Chemistry Paper 3 June 2020

1. Arate = $k[RCHO][H^+][CN^-]$
2. Brate = $k[RCHO]^2[CN^-]^2$
3. Crate = $k[RCHO][CN^-]$
4. Drate = $k[RCHO][H^+]$

1. A166.7 mol⁻¹dm³s⁻¹
2. B500.0 mol⁻¹dm³s⁻¹
3. C16.7 mol⁻¹dm³s⁻¹
4. D55.6 mol⁻¹dm³s⁻¹

1. A0.30
2. B2.25
3. C4.20
4. D4.50

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1. Athe rate is unchanged because [RCHO] and [CN⁻] were both altered to cancel out
2. Bthe comparison is invalid because three concentrations differ between the experiments
3. Cthe rate is unchanged although [H⁺] changes, so the reaction is zero order in H⁺
4. Dthe rate doubles when [H⁺] is halved, so the reaction is first order in H⁺

1. Abenzoic acid crystallises around individual microorganisms on the surface of the food, physically trapping them and starving them of the nutrients they need to grow
2. Bit lowers the food to boiling point during storage, which denatures the proteins and enzymes of any bacteria, fungi and moulds present in it, sterilising the food completely
3. Cthe mixture buffers the food at a low, acidic pH that inhibits microbial growth, and the undissociated benzoic acid molecules can also penetrate microbial cell membranes and disrupt their internal processes
4. Dsodium benzoate reacts with water absorbed by the food to release oxygen gas, which kills anaerobic bacteria that cannot survive in an oxygen-rich environment inside the packaging

1. Airon filings gradually dissolve and decrease in mass
2. Biron filings become coated with a black insoluble layer
3. Csolution turns from colourless to a deep purple colour
4. Dsolution effervesces vigorously, releasing a colourless gas

1. Airon is more reactive (has a more negative electrode potential) than tin, so it reduces Sn4+ to Sn2+ while itself being oxidised to Fe2+: Fe + Sn4+ → Fe2+ + Sn2+
2. Btin(IV) ions act as stronger reducing agents than metallic iron, donating electrons to the iron filings and reducing them, while the tin ions are themselves oxidised in the process
3. Ciron and tin(IV) ions do not react at all, because both iron and tin lie in the same group of the reactivity series and therefore have very similar, closely matched electrode potentials
4. Diron filings simply adsorb tin(IV) ions onto their surface by physical attraction, with no electron transfer or change in the oxidation state of either metal taking place

1. Athe number of electron shells decreases
2. Bshielding by the inner shells increases
3. Cthe atoms gain electrons into new shells
4. Dthe nuclear charge increases while electrons enter the same shell

1. A0.08 atm
2. B0.16 atm
3. C0.24 atm
4. D0.36 atm

1. Aargon has a smaller nuclear charge than chlorine
2. Bthe radius of argon is a van der Waals radius, not a covalent radius
3. Cargon atoms have more shielding electrons than chlorine atoms
4. Dargon has an extra electron shell

1. Athe distance from the nucleus to the outermost occupied electron shell of a free, isolated atom
2. Bhalf the distance between the nuclei of two ionically bonded atoms of different elements
3. Cthe average distance between the nucleus and all the electrons in every shell of the atom
4. Dhalf the distance between the nuclei of two covalently bonded atoms of the same element

1. Athe increase in nuclear attraction on outer electrons caused by additional protons being added to the nucleus
2. Bthe reduction of the nuclear attraction on outer electrons caused by inner electron shells lying between them and the nucleus
3. Cthe repulsion between electrons in the same shell that pushes them further away from the nucleus
4. Dthe complete blocking of the nuclear charge by the innermost electron shell only, regardless of how many shells exist

1. Athe ions are isoelectronic (same electron arrangement) but the nuclear charge increases from 11 to 13, pulling the same number of electrons in more strongly
2. Bthe shielding effect from inner electrons increases faster than the rising nuclear charge from Na⁺ to Al³⁺, weakening the overall pull on the outer electrons
3. Ceach of these ions has a different number of occupied electron shells altogether, so their ionic radii simply shrink at random with no consistent pattern across the row
4. Dsodium, magnesium and aluminium ions actually gain an increasing number of electron shells as the period is crossed, which is what causes their radii to fall

1. Aan anion always gains one whole additional electron shell compared with its corresponding cation, and this extra shell alone accounts for its larger size
2. Bcations always have a larger nuclear charge than the anion formed from the same element, so the nucleus attracts the cation's remaining electrons far more weakly
3. Canions are always formed only from metallic elements, and metal atoms are inherently larger than non-metal atoms regardless of any change in electron number
4. Dan anion has gained electrons, increasing electron-electron repulsion and reducing the effective nuclear charge felt by each electron, so the electron cloud expands

1. AP³⁻ has three extra electrons compared with the phosphorus atom, increasing electron repulsion and reducing the effective nuclear charge per electron
2. BP³⁻ has exactly three fewer occupied electron shells than a neutral phosphorus atom, and this alone would make the ion noticeably smaller, not larger
3. Cphosphorus atoms are inherently unstable in air and physically swell in size the longer they are left standing before being weighed or measured
4. Dthe nuclear charge of the P³⁻ ion is actually greater than that of a neutral phosphorus atom, pulling its electrons in more strongly and shrinking it

1. A$PCl_5$ molecules are held together by hydrogen bonds between neighbouring molecules, a force that $PCl_3$ completely lacks
2. Bsolid $PCl_5$ molecules become covalently bonded to their neighbours throughout the entire crystal, forming a giant covalent lattice similar to diamond
3. C$PCl_5$ exists as $[PCl_4]^+[PCl_6]^-$ ions held together by strong ionic bonding in the solid state, whereas $PCl_3$ is a simple covalent molecule held together only by weak van der Waals forces
4. D$PCl_3$ actually has a much higher relative molecular mass than $PCl_5$, so its molecules pack together far more strongly in the solid state

1. Aphosphorus is actually more electronegative than nitrogen, and it is this higher electronegativity that weakens and lengthens the P-H bond
2. Bnitrogen is unable to form a dative covalent bond with $H^+$ at all, unlike phosphorus which readily accepts a proton to form $PH_4^+$
3. Cthe P-H bond is weaker and longer than the N-H bond, since phosphorus is a larger atom than nitrogen
4. D$PH_4^+$ carries a smaller overall positive charge than $NH_4^+$, so weaker electrostatic attraction holds its four hydrogens in place

1. A$5.00 \times 10^{-3}$
2. B$6.25 \times 10^{-4}$
3. C$1.25 \times 10^{-3}$
4. D$2.50 \times 10^{-3}$

1. A39.7 %
2. B63.5 %
3. C79.4 %
4. D158.8 %

1. A−4 875 kJmol⁻¹
2. B−221 kJmol⁻¹
3. C+221 kJmol⁻¹
4. D+2 327 kJmol⁻¹

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1. Athe enthalpy change when one mole of a solid ionic lattice is broken up completely into its separate gaseous ions
2. Bthe enthalpy change when one mole of a solid dissolves completely in a small, fixed volume of water
3. Cthe enthalpy change released when one mole of water molecules freezes solid around a dissolved ion
4. Dthe enthalpy change when one mole of gaseous ions is dissolved in water to form an infinitely dilute solution

1. Ais chemically unreactive (inert) towards the acid and, unlike glass, does not crack or shatter from the heat released as the concentrated acid dilutes
2. Bis far more transparent than glass, which allows any colour change in the acid during dilution to be observed clearly through the container wall
3. Creacts with the concentrated acid on contact, forming a thin protective and completely unreactive coating on the inside of the container
4. Dis a much better conductor of heat than glass, so it dissipates the heat released during dilution far more quickly and safely

1. A$2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2$
2. B$Cu^{2+} + I^- \rightarrow CuI$
3. C$2Cu^{2+} + 2I^- \rightarrow 2Cu^+ + I_2$
4. D$Cu^{2+} + 2I^- \rightarrow CuI_2$

1. Athe high pressure inside the sealed container physically destroys any microorganisms present in the drink before it is opened
2. Bby Le Chatelier's principle, the high pressure of CO₂ above the liquid shifts the equilibrium to the right, keeping more CO₂ dissolved as $H_2CO_3$
3. Chigh pressure inside the sealed container simply prevents any water from evaporating out of the drink while it is stored
4. Dhigh pressure lowers the overall temperature of the sealed drink, and this drop in temperature alone slows down the forward reaction considerably

1. Athe sudden exposure to light when the container is opened catalyses the decomposition of $H_2CO_3$ into $CO_2$ gas and water
2. Bopening the container suddenly lowers the pressure above the liquid, so the equilibrium $CO_{2(g)} + H_2O_{(l)} \rightleftharpoons H_2CO_{3(aq)}$ shifts left, releasing dissolved $CO_2$ gas rapidly
3. Copening the container sharply raises the temperature of the drink, and this sudden rise in temperature alone decomposes the dissolved $H_2CO_3$
4. Dair rushing into the open container reacts directly with the dissolved $H_2CO_3$ already present, releasing $CO_2$ gas as a by-product of that reaction

1. Anitrogen is less electronegative than oxygen
2. Bnitrogen atoms have a complete outer shell of electrons
3. Cnitrogen molecules are non-polar
4. Dthe $N \equiv N$ bond is very strong and needs a high activation energy to break

1. A$SO_2 + NO_2 \rightarrow SO_3 + NO$ then $2NO + O_2 \rightarrow 2NO_2$
2. B$3NO_2 + H_2O \rightarrow 2HNO_3 + NO$ then $SO_2 + H_2O \rightarrow H_2SO_3$
3. C$N_2 + O_2 \rightarrow 2NO$ then $2NO + O_2 \rightarrow 2NO_2$
4. D$2SO_2 + O_2 \rightarrow 2SO_3$ then $SO_3 + H_2O \rightarrow H_2SO_4$

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1. Athe melting point is much higher
2. Bthe layers are held together by van der Waals forces
3. Call four valence electrons of each atom are used in localised covalent bonds
4. Dthe bonds between silicon and carbon are ionic

1. Amoisture released during the storm acts as a catalyst for the reaction, completely removing the need for the strong $N \equiv N$ and $O=O$ bonds to be broken
2. Bthe sudden drop in atmospheric pressure during a storm permanently weakens the strong $N \equiv N$ bond in nitrogen molecules
3. Cthe high temperature of the lightning discharge supplies enough energy to break the strong $N \equiv N$ and $O=O$ bonds, allowing $N_2 + O_2 \rightarrow 2NO$ to occur
4. Dlightning permanently ionises nitrogen atoms in the air, and once ionised they remain reactive even at normal room temperature afterwards

1. Athe high temperature and pressure of combustion allow atmospheric $N_2$ and $O_2$ to combine directly: $N_2 + O_2 \rightarrow 2NO$
2. Bthe engine's spark plug ionises oxygen molecules, which then travel outside the cylinder and react with cool atmospheric nitrogen there
3. Cunburnt hydrocarbon vapour reacts directly with atmospheric nitrogen at ordinary room temperature once it leaves the exhaust pipe
4. Dnitrogen present within the fuel itself decomposes directly during combustion to release $NO_2$ gas straight into the exhaust

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1. Adissolves readily in water at room temperature, allowing it to be poured and moulded into any cutting tool shape required
2. Bis naturally a soft, mouldable material that can be shaped easily into cutting tool blades before being hardened afterwards by rapid cooling
3. Cis an excellent conductor of electricity, which allows a thin layer of it to be electroplated directly onto cutting tool surfaces
4. Dhas a giant covalent (macromolecular) structure with strong covalent bonds extending throughout the crystal, making it extremely hard and resistant to wear

1. A$X_2O_3 + 6NaOH \rightarrow 2Na_3XO_3 + 3H_2O$
2. B$X_2O_3 + 2NaOH \rightarrow 2NaXO_2 + H_2O$ (on fusion with solid NaOH)
3. C$X_2O_3$ does not react with sodium hydroxide
4. D$X_2O_3 + 2NaOH + 3H_2O \rightarrow 2NaX(OH)_4$

1. Ais the same for all the carbonates in the group
2. Bincreases down the group because the lattice energy increases
3. Cdecreases down the group as the cations get larger
4. Dincreases down the group as the polarising power of the cation falls

1. A$AgX + 4NH_3 \rightarrow [Ag(NH_3)_4]^+ + X^-$
2. B$AgX + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + X^-$
3. C$AgX + NH_4OH \rightarrow AgOH + NH_4X$
4. D$AgX + NH_3 \rightarrow Ag(NH_3)X$

1. A$Al_2O_3$ does not react with dilute sulphuric acid
2. B$Al_2O_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2O$
3. C$Al_2O_3 + H_2SO_4 \rightarrow Al_2SO_4 + 3H_2O$
4. D$Al_2O_3 + 6H_2SO_4 \rightarrow 2Al(SO_4)_3 + 3H_2O$

1. A$AlCl_3$ behaves as a strong base in water, reacting with it to release hydroxide ions and raise the pH of the solution
2. B$AlCl_3$ simply dissolves physically in water without undergoing any chemical reaction with the water molecules at all
3. Cchloride ions released from $AlCl_3$ react with dissolved oxygen in the water to slowly form hydrochloric acid over time
4. Dthe small, highly charged $Al^{3+}$ ion polarises coordinated water molecules, releasing $H^+$ ions: $AlCl_3 + 3H_2O \rightarrow Al(OH)_3 + 3HCl$

1. Aa known excess of standard hydrochloric acid is reacted with the soil sample, and the unreacted acid remaining is then titrated against standard sodium hydroxide
2. Bthe soil sample is simply dissolved directly in distilled water and its pH is then measured using a calibrated pH meter
3. Cthe soil sample is heated strongly in a crucible and the total mass lost on ignition is measured and recorded directly
4. Dthe soil sample is titrated directly against standard sodium hydroxide solution until a suitable indicator suddenly changes colour

1. A$Y_2O_3^{2-} + Br_2 + H_2O \rightarrow 2YO_4^{2-} + 2Br^- + 2H^+$
2. B$Y_2O_3^{2-} + 4Br_2 + 5H_2O \rightarrow 2YO_4^{2-} + 8Br^- + 10H^+$
3. C$Y_2O_3^{2-} + 2Br_2 + 5H_2O \rightarrow 2YO_4^{2-} + 4Br^- + 10H^+$
4. D$Y_2O_3^{2-}$ cannot be oxidised by bromine under any laboratory conditions

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1. A$2NaI + H_2SO_4 \rightarrow Na_2SO_4 + 2HI$ (simple acid-base reaction only)
2. B$NaI + H_2SO_4 \rightarrow NaHSO_4 + HI$ (no redox change occurring)
3. C$8NaI + H_2SO_4 \rightarrow 8NaHSO_4 + 4I_2 + H_2S$ (unbalanced oxygen and hydrogen)
4. D$8NaI + 5H_2SO_4 \rightarrow 4Na_2SO_4 + H_2S + 4I_2 + 4H_2O$

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1. AKCN in ethanol, heat under reflux
2. Baqueous NaOH, warm
3. Cconcentrated $H_2SO_4$ at 170 °C
4. D$NH_3$ in ethanol, heated in a sealed tube

1. Abut-1-ene, buta-1,3-diene and but-2-ene
2. Bbut-1-ene, cis-but-2-ene and trans-but-2-ene
3. Cbut-1-ene, but-2-ene and 2-methylpropene
4. Dcis-but-2-ene, trans-but-2-ene and 2-methylpropene

1. A2-methylpropan-2-ol
2. B2-methylpropane-1,2-diol
3. C2-methylpropanoic acid
4. Dpropanone and carbon dioxide

1. A$LiAlH_4$ in dry ether, added under anhydrous conditions (or $H_2$/Ni catalyst as an alternative)
2. Baqueous NaOH, warmed gently under reflux for several minutes (hydrolysis conditions)
3. Cconcentrated $H_2SO_4$, heated strongly to around 170°C (dehydration conditions)
4. D$KMnO_4$ solution, dilute and kept cold in an ice bath throughout

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1. A$Br_2$ water, added dropwise at room temperature with no heating
2. Baqueous NaOH, warmed gently for a short time only (substitution conditions)
3. Cethanolic (alcoholic) NaOH or KOH, heated under reflux to favour elimination over substitution
4. Dconcentrated $H_2SO_4$, heated strongly to around 170°C in the absence of a base

1. Apropanone, $(CH_3)_2C=O$, and carbon dioxide, $CO_2$
2. B2-methylpropan-2-ol only, with no C=C cleavage
3. C2-methylpropane-1,2-diol, $(CH_3)_2C(OH)CH_2OH$
4. Dbutan-2-one and methanoic acid (from C=C cleavage)

1. Athe purple colour of $KMnO_4$ actually intensifies further in both reactions as a stable purple diol complex slowly forms
2. Bno colour change occurs at all with cold $KMnO_4$, but the hot reaction instead turns the whole mixture a deep blue colour
3. Ca bright yellow precipitate forms during the cold reaction, and this precipitate then redissolves completely on heating
4. Dthe purple $KMnO_4$ is decolourised in both reactions, and the hot reaction also produces effervescence of a colourless gas ($CO_2$)

1. A$H_2NCH(CH_3)COO^-$
2. B$^+H_3NCH(CH_3)COO^-$
3. C$^+H_3NCH(CH_3)COOH$
4. D$H_2NCH(CH_3)COOH$

1. Astructural isomerism
2. Bgeometrical isomerism
3. Ccis-trans isomerism
4. Doptical isomerism

1. Apropylamine has only a basic group whereas the amino acid also has an acidic group
2. Bboth solutions have the same pH
3. C2-aminopropanoic acid is more alkaline owing to its $-NH_2$ group
4. Dpropylamine is acidic owing to its $-NH_2$ group

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1. Athe carbon chain of propylamine reacts with water to release $OH^-$ ions
2. Bthe lone pair of electrons on the nitrogen atom can accept a proton from water
3. Cthe $-NH_2$ group readily loses a proton to water, releasing $OH^-$ ions
4. Dpropylamine ionises completely to release $H^+$ ions in water

1. Ahas a plane of symmetry passing directly through its central carbon atom, dividing it into two identical halves
2. Bexists as two or more structural isomers that each have a genuinely different carbon skeleton
3. Ccontains a carbon-carbon double bond that is able to exist in separate cis and trans geometric forms
4. Dis not superimposable on its own mirror image, usually because it contains a carbon atom bonded to four different groups

1. Acarries both a positive and a negative charge on different atoms of the same molecule, but is overall electrically neutral
2. Bhas no electrical charge anywhere on the molecule under any solution conditions
3. Ccarries only a single positive charge spread evenly across the whole molecule
4. Dis formed only when an amino acid is dissolved in a strongly acidic solution at low pH

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1. Ait reacts completely and irreversibly with any acid or base added to the solution, destroying the added species entirely
2. Bit contains only an acidic $-COOH$ group overall, which can neutralise added base but is unable to neutralise added acid
3. Cthe zwitterion form is chemically inert and does not react with either added acids or added bases
4. Dits $-COOH/-COO^-$ pair can neutralise added base while its $-NH_2/-NH_3^+$ pair can neutralise added acid, so both small additions of acid and of base leave the pH almost unchanged

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1. Aelectrophilic addition
2. Bnucleophilic addition
3. Cnucleophilic substitution
4. Dcondensation

1. Aesterification
2. Belimination
3. Coxidation
4. Dhydrolysis

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1. Aelectrophilic addition, giving a vicinal dibromide across the ring's C=C bond
2. Bnucleophilic substitution, giving a mono-brominated ring with loss of HBr
3. Ccondensation, giving an ester linkage with loss of HBr
4. Dfree radical substitution, giving a brominated side chain

1. Aoxidation, in which the ketone group is converted directly into a carboxylic acid group
2. Baddition, in which $PCl_5$ molecules add directly across the ring's C=C double bond
3. Csubstitution, in which any -OH group present is replaced by -Cl with the evolution of steamy HCl fumes
4. Dhydrolysis, in which the ester link is broken apart to give a free alcohol and a carboxylic acid

1. Aesterification, forming an entirely new ester linkage directly with the sodium hydroxide
2. Breduction, converting the ester group directly into a primary alcohol only
3. Chydrolysis, breaking the ester link to give a carboxylate salt and an alcohol/phenol
4. Dno reaction at all takes place, since lactone rings are completely unreactive towards hot alkali

1. Anucleophilic substitution, in which the cyanide ion directly replaces the oxygen atom
2. Bcondensation, releasing a molecule of water as the cyanohydrin product forms
3. Cnucleophilic addition, with the cyanide ion adding across the C=O bond to give a cyanohydrin
4. Delectrophilic addition, with a hydrogen ion adding first across the carbonyl C=O bond

1. Athe cyanide complex has no d electrons
2. B$CN^-$ is a strong field ligand, so all the d electrons become paired
3. C$CN^-$ has no lone pair to donate to the iron
4. Diron is in a different oxidation state in the two complexes

1. A$H_2O$
2. B$Cl^-$
3. Cethane-1,2-diamine
4. D$NH_3$

1. Aa solvent rising up a plate
2. Ban inert carrier gas
3. Ca liquid coating the inside of the column
4. Da solid adsorbent packed in the column

1. Adonates only a single lone pair to the central metal ion, whereas a polydentate ligand always donates exactly two lone pairs
2. Bmust always carry an overall negative charge, whereas a polydentate ligand is always electrically neutral
3. Ccan only form coordinate bonds with transition metals, whereas polydentate ligands can bond to any metal ion
4. Ddonates two lone pairs (forms two coordinate bonds) to the central metal ion, whereas a polydentate ligand donates three or more lone pairs

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1. Areplacing $H_2O$ ligands with $NH_3$ increases the oxidation state of the copper ion from +2 to +3
2. B$NH_3$ ligands are themselves intrinsically blue in colour, and simply tint the whole complex blue
3. C$NH_3$ is a stronger field ligand than $H_2O$, causing greater d-orbital splitting, which shifts the wavelength of light absorbed and deepens the colour
4. D$[Cu(NH_3)_4(H_2O)_2]^{2+}$ actually contains a greater number of copper ions per formula unit than $[Cu(H_2O)_6]^{2+}$

1. ATLC uses a solid stationary phase on a plate with a liquid mobile phase and separates by adsorption/$R_f$, whereas GLC uses a liquid stationary phase in a column with an inert carrier gas as the mobile phase, separating volatile components by retention time
2. BGLC uses a liquid mobile phase flowing over a solid stationary phase on a plate, whereas TLC uses an inert carrier gas flowing through a packed column, the reverse of their true set-ups
3. CTLC and GLC both use a solid stationary phase packed into a column, differing only in that TLC passes a liquid mobile phase through it while GLC passes an inert carrier gas through the same solid packing
4. DTLC separates compounds only by their colour under UV light, whereas GLC separates compounds only by their boiling point measured directly on the plate before it is placed in the column

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1. Aonly be used to deliver drugs that are already fully effective when given by conventional methods anyway
2. Bcompletely eliminate the need for any drug at all to ever be administered to the patient
3. Ctarget the drug to specific sites (e.g. a tumour), deliver it in controlled amounts, and cross biological barriers, reducing side effects and the dose needed
4. Dincrease the total overall dose of drug needed to achieve the same therapeutic effect as conventional methods

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1. Aelectrophoresis
2. Bthin layer chromatography
3. Cfractional distillation
4. Dgas liquid chromatography

1. Ahas the largest relative molecular mass
2. Bcarries a net positive charge at pH 7
3. Cis at its isoelectric point and exists as a zwitterion
4. Dis insoluble in the buffer used

1. AFehling's solution
2. Bacidified potassium dichromate(VI)
3. Cbromine water
4. Dninhydrin

1. AS simply has a much larger relative molecular mass than Q, and this alone is why it migrated further
2. BS carries a net positive charge (its isoelectric point is above 7), while Q is close to its isoelectric point and carries little or no net charge
3. CQ and S actually share identical isoelectric points and therefore carry exactly the same net charge at pH 7
4. DQ carries a net positive charge at pH 7, while S remains completely neutral and does not migrate at all

1. ApH only affects how well the buffer conducts electricity during the run, and has no influence at all on the net charge carried by any of the amino acid molecules themselves
2. Bat a given pH, each amino acid has a different net charge depending on its own isoelectric point, so below its isoelectric point it moves to the cathode and above it, it moves to the anode
3. Camino acids carry no electrical charge at any pH value, so changing the pH of the buffer has no effect whatsoever on how far or which way they migrate
4. Draising the pH of the buffer always makes every amino acid present migrate towards the anode at exactly the same rate, regardless of its own isoelectric point

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1. Ais completely unreactive once released into the atmosphere and has no measurable effect on the environment at all
2. Bdissolves in rainwater to form acid rain, which lowers the pH of soils and water bodies, damages vegetation and aquatic life, and corrodes buildings and metals
3. Conly affects marine ecosystems directly and has no measurable impact on land-based vegetation or soils
4. Dreacts directly with atmospheric oxygen high in the stratosphere to form ozone, which protects the earth from UV radiation

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