) done by Fext on A is W= Fexta. This work will change gravitational potential energy of A by Egrav, and kinetic energy of A by Ekin as per the following equation. Fexta= Egrav+ Ekin Since Fext is just more than the static frictional force, we take the limiting case of Ekin = 0, when Fext= μMg. 4.6 Determine (i) increase in height $\text{Deltaz}$ of atom P as it moves from x= 0 to x= a; and (ii) μ in terms of rAB and a showing all steps clearly.

ii) Morse Model- We include Morse interactions (potential energy) between atom P and B atoms such that rAB= re (rAB is distance without considering gravitational effects). In this model, due to gravitational force, equilibrium distance r between centers of atoms P and B will be lower than rAB. Further, in this model,

Fexta= Egrav+ EMorse+ Ekin

rAB z P (0, 0, z) zH rAB -a a x 0 zH -a a x 0 r(0) z P (0, 0, z) z(0) -a a x 0 z (a) -a a x 0 z z Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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The difference in equilibrium position δr= $r-re$ due to the gravitational force is much smaller compared to re. Thus, we can take $e-\alpha (r-re$) $\approx 1-\alpha (r-re$), i.e. VMorse(r) $\text{approxD}\alpha 2(r-re)2$, and z(0) r(0) $\text{approxzeq}$ re

where zeq is the value of z without gravitational effects. As atom P moves from x= 0 to x= a, its Morse interaction with atom at (–a, 0, 0) becomes very weak that we can consider that bond to be broken. Thus for movement of n equivalent contact atoms of the surface A, the minimum energy EMorse to break one of the two bonds with contact atoms of B for n atoms would be nε. 4.7 Derive an expression for equilibrium value of z at (i) x= 0 and (ii) x= a with Morse Model. At x= a, assume atom P interacts only with (a, 0, 0) atom. Use the approximations $r-re$ << re to get the simplified expression.

For many surfaces, static frictional force is related to M as Fstatic= μMg+ K, which can be explained by the Morse model. 4.8 Taking the limiting case of Ekin = 0, derive μ and K in terms of D, α, re, a, and n showing all steps clearly. [Note that μ will also have a small component which is dependent on M] Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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4.9 For the Morse model, estimate the value of μ, given M = 1 kg, g =10 m $s-2$, n= 1010, D= 6.4 $\times$ $10-19$ J, α= 1010 $m-1$, re= 2.5 Å, a =1.5 Å. Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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Topic: Newtonian Mechanics, Conservation of Energy, Electrostatics Metodi: Free-Body Diagram, Energy Conservation Method, Approximation & Series Expansion, Calculus-Integration Competenze: Mathematical Modeling, Physical Reasoning, Diagrammatic Reasoning Fonte: Testo (PDF) — p.15

Problema 5

Problem 5

14 marks Analysis of a solid mixture containing iron and iron oxides Triplicate samples (each with mass 4.72 g) of a homogeneous mixture containing iron (Fe) and oxides of Fe (II) and Fe (III) were received in a lab for analysis. The samples had to be analyzed first to determine the exact moles of each component present in the mixture. To determine the same, two methods (A and B) described below were used by the lab expert A: 4.72 g of the sample was taken in a flask and hydrogen gas was filled in the flask. The flask was sealed and heated. At the end of reaction. 3.92 g iron and 0.90 g of water were obtained. B: An excess of aqueous CuSO4 solution was added to the second sample of mass 4.72 g. At the end of the reaction, 4.96 g of a solid mixture was obtained. Eo values of reducing half-cell equations 2H+(aq) + 2e– H2(g) 0.00 V Fe2+(aq) + 2e– Fe(s) – 0.44 V Fe3+(aq) + 3e– Fe(s) – 0.04 V

5.1 i) Write balanced chemical equations for the reactions involved in methods A and B.

ii) Calculate the moles of Fe (metal), Fe (II) and Fe (III) oxides present in the sample using method A and B. Show all steps in the calculations.

Method A:

Method B: Indian National Chemistry Olympiad 2024 ©HBCSE February 3, 2024

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For another analysis, the third sample weighing 4.72 g was dissolved completely in 7.3% (w/w) aqueous HCl.

5.2 i) Write balanced chemical equations for all reactions involved in the dissolution of the sample in HCl.

ii) Determine the minimum volume of 7.3% (w/w) aqueous HCl $(\rho$ = 1.03 g cm–3) required in milliliters (mL) for complete dissolution of the sample. Show all steps in the calculations.

iii) Calculate the volume of gas released in the above reaction at 1 atm and 25 C. Indian National Chemistry Olympiad 2024

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ADDITIONAL SPACE FOR ANSWERS Indian National Chemistry Olympiad 2024

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ROUGH WORK Indian National Chemistry Olympiad 2024

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ROUGH WORK 1 H hydrogen 1.008 [1.0078, 1.0082] 1 18 3 Li lithium 6.94 [6.938, 6.997] 4 Be beryllium 9.0122 11 Na sodium 22.990 12 Mg magnesium 24.305 [24.304, 24.307] 19 K potassium 39.098 20 Ca calcium 40.078(4) 37 Rb rubidium 85.468 38 Sr strontium 87.62 38 Sr strontium 87.62 55 Cs caesium 132.91 55 Cs caesium 132.91 56 Ba barium 137.33 87 Fr francium 88 Ra radium 5 B boron 10.81 [10.806, 10.821] 13 Al aluminium 26.982 31 Ga gallium 69.723 49 In indium 114.82 81 Tl thallium 204.38 [204.38, 204.39] 6 C carbon 12.011 [12.009, 12.012] 14 Si silicon 28.085 [28.084, 28.086] 32 Ge germanium 72.630(8) 50 Sn tin 118.71 82 Pb lead 207.2 7 N nitrogen 14.007 [14.006, 14.008] 15 P phosphorus

30.974 33 As arsenic 74.922 51 Sb antimony 121.76 83 Bi bismuth

208.98 8 O oxygen 15.999 [15.999, 16.000] 16 S sulfur 32.06 [32.059, 32.076] 34 Se selenium 78.971(8) 52 Te tellurium 127.60(3) 84 Po polonium 9 F fluorine 18.998 17 Cl chlorine 35.45 [35.446, 35.457] 35 Br bromine 79.904 [79.901, 79.907] 53 I iodine 126.90 85 At astatine 10 Ne neon 20.180 2 He helium 4.0026 18 Ar argon 39.95 [39.792, 39.963] 36 Kr krypton 83.798(2) 54 Xe xenon 131.29 86 Rn radon 22 Ti titanium 47.867 22 Ti titanium 47.867 40 Zr zirconium 91.224(2) 72 Hf hafnium 178.49(2) 104 Rf rutherfordium

23 V vanadium 50.942 41 Nb niobium 92.906 73 Ta tantalum 180.95 105 Db dubnium 24 Cr chromium 51.996 24 Cr chromium 51.996 42 Mo molybdenum 95.95 74 W tungsten 183.84 106 Sg seaborgium 25 Mn manganese 54.938 43 Tc technetium 75 Re rhenium 186.21 107 Bh bohrium 26 Fe iron 55.845(2) 44 Ru ruthenium 101.07(2) 76 Os osmium 190.23(3) 108 Hs hassium 27 Co cobalt 58.933 45 Rh rhodium 102.91 77 Ir iridium 192.22 109 Mt meitnerium 28 Ni nickel 58.693 46 Pd palladium 106.42 78 Pt platinum 195.08 110 Ds darmstadtium 29 Cu copper 63.546(3) 47 Ag silver 107.87 79 Au gold 196.97 30 Zn zinc 65.38(2) 48 Cd cadmium 112.41 80 Hg mercury 200.59 111 Rg roentgenium 112 Cn copernicium 114 Fl flerovium 113 Nh nihonium 115 Mc moscovium 117 Ts tennessine 118 Og oganesson 116 Lv livermorium 57 La lanthanum 138.91 58 Ce cerium 140.12 59 Pr praseodymium 140.91 60 Nd neodymium 144.24 61 Pm promethium 62 Sm samarium 150.36(2) 63 Eu europium 151.96 64 Gd gadolinium 157.25(3) 65 Tb terbium 158.93 66 Dy dysprosium 162.50 67 Ho holmium 164.93 68 Er erbium 167.26 69 Tm thulium 168.93 70 Yb ytterbium 173.05 71 Lu lutetium 174.97 89 Ac actinium 90 Th thorium 232.04 91 Pa protactinium 231.04 92 U uranium 238.03 93 Np neptunium 94 Pu plutonium 95 Am americium 96 Cm curium 97 Bk berkelium 98 Cf californium 99 Es einsteinium 100 Fm fermium 101 Md mendelevium 102 No nobelium 103 Lr lawrencium 21 Sc scandium 44.956 39 Y yttrium 88.906 57-71

lanthanoids 89-103

actinoids atomic number Symbol name conventional atomic weight standard atomic weight 2 13 14 15 16 17 Key: 3 4 5 6 7 8 9 10 11 12 For notes and updates to this table, see www.iupac.org. This version is dated 1 December 2018. Copyright © 2018 IUPAC, the International Union of Pure and Applied Chemistry. IUPAC Periodic Table of the Elements

Topic: Conservation of Energy, Thermodynamics, Modern-Quantum Physics Metodi: Energy Conservation Method, Conservation Laws, First Law of Thermodynamics, Thermodynamic Cycle Analysis Competenze: Physical Reasoning, Mathematical Modeling Fonte: Testo (PDF) — p.21