Exam 5: General Chemistry

Passage Living organisms that require oxygen to respire can build up hydrogen peroxide (H₂O₂) as a byproduct of respiration.  Because hydrogen peroxide can cause oxidative damage to cells, the amount of H₂O₂ in cells must be closely regulated.  Although H₂O₂ decomposes spontaneously, a catalyst is needed for the reaction to occur at a sufficiently rapid rate suitable for biological purposes.  In biological systems, an enzyme called a catalase accelerates the catalytic decomposition of H₂O₂ into water and oxygen (Reaction 1). Reaction 1In a laboratory setting, several different inorganic compounds can also serve as catalysts for H₂O₂ decomposition, allowing researchers to compare how different catalysts affect the efficiency of the reaction.  To compare Fe(NO₃)₃ (a metal homogeneous catalyst) and NaI (a halogen homogeneous catalyst), researchers measured the amount of heat evolved after placing each catalyst in a fresh solution of 3% hydrogen peroxide.To perform the measurements, the researchers placed 50 mL of 3% H₂O₂ in an insulated coffee cup to be used as a calorimeter, as shown in Figure 1.  Before adding any catalyst, the initial, baseline temperature of the H₂O₂ solution was determined by recording the temperature every 30 seconds for 5 minutes.  Then 10 mL of 0.10 M Fe(NO₃)₃(aq) catalyst was added to the H₂O₂ solution.  The temperature was recorded every 30 seconds for another 15 minutes.  The experiment was then repeated using 10 mL of 0.50 M sodium iodide in 0.01 M NaOH as the catalyst.  The results for each reaction are shown in Figure 2. Figure 1  Coffee cup calorimeter Figure 2  Temperature vs. time measurements for reactions catalyzed by Fe(NO₃)₃ and NaI Adapted from: C. Marzzacco, "The effect of a change in catalyst on the enthalpy of decomposition of hydrogen peroxide." Chem13 News Magazine. ©2008 University of Waterloo. -Based on the results in Figure 2, the magnitude of ΔT for the decomposition of H₂O₂ using Fe(NO₃)₃ is closest to which of the following?

Passage Within the renal system, two buffers operate to counter the production of excess H⁺ from luminal H₂CO₃.  They are located within the tubular lumen, and maintain urinary pH levels between 4.5 and 8.0.The first of these is a phosphate buffer, which serves as the primary buffer for urine because its pK_a is very close to extracellular fluid pH of 7.4.  The phosphate buffer operates according to Reaction 1.  H₂PO₄⁻ is excreted in urine. Reaction 1During acidosis when tubular pH is low, the phosphate buffer system becomes saturated.  An ammonia buffer system then becomes the primary buffer, and NH₄⁺ is excreted in urine. Reaction 2Titratable acid (TA) is the total acid in urine, and it can be found using Folin's method.  Under normal physiological conditions, TA is approximately equal to the amount of phosphate in urine, which is a product of serum phosphate concentration and glomerular filtration rate.  This makes TA a good indicator of kidney function.Folin's method involves back titration of a urine sample to a pH of 7.4 using a strong base.  In a typical procedure, potassium oxalate is added to remove calcium from the sample prior to titration with 1.0 M NaOH.  The titration continues until the solution reaches a pH of 7.4 (measured by a pH meter).  The amount of base required for the titration can then be used to calculate the amount of H₂PO₄⁻ in the urine sample. -What are the predominant species in the phosphate buffer system and in the ammonia buffer system in an aqueous solution at a pH of 8?

Passage The tendency of a chemical species to undergo reduction during an electrochemical reaction is indicated by the standard potential E° for the reaction.  As represented generally by Reaction 1, E° measures the potential Z at which element X with oxidation number N accepts N electrons and is reduced to its elemental state.X(N) + Ne⁻ → X(0),    E° = Z voltsReaction 1In a galvanic cell, E° > 0 indicates a spontaneous reaction.  Therefore, chemical species with higher positive E° values are reduced more easily than those with lower E° values.Using E° as a basis of comparison, the relative thermodynamic stabilities of different chemical species involving the same element at different oxidation states can be presented graphically using a Frost diagram.  For a series of compounds containing element X, a Frost diagram plots the value of NE° for each compound against the corresponding oxidation number N of element X within the species.  Frost diagrams for several species containing manganese or chlorine are shown in Figure 1. Figure 1  Frost diagrams for manganese and chlorineOn a Frost diagram, NE° is proportional to the standard Gibbs free energy ΔG° according to the relationshipΔG° = −FNE° = −nFE°Equation 1where n is the number of moles of electrons transferred during the electrochemical process and F is the Faraday constant.  Free energy considerations also show that a species is prone to disproportionation if its position on the Frost diagram lies above a line connecting the points of two adjacent species, as seen for species B in Figure 2. Figure 2  General Frost diagram for an element forming species A, B, C, and D.The slope of a line segment joining two species on a Frost diagram is equal to the standard reduction potential for the couple.  A greater slope indicates a higher corresponding reduction potential.  As a result, in Figure 2, the reduction potential for B to A is lower than that for D to C. -Based on the passage, which of the following species is LEAST likely to undergo a disproportionation reaction?

Passage Amino acids and copper both serve vital roles in many biological processes of living organisms.  Beyond their separate biological roles, however, chemists have also discovered that amino acids are able to function as chelates with Cu²⁺ ions to form blue complexes with potential medicinal applications.  One such complex, copper(II) glycinate, has shown to be effective against some oral diseases.  The reaction to form copper(II) glycinate is shown in Figure 1. Figure 1  Synthesis of two isomers of copper(II) glycinateTwo different isomers can form depending on the reaction conditions.  Mixing hot aqueous glycine (70 °C) with a hot solution of copper(II) acetate dissolved in 50% aqueous ethanol, followed by cooling the mixture on ice, yields a precipitate of only the cis isomer.  If a suspension of the cis isomer is held at boiling for an hour or longer in a solution of glycine and copper(II) acetate, the trans isomer is obtained.Because of the geometric differences between the two isomers, the stretching vibrations of the central Cu-N and Cu-O bonds in each isomer are different.  Symmetric and asymmetric bond stretching give four possible vibrational modes for each isomer, as shown in Figure 2. Figure 2  Vibrational modes of the coordinate bonds in copper(II) glycinateIf the stretching of the bonds during vibration causes a transient, unequal distribution of electron density between the atoms within the bond, an oscillating dipole will form.  Bonds with dipoles that are not cancelled out by another dipole in the molecule will absorb infrared light at particular frequencies (often expressed in wave numbers) and are designated as IR-active.  As a result, the vibrational differences between the two isomers can be analyzed by far-infrared (far-IR) spectroscopy, as shown in Figure 3. Figure 3  Far-IR spectra of the cis and trans isomers of copper(II) glycinateBoth isomers show a skeletal absorbance (unrelated to coordinate bond vibration) near 385 cm⁻¹.  Additional absorbance bands are seen that specifically relate to the vibrational modes of the bonds in each isomer. -In the reaction to form the cis isomer shown in Figure 1, which of the following structural sites function(s) as a Lewis base?

Passage The alkali metals are a highly reactive family of metals occupying Group 1 (1A) of the periodic table.  In their elemental form, the alkali metals have a shiny, metallic appearance, but they are also very soft and can be easily cut by hand.  Due to their high reactivity, however, the alkali metals do not exist in their elemental form in nature.Because the alkali metals possess only a single valence electron with a low first ionization energy, the alkali metals (M) readily react with nonmetals and lose this electron to form cations with a +1 charge (M⁺) that are incorporated into ionic compounds.  Consequently, elemental samples of alkali metals will rapidly tarnish in oxygen by forming oxides (Reaction 1) and vigorously react with elemental halogens (X₂) to form halide salts (Reaction 2).2 M(s) + O₂(g) → M₂O(s)Reaction 12 M(s) + X₂(g) → 2 MX(s)Reaction 2Alkali metals also react violently with water to form very basic metal hydroxide (alkali) solutions while releasing hydrogen gas in the process (Reaction 3).2 M(s) + 2 H₂O(l) → 2 MOH(aq) + H₂(g)Reaction 3As demonstrated by these reactions, alkali metals possess a low electron affinity that makes accepting an additional electron to form an anion less favorable than losing the valence electron to form a cation.Of the alkali metals, francium is the rarest since it is a trace radioactive byproduct formed from the decay of uranium ores and has a very short half-life.  As a result, any francium formed will quickly decay into radium or astatine by beta decay or by alpha decay, respectively.Whereas hydrogen is a part of Group 1 based on its number of valence electrons, hydrogen is not an alkali metal.  Interestingly, some research has suggested that under extreme pressures and very low temperatures, hydrogen may form a solid with metallic properties. -If elemental hydrogen forms a metallic solid under extreme pressure, as mentioned in the passage, the chemical properties of the metallic hydrogen formed should be similar to those of which element?

Passage Gastric acid is one of many acids important to biological function in the human body.  Production of the main component of gastric acid, hydrochloric acid (HCl), is accomplished in the stomach by the parietal cells, as shown in Figure 1. Figure 1  Formation of gastric acid in the stomachFacilitated by the carbonate dehydratase enzyme (1), the parietal cells utilize the interaction between aqueous carbon dioxide and water in the blood as a source of bicarbonate ions and acidic H⁺ ions (protons).  The H⁺ ions produced in this process are then exchanged for potassium ions through the cell membrane by a proton pump, the H⁺/K⁺-ATPase (2).  In conjunction with this proton exchange, bicarbonate ions are also exchanged for chloride ions from the interstitial fluid.  As a result, H⁺ and Cl⁻ ions are concentrated in the lumen of the stomach to form aqueous hydrochloric acid.Because the production of the acid proceeds with a drop in pH from 7.0 in the cell cytosol to 1.0 in the stomach lumen, the process occurs against a concentration gradient and is endergonic, requiring the input of energy supplied by the hydrolysis of ATP.  Ignoring effects from ion charge, the free energy change for ion transport across the cell membrane (ΔG_t) can be estimated by Equation 1: ∆Gt=kRT logCfCiEquation 1where R is the gas constant (8.32 J/mol·K), k is a logarithm conversion constant, T is the absolute temperature, and C_i and C_f are the initial and final molar concentrations across the gradient, respectively.Although dilute, the gastric (hydrochloric) acid released into the lumen yields a strongly acidic environment crucial to protein digestion and the destruction of harmful micro-organisms.  A comparison of its acid dissociation constant (K_a) to those of other biologically relevant organic acids is given in Table 1.Table 1  Comparison of Strengths of Selected 0.10 M Acids at 25°C Adapted from Koolman J, Roehm KH. Color Atlas of Biochemistry. Thieme; 2011: 268-271; and Lehninger AL, Nelson DL, Cox MM. Lehninger Principles of Biochemistry. Macmillan; 2005: 397-398, 419, 520. -Based on the acid comparison in Table 1, which of the following organic acids will give a 0.10 M aqueous solution with the lowest pH?

Passage The alkaline-earth metals are a highly reactive family of metals occupying Group 2 (IIA) of the periodic table.  In their elemental form, the alkaline-earth metals have a shiny, silvery-white to slightly yellowish appearance.  The alkaline-earth metals exhibit some similarities to the chemistry of the alkali metals (Group 1) but are a little less reactive.Atoms of the alkaline-earth metals (M) have only two valence electrons and a low first and second ionization energy.  Consequently, these metals readily lose both electrons to produce cations with a +2 charge (M⁺²) during reactions with nonmetals.  As a result, the alkaline-earth metals do not exist in their elemental metallic form in nature.  Instead, the Group 2 elements are found primarily as metal oxides, which form by reacting oxygen (Reaction 1). Similarly, elemental samples of alkaline-earth metals will react with elemental halogens (X₂) to form metal halides (Reaction 2), all of which are ionic except for BeCl₂. Except for beryllium, the alkaline-earth metals react with water to produce very basic metal hydroxide solutions and hydrogen gas (Reaction 3). The most difficult alkaline-earth metal to study is radium, because it is a rare radioactive element that is produced from the nuclear decay of uranium.  The most common natural isotope, radium-226, has a half-life of 1,600 years and decays by alpha emission into radon-222 (Reaction 4). -The reaction responsible for the conversion of Compound 1 into Compound 3 is:

Passage Gastric acid is one of many acids important to biological function in the human body.  Production of the main component of gastric acid, hydrochloric acid (HCl), is accomplished in the stomach by the parietal cells, as shown in Figure 1. Figure 1  Formation of gastric acid in the stomachFacilitated by the carbonate dehydratase enzyme (1), the parietal cells utilize the interaction between aqueous carbon dioxide and water in the blood as a source of bicarbonate ions and acidic H⁺ ions (protons).  The H⁺ ions produced in this process are then exchanged for potassium ions through the cell membrane by a proton pump, the H⁺/K⁺-ATPase (2).  In conjunction with this proton exchange, bicarbonate ions are also exchanged for chloride ions from the interstitial fluid.  As a result, H⁺ and Cl⁻ ions are concentrated in the lumen of the stomach to form aqueous hydrochloric acid.Because the production of the acid proceeds with a drop in pH from 7.0 in the cell cytosol to 1.0 in the stomach lumen, the process occurs against a concentration gradient and is endergonic, requiring the input of energy supplied by the hydrolysis of ATP.  Ignoring effects from ion charge, the free energy change for ion transport across the cell membrane (ΔG_t) can be estimated by Equation 1: ∆Gt=kRT logCfCiEquation 1where R is the gas constant (8.32 J/mol·K), k is a logarithm conversion constant, T is the absolute temperature, and C_i and C_f are the initial and final molar concentrations across the gradient, respectively.Although dilute, the gastric (hydrochloric) acid released into the lumen yields a strongly acidic environment crucial to protein digestion and the destruction of harmful micro-organisms.  A comparison of its acid dissociation constant (K_a) to those of other biologically relevant organic acids is given in Table 1.Table 1  Comparison of Strengths of Selected 0.10 M Acids at 25°C Adapted from Koolman J, Roehm KH. Color Atlas of Biochemistry. Thieme; 2011: 268-271; and Lehninger AL, Nelson DL, Cox MM. Lehninger Principles of Biochemistry. Macmillan; 2005: 397-398, 419, 520. -Based on the biological system described in the passage, how many millimoles of hydrogen ions are present in 10.0 mL of gastric acid?

Osmolarity, the number of moles of osmotically active species (osmoles) per liter of solution (Osmol/L), equals the molar solute concentration M multiplied by the van 't Hoff factor i.  Osmolarity is calculated by dividing osmotic pressure Π by RT, where R is the gas constant and T is the absolute temperature.If the osmotic pressure of blood plasma is 6.9 atm at 37 °C, what is its osmolarity?  (Note:  Assume R = 0.08 L·atm/mol·K, and i = 2.)

Passage An automated external defibrillator (AED) is a medical device used to send an electric shock to the heart after cardiac arrest.  A key component of the AED is the power source, or battery.  Batteries used in AEDs need to have a good charge-to-weight ratio; they must be safe and reliable as well as rechargeable.Two types of rechargeable batteries used in early models of AEDs are the lead storage (also called lead-acid) battery and the nickel-cadmium (NiCd) battery.  These batteries consist of multiple electrochemical cells that are connected in series to deliver a potential between 9 V and 18 V.  A capacitor allows the AED to accumulate charge so that it can deliver between 300 V and 1,000 V.Some AEDs use a sealed lead storage battery.  Lead storage batteries are robust and hold a charge for a long time.  However, they have a low energy-to-weight ratio.  Each lead storage cell delivers approximately 2.0 V, and a battery of four cells weighing 1000 g can provide 30 W∙h of energy. Figure 1  Electron flow in a lead storage battery when discharging and chargingThe half reactions for the anode and cathode of a lead storage battery in 4 M of sulfuric acid (H₂SO₄) are shown in Reactions 1 and 2: Reaction 1 Reaction 2 NiCd batteries have a higher energy-to-weight ratio than lead storage batteries but cannot hold as much charge.  Each NiCd cell delivers approximately 1.3 V, and a single-cell battery weighing 120 g can provide 7.2 W∙h of energy.The half reactions at the anode and cathode for a NiCd battery in KOH are shown in Reactions 3 and 4: Reaction 3 Reaction 4 -Which statement correctly describes the Pb electrode when the lead storage battery is charging as shown in Figure 1?

A reaction was performed in which compounds A and B were combined to form compound C.  The table below shows an analysis of the kinetics for the reaction with changing reagent concentration. Based on the kinetic data shown in the table, which overall reaction order and rate law best describe this reaction?

Passage Heme is found in several proteins, including hemoglobin in blood and myoglobin in muscle.  It has a four-coordinate iron-bound porphyrin structure where ferrous iron (Fe²⁺) is bound to four nitrogen atoms through coordinate covalent bonds.  The most common heme found in the body is heme b in hemoglobin (Figure 1). Figure 1  Structure of heme b found in hemoglobinAlong with the four nitrogen ligands in the porphyrin structure, His F8 also binds to Fe²⁺ in the axial position, leaving the other axial position available to bind oxygen.  In this setting, iron's atomic d orbitals are no longer degenerate.  The energetic difference between the d orbitals depends on the ligands.In deoxygenated heme, iron sits out of the plane formed by the nitrogen ligands in the porphyrin because its d orbitals make it slightly too large to fit in the porphyrin ring.  However, when O₂ binds to heme, iron's electrons re-arrange themselves.  The d orbitals become smaller, and iron sits within the plane of the nitrogen atoms.When iron is scarce in the body, Zn²⁺ will coordinate to the ligands in the protoporphyrin structure through a nonenzymatic process, forming Zn-protoporphyrin IX (ZPP).  ZPP and protoporphyrin IX (PPIX) have distinct emission spectra when excited in the blue range whereas heme does not.  Hematofluorometry can be used as a simple, inexpensive indicator for iron deficiency in whole blood by placing a blood sample in a fluorometer and measuring the amount of ZPP relative to the amount of heme.  Figure 2 illustrates the fluorescence emission spectra for ZPP and protoporphyrin IX measured from a blood sample when ZPP is excited at 425 nm and PPIX is excited at 407 nm. Figure 2  ZPP and PPIX fluorescence emission spectra Adapted from Hennig G, Gruber C, Vogeser M, et al. Dual-wavelength excitation for fluorescence-based quantification of zinc protoporphyrin IX and protoporphyrin IX in whole blood. J Biophotonics. 2014;7(7):514-24. -Which of the following statements best explains why hemoglobin is red when it binds to O₂?

Passage Umami is one of five basic tastes that humans recognize, and it is triggered by L-amino acids as well as some synthetic ingredients and organic acids.  Three common umami tastant molecules are shown in Figure 1. Figure 1  Three common tastant molecules known to trigger the umami taste sensationUmami taste signals are triggered upon binding of tastant molecules to the G protein-coupled receptor (GPCR) heterodimer T1R1-T1R3, which ultimately causes the breakdown of phosphatidylinositol 4,5-bisphosphate (PIP₂) into the messenger molecules inositol-1,4,5-trisphosphate (IP₃) and diacylglycerol (DAG).  IP₃ triggers the release of Ca²⁺ cations and the subsequent gating of the taste-transduction channels (Figure 2). Figure 2  Signal pathway for transduction of umami taste sensationInosine monophosphate (IMP) has been suggested as enhancing (potentiating) the response of T1R1+T1R3 GPCRs to umami tastant molecules.  Scientists tested this hypothesis by culturing receptors from human embryonic kidney (HEK) cells containing promiscuous G proteins.  The HEK cells were then stimulated with umami tastants in the presence or absence of IMP.  In each test, the cells were exposed to the same amounts of amino acid tastant and IMP.  The number of cells that responded to each tastant was measured using a calcium indicator dye (FURA-2AM) in conjunction with fluorescence microscopy.  By comparing the cellular response ratio of the stimulated to unstimulated HEK cells, the extent of potentiation can be assessed.  The results are shown in Figure 3. Figure 3  Quantification of amino-acid responses for T1R1+T1R3 with and without IMP Adapted from Nelson G, Chandrashekar J, Hoon MA, et al. An amino-acid taste receptor. Nature. 2002;416(6877):199-202. -What is the hybridization of the numbered atoms in the tastant molecule L-AP4 shown?

An element that exists in its standard state as a diatomic gas and contains a double covalent bond is:

Passage Nuclear medicine uses radiopharmaceuticals for disease treatment and as tracers in medical imaging studies.  Yttrium-90 (⁹⁰Y) is a radiopharmaceutical agent used to treat overgrown joint lining known as pigmented villonodular synovitis, as well as some forms of liver cancer.  It has a half-life of about 64 hours.Although the relatively short half-life of ⁹⁰Y is optimal for use in medical applications, transportation and storage of the radiopharmaceutical are not feasible.  As a result, strontium-90 (⁹⁰Sr), with a half-life of 28.8 years, is the most common source of ⁹⁰Y.  ⁹⁰Sr is created by a nuclear fission process that begins with uranium-235 (²³⁵U) in a nuclear reactor as shown in Reaction 1.²³⁵U → ⁹⁰Sr + ZReaction 1Fission of ²³⁵U produces ⁹⁰Sr as well as additional fission products (Z) including, but not limited to, technetium-99 (⁹⁹Tc), iodine-129 (¹²⁹I), and zirconium-93 (⁹³Zr).  Figure 1 shows the distribution of fission products. Figure 1  Distribution of ²³⁵U fission products by atomic weightTo create the final ⁹⁰Y required for nuclear medicine studies, ⁹⁰Sr decay is carried out in a controlled ⁹⁰Y generator.  ⁹⁰Y is then separated from residual ⁹⁰Sr for use in clinical applications. Adapted from Wheeler CE. Comments on vaccines, August 1987. J Am Acad Dermatol. 1988;18(1 Pt 2):232-4. -Fission of ²³⁵U produces multiple elements.  Based on Figure 1, which of the following is the atomic number of the most common fission product?

Passage Combustion occurs when an oxidation-reduction reaction takes place between a reduced fuel source and an oxidizer, most frequently oxygen.  The reaction between hydrogen and oxygen forms water (Reaction 1) and is highly exothermic and thermodynamically favorable.  At the completion of hydrogen combustion, essentially all H₂ molecules have been converted to water.  Nevertheless, at room temperature and atmospheric pressure, molecular hydrogen and oxygen can coexist quite stably.2 H₂ + O₂ → 2 H₂OReaction 1A group of researchers designed two prototype internal combustion engines, each of which could inject a hydrogen/oxygen mixture into a 5-mL combustion chamber at a total pressure of 1 atm and could repeat this process for several cycles.  In one engine design, a spark was generated to initiate combustion as hydrogen and oxygen were injected.  In another method, the combustion chamber was coated with a small amount of platinum powder.  Platinum dramatically increases the reaction rate at room temperature without the need for a spark.  The platinum itself is not altered by the reaction and can be reused.  Researchers measured the maximum power output of both engine designs under identical injection and temperature conditions (Figure 1). Figure 1  Relative power outputs of hydrogen engines with different ignition sources at 298 KThe power output of an engine is directly related to the rate of combustion.  In the presence of platinum, researchers measured the engine's rate of water production at various temperatures and H₂/O₂ mixture compositions.  The results are shown in Figure 2. Figure 2  Rate of combustion in platinum-coated engines under various conditions Adapted from Schultze M, Mantzaras J. Hetero-/homogeneous combustion of hydrogen/air mixtures over platinum: Fuel-lean versus fuel-rich combustion modes doi: 10.1016/j.ijhydene.2013.06.069 -At a given temperature and injection composition, if researchers allowed each engine to carry out only one complete combustion reaction, which engine would produce more water?

Passage Dmitri Mendeleev is credited with grouping the elements by atomic mass and observing that similar properties emerge periodically.  Henry Moseley then determined that a fundamental property, which Ernest Rutherford later called the atomic number, is more important than atomic mass for ordering the elements.As the periodic table took shape, scientists noted certain properties and trends.  For example, Rayleigh and Ramsay observed that some elements are unreactive and gave them the name noble gases.  But it was not until Niels Bohr described his model of the atom that scientists understood why noble gases are stable.Another important trend was discovered by Linus Pauling.  He was the first to identify electronegativity as a way to describe bonds that are neither completely ionic nor completely covalent.  Pauling assigned fluorine the highest electronegativity value of 4.0 because of its strong tendency to attract electrons when bonded to other atoms.  All other atoms were assigned values relative to fluorine.  This became known as the Pauling electronegativity scale.Scientists identified other trends based on x-ray crystallography and reactivity studies.  Figure 1 shows the first ionization energy of the elements, and Figure 2 shows electron affinity. Figure 1  First ionization energy of elements Figure 2  Electron affinity of several elements (larger negative values indicate greater electron affinity) -Oxygen is the most electronegative element of Group 6A.  However, it also has the lowest electron affinity in the group.  What factor can be used to explain this discrepancy?

Passage The alkali metals are a highly reactive family of metals occupying Group 1 (1A) of the periodic table.  In their elemental form, the alkali metals have a shiny, metallic appearance, but they are also very soft and can be easily cut by hand.  Due to their high reactivity, however, the alkali metals do not exist in their elemental form in nature.Because the alkali metals possess only a single valence electron with a low first ionization energy, the alkali metals (M) readily react with nonmetals and lose this electron to form cations with a +1 charge (M⁺) that are incorporated into ionic compounds.  Consequently, elemental samples of alkali metals will rapidly tarnish in oxygen by forming oxides (Reaction 1) and vigorously react with elemental halogens (X₂) to form halide salts (Reaction 2).2 M(s) + O₂(g) → M₂O(s)Reaction 12 M(s) + X₂(g) → 2 MX(s)Reaction 2Alkali metals also react violently with water to form very basic metal hydroxide (alkali) solutions while releasing hydrogen gas in the process (Reaction 3).2 M(s) + 2 H₂O(l) → 2 MOH(aq) + H₂(g)Reaction 3As demonstrated by these reactions, alkali metals possess a low electron affinity that makes accepting an additional electron to form an anion less favorable than losing the valence electron to form a cation.Of the alkali metals, francium is the rarest since it is a trace radioactive byproduct formed from the decay of uranium ores and has a very short half-life.  As a result, any francium formed will quickly decay into radium or astatine by beta decay or by alpha decay, respectively.Whereas hydrogen is a part of Group 1 based on its number of valence electrons, hydrogen is not an alkali metal.  Interestingly, some research has suggested that under extreme pressures and very low temperatures, hydrogen may form a solid with metallic properties. -The graph below illustrates the radioactive decay of a population of francium-223 atoms.  How long did it take for two half-lives of francium-223 to elapse?

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)For the balanced reaction above, how many grams of nitrogen monoxide are produced when 16.0 g of oxygen react?

A sodium atom in an excited state has an electron configuration of 1s²2s²2p⁶3p¹.  The excited electron within the configuration was excited from the: