The Density of Gas - Assignment Example

Written Assignment 6 A gas occupying a volume of 725 mL at a pressure of 0.970 atm is allowed to expand at constant temperature until its pressure reaches 0.541 atm. What is its final volume? Solution: Using Boyle’s Law = (assuming ideal gas) = = = 1,299.91 mL ≈ 1.30 * 103 mL (2) The volume of a gas is 5.80 L, measured at 1.00 atm. What is the pressure of the gas in mmHg if the volume is changed to 9.65 L? (The temperature remains constant). Solution: Using Boyle’s Law = (assuming ideal gas) = = = 0.601 atm or 457 mmHg (3) Why is the density of a gas much lower than that of a liquid or solid under atmospheric conditions? What units are normally used to express the density of gases? Density generally depends on temperature and gas, having more dispersed molecules will weigh lighter compared to either liquid or solid for density is directly proportional to mass. Since density refers to mass per unit volume, density of gases is normally expressed in units g/mL, g/cm3, kg/m3, etc. (4) A sample of nitrogen gas kept in a container of volume 2.3 L and at a temperature of 32˚C exerts a pressure 4.7 atm. Calculate the number of moles of gas present. Solution: Using ideal gas equation PV = nRT, number of moles, n = = = 0.43 mole (5) What volume will 5.6 moles of sulfur hexafluoride () gas occupy if the temperature and pressure of the gas are 128˚C and 9.4 atm? Solution: Using ideal gas equation PV = nRT, volume, V = = = 19.6 L (6) A gas-filled balloon having a volume of 2.50 L at 1.2 atm and 25˚C is allowed to rise to the stratosphere (about 30 km above the surface of Earth), where the temperature and pressure are -23˚C and 3.00 x 10-3 atm, respectively. Calculate the final volume of the balloon. Solution: By combined gas law, = = = = 839 L or 840 L (2 sig.figs) (7) A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0˚C. (a) Calculate the density of the gas in grams per liter. (b) What is the molar mass of the gas? Solution: (a) Density = = = 2.21 x 10-3 g/mL (b) ρ = MM·P/(RT) --- MM = MM = = 54.6 g/mol (8) A certain anaesthetic contains 64.9 percent carbon, 13.5 percent hydrogen, and 21.6 percent oxygen by mass. At 120˚C and 750 mmHg, 1.00 L of the gaseous compound weighs 6.90 g. What is the molecular formula of the compound? Solution: molecular weight = MW = 225.685 g/mol C = 64.9 / 12.00 = 5.41 moles ; H = 13.5 / 1.00 = 13.5 moles ; O = 21.6 / 16.00 = 1.35 moles Dividing by the least #moles, C = 5.41 / 1.35 ≈ 4 ; H = 13.5 / 1.35 = 10 ; O = 1.35 / 1.35 = 1 Thus, empirical formula is whose empirical mass is about 74 g/mol N = MW / FW = 225.685 / 74 ≈ 3 --- to be multiplied by 4:10:1 So that, molecular formula must be (9) What is the mass of the solid NH4Cl formed when 73.0 g of NH3(g) are mixed with an equal mass of gaseous HCl? What is the volume and identity of the gas remaining, measured at 14.0˚C and 752 mmHg? ( NH3 (g) + HCl (g) --- NH4Cl ) Solution: 73.0 g NH3 * * * = 156.3 g HCl needed and this is greater than 73.0 g HCl actually present, so that HCl is the limiting reactant. amount of NH3 (in excess) used = 73.0 g HCl * * * = 34.1 grams Remaining amount of NH3 = 73.0 g - 34.1 g = 38.9 grams NH3 remaining moles NH3 = 38.9 g NH3 * = 2.284 moles NH3 volume of NH3 = = = 54.4 L NH3 (10) A mixture of gases contains 0.31 mol CH4, 0.25 mol C2H6, and 0.29 mol C3H8. The total pressure is 1.50 atm. Calculate the partial pressures of the gases. Solution: total # moles = 0.31 + 0.25 + 0.29 = 0.85 mole Partial pressure of a gas component = PT * P (CH4) = 1.50 * = 0.55 atm ; P (C2H6) = 1.50 * = 0.44 atm and P (C3H8) = 1.50 * = 0.51 atm (11) Propane (C3H8) burns in oxygen to produce carbon dioxide gas and water vapour. (a) Write a balanced equation for this reaction. (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from 7.45 g propane. Solution: (a) + --- + (b) 7.45 g C3H8 * * = 0.5068 mole C3H8 V = = = 11.37 L (12) A 10.0 g-piece of pure aluminium is placed in 75.0 mL of 0.54 M HCl at STP condition. They react as follows: 2 Al + 6 HCl --- 3 H2 (g) + 2 AlCl3 Calculate the following: (a) Volume in liters of H2 HCl is the limiting reactant, so 0.075 L * 0.54 M = 0.0405 mole HCl 0.0405 mole HCl * = 0.02025 mole H2 and by ideal gas equation, V = = 0.454 L (b) Molarity of Al+3. (Assume 75.0 mL solution) 0.0405 mole HCl * = 0.0135 mole --- molarity of Al+3 = 0.0135 mole / 0.075 L = 0.18 M (c) Molarity of Cl-. (Assume 75.0 mL solution) 0.0405 mole HCl * = 0.0405 mole Cl- Molarity of Cl- = 0.0405 mole / 0.075 L = 0.54 M Written Assignment 7 (1) Consider this reaction (a) The reaction is exothermic (b) 2 * (-1452.8 kJ/mol) = - 2,905.6 kJ/mol (c) The value would be +1452.8 kJ/mol (d) ΔH˚f (H2O, g) = -241.818 kJ/mol while ΔH˚f (H2O, l) = -285.8 kJ/mol --- -1452.8 + 4(285.8) - 4(241.818) = -1,276.9 kJ/mol (2) The first step in the industrial recovery of zinc from the zinc sulphide ore is roasting, that is, the conversion of ZnS to ZnO by heating: --- ΔH = -879 kJ/mol Calculate the heat evolved (in kJ per gram) of ZnS roasted. Solution: molar mass of ZnS = 97.474 g/mol heat evolved = -879 * = -9.02 kJ/g (3) A 6.22 kg piece of copper metal is heated from 20.5˚C to 324.3˚C. Calculate the heat absorbed (in kilojoules) by the metal. Specific heat Cu = 0.385 J/g˚C. Heat absorbed, Q = m * s * ΔT Q = 6.22 * 103 g * (0.385 J/g˚C) * (324.3 - 20.5)˚C * (1 kJ/1000 J) = 728 kJ (4) A sheet of gold weighing 10.0 g and at a temperature of 18.0 ˚C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 ˚C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings. Specific heat Au = 0.129 J/g˚C and Specific heat Fe = 0.444 J/g˚C Solution: Q(system) + Q(surroundings) = 0 --- Q(Au) + Q(Fe) + 0 = 0 (10.0 g) (0.129 J/g-C) (TF - 18.0˚C) + (20.0 g) (0.444 J/g-C) (TF - 55.6˚C) = 0 Arranging the equation and solving for final temperature TF, TF = 50.8˚C (5) A 0.1375 g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/˚C. The temperature increases by 1.126˚C. Calculate the heat given off by the burning Mg, in kJ/g and in kJ/mol. Solution: Heat given off = = 24.76 kJ/g And 24.76 kJ/g * (24.305 g / 1 mole) = 1.019 kJ/mole (6) Calculate the heat of decomposition for this process at constant pressure and 25˚C. --- CaO (s) + CO2 (g) Solution: = -635.6 kJ/mol + (-393.5 kJ/mol) - (-1206.9 kJ/mol) = 177.8 kJ/mol (7) Methanol, ethanol, and n-propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is liberated as shown by the following data (). Calculate the heats of combustion of these alcohols in kJ/mol. (a) Methanol (CH3OH), -22.6 kJ/g ; molar mass (CH3OH) = 32.04 g/mol Heat of combustion = -22.6 kJ/g * 32.04 g/mol = -724 kJ/mol (b) Ethanol (C2H5OH), -29.7 kJ/g ; molar mass (C2H5OH) = 46.07 g/mol Heat of combustion = -29.7 kJ/g * 46.07 g/mol = -1,368 kJ/mol (c) N-propanol (C3H7OH), -33.4 kJ/g ; molar mass (C3H7OH) = 60.095 g/mol Heat of combustion = -33.4 kJ/g * 60.095 g/mol = -2,007 kJ/mol (8) From the standard enthalpies of formation, calculate ΔH˚rxn for the reaction C6H12 (l) + 9 O2 (g) --- 6 CO2 (g) + 6 H2O (l) ΔH˚rxn = 6*(-393.5 kJ/mol) + 6*(-285.8 kJ/mol) - [ 0 + (-151.9 kJ/mol) ] ΔH˚rxn = -3,924 kJ/mol (9) Determine the amount of heat (in kJ) given off when 1.26 * 104 g of ammonia are produced according to the equation N2 (g) + 3 H2 (g) --- 2 NH3 (g) ΔH˚ = -92.6 kJ/mol Assume that the reaction takes place under standard conditions at 25˚C. Solution: 1.26 * 104 g NH3 * = 739.8 moles Thus, amount of heat given off = -92.6 kJ/mol * (739.8 moles) = -6.85 * 104 kJ (10) From the following heats of combustion, calculate the enthalpy of formation of methanol (CH3OH) from its elements C (graphite) + 2 H2 (g) + ½ O2 (g) --- CH3OH (l) Solution: Reverse 1st equation: CO2 (g) + 2 H2O (l) --- CH3OH (l) + 3/2 O2 (g) ΔH˚rxn = 726.4 kJ/mol Retain 2nd equation: C (graphite) + O2 (g) --- CO2 (g) ΔH˚rxn = -393.5 kJ/mol Multiply 3rd equation by 2: 2H2 (g) + O2 (g) --- 2 H2O (l) ΔH˚rxn = -571.6 kJ/mol Therefore, ΔHf (CH3OH) = 726.4 kJ/mol + (-393.5 kJ/mol) + (-571.6 kJ/mol) ΔHf (CH3OH) = -238.7 kJ/mol (11) Methanol (CH3OH) is an organic solvent and is also used as a fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol. 2 CH3OH (l) + 3 O2 (g) --- 2 CO2 (g) + 4 H2O (l) ΔH˚rxn = -1,452.8 kJ/mol Solution: 2(-393.5 kJ/mol) + 4(-285.8 kJ/mol) - [ 2*ΔH˚f (CH3OH) + 3*(0) ] = 1,452.8 kJ/mol --- ΔH˚f (CH3OH) = -1,691.5 kJ/mol (12) A 44.0 g sample of unknown metal at 99.0˚C was placed in a constant-pressure calorimeter, with a heat capacity of 12.4 J/˚C, containing 80.0 g of water at 24.0˚C. The final temperature of the system was found to be 28.4˚C. Calculate the specific heat of the metal. Solution: Q(metal) = - [ Q(H2O) + Qcal ] (m * s * ΔT)metal = - [ (m * s * ΔT)H2O + Ccal * ΔT ] (44.0 g)(s)(28.4˚C - 99.0˚C) = - [(80.0 g)(4.184 J/g˚C) + 12.4 J/˚C](28.4˚C - 24.0˚C) Solving for the metal’s specific heat ‘s’ --- s = 0.492 J/g˚C (13) A person ate 0.50 pound of cheese (an energy intake of 4000 kJ). Suppose that none of the energy was stored in his body. What mass (in grams) of water would he need to perspire in order to maintain his original temperature? (It takes 44.0 kJ to vaporize 1 mole of water). Solution: 4,000 kJ * * = 1,637.75 grams ≈ 1.6 * 103 grams (14) The total volume of the Pacific Ocean is estimated to be 7.2 x 108 km3. A medium-sized atomic bomb produces 1.0 x 1015 J of energy upon explosion. Calculate the number of number of atomic bombs needed to release enough energy to raise the temperature of the water in the Pacific Ocean by 1˚C. Solution: Let ‘N’ = number of atomic bombs needed specific heat of seawater = 3.85 J/g˚C ; density of seawater = 1,025 kg/m3 N * (1.0 x 1015 J) = 7.2 x 108 km3 * * * * 3.85 J/g˚C * (1 ˚C) N = 2,841,300,000 atomic bombs Written Assignment 8 (1) The blue color of the sky results from the scattering of sunlight by air molecules. The blue light has a frequency of about 7.5 x 1014 Hz. (a) Calculate the wavelength in nm, associated with this radiation wavelength = = * = 400 nm (b) Calculate the energy, in joules, of a single photon associated with this frequency. photon energy, E = hf = (7.5 x 1014 s-1) = 4.97 x 10-19 J (2) (a) What is energy level? Explain the difference between ground state and excited state. An energy level refers to a definite stable energy which electrons in an atom can have or one of the stable states of constant energy which may be assumed by a physical system as either a ground state or an excited state, and based on quantum theory, only certain energy levels are possible. A ground state is the lowest energy state or the state in which the total energy of the electrons cannot be lowered by transferring one or more electrons to different orbitals. On the other hand, excited state pertains to the state of energy in a physical system in which electrons are brought to obtain energy higher than that of the ground level or into a higher-energy orbital. (3) Explain the statement: matter and radiation have “dual nature”. Considering ‘light’ which has a dual nature described by the wave model and the particle model, light as a matter consists of particles that undergo a change of state (solid, liquid, and gas) and exert pressure or force with associated properties as mass, volume, density, temperature, etc. On the other hand, light as a radiation possesses properties similar to that of a wave which has wavelength, frequency, amplitude, and energy either by absorption or emission. (4) Which of the four quantum numbers ( ) determine (a) The energy of an electron in a hydrogen atom and in a many-electron atom  n (b) The size of an orbital  n (principal quantum number) (c) The shape of an orbital  l (azimuthal quantum number) (d) The orientation of an orbital in space  (magnetic quantum number) (5) Determine the maximum number of electrons that can be found in each of the following subshells: 3s, 3d, 4p, 4f. 3s  2 electrons, 3d  10 electrons, 4p  6 electrons, 4f  14 electrons (6) How many electrons would fill the third energy level (n = 3) ? = = 18 electrons (7) Explain the meaning of the symbol . In an electronic configuration, this refers to the 4th energy level where the d-suborbital has 6 electrons at that level. (8) For each of the following pairs of subshells, indicate which is higher in energy: (a) 2p is higher than 2s (b) 4p is higher than 3d (c) 4p is higher than 4s (d) 3d is higher than 4s (9) Explain the meaning of diamagnetic and paramagnetic. Give an example of an element that is diamagnetic and one that is paramagnetic. What does it mean when we say that electrons are paired? Diamagnetic describes an element that is unaffected by magnetic fields and whose subshells are completed with both spins present (paired electrons only) whereas ‘paramagnetic’ describes an element influenced by magnetic fields so that its subshells are not completely filled (at least one unpaired electrons). Examples are Be (diamagnetic) and Li (paramagnetic). By Hund’s Rule of Multiplicity, pairing of electrons takes place when more electrons are allowed to occupy orbitals (of equal energy) that have been singly filled in with electrons prior. (10) Write the ground-state electron configuration for the following elements. (a) B  or [He] 2s2 2p1 (b) V  [Ar] 3d3 4s2 (c) Ni  [Ar] 3d8 4s2 (d) I  [Kr] 4d10 5s2 5p5 (11) The electron configuration of a neutral atom is . Name the element. The element is Magnesium (Mg). (12) A 368 g sample of water absorbs infrared radiation at 1.06 x 104 nm from a carbon dioxide laser. Suppose all the absorbed radiation is converted to heat. Calculate the number of photons at this wavelength required to raise the temperature of the water by 5.00˚C. Solution: frequency = c / λ = = 2.83 x 1013 Hz Q = m * s * ΔT = (368 mg) (4.18 J/g˚C) (5.00˚C) = 7,691.2 J E = hf = (6.626 x 10-34 J*s) (2.83 x 1013 s-1) = 1.875 x 10-20 J If n  number of photons, then Q = n*E N = = = 4.10 x 1023 photons (13) Draw the orbital diagrams for atoms with the following electron configurations. (a)  ____ ____ ____ ____ ____ (b)  ____ ____ ____ ____ ____ ____ ____ ____ ____ (c)  ____ ____ ____ ____ ____ ____ ____ ____ ____ ____ ____ ____ ____ ____ ____ (14) Define the following terms and principles: (a) Heisenberg Uncertainty Principle -- is the basis for the initial realization of fundamental uncertainties in the ability of an experimenter to measure more than one quantum variable at a time. Mathematically, it may be expressed as E  t    h / 4      or       x  p    h / 4  (b) Aufbau Principle -- refers to the principle used to determine the electron configuration of an atom or element as well as the distribution of electrons among energy levels in the ground (most stable) states of atoms. (c) Pauli’s Exclusion Principle -- is a principle stating that “no two electrons in an atom can have identical quantum numbers.” (d) Hund’s Rule -- pertains to the rule in which every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin. (15) What are the valence electrons? How many valence electrons do each of the following elements have: H  1 ; Ca  2 ; Al  3 ; N  5 ; O  6 ; Cl  7 ; He  2 ; Ne  8 (16) In the periodic table, the element hydrogen is sometimes grouped with the alkali metals and sometimes with the halogens. Explain why hydrogen can resemble the Group IA and the Group 7A elements. Hydrogen can be like a Group IA element for being capable of giving up a single electron in the outer shell to have a valence of +1. In the similar manner, hydrogen can also resemble a Group 7A element for its capacity to accept a single electron in the outer shell, forming -1 charge. (17) Group the following electron configurations in pairs that would represent similar chemical properties of their atoms. Groups: (a) and (d) (b) and (f) (c) and (e) (18) Write the ground-state electron configurations of the following ions: (a) Li+  or [He] (b) N-3  or [Ne] (c) S-2  or [Ar] (d) Al+3  or [Ne] (e) Ba+2  [Xe] (f) Pb+2  (19) Which of the following species are isoelectric with each other? C, Cl-, Mn+2, B-, Ar, Zn, Fe+3, Ge+2. Ar and Cl- ; Zn and Ge+2 ; C and B- ; Mn+2 and Fe+3 (20) On the basis of their positions in the periodic table, select the atom with the larger atomic radius in each of the following pairs: (a) Na, Cs --- Cs has larger atomic radius (b) Be, Ba --- Ba has larger atomic radius (c) N, Sb --- Sb has larger atomic radius (d) F, Br --- Br has larger atomic radius (21) Why do elements that have high ionization energies also have more positive electron affinities? Generally, which elements have the highest ionization energies? Which elements would have the lowest? Elements that have high ionization energies are apparently more stable and on this ground, thus, they have more positive electron affinities as well. Generally, the non-metallic elements or those lying at the upper right of the periodic table yield the highest ionization energies while the elements in the lower left of the periodic table yield the lowest ionization energies. Written Assignment 9 (1) Use the second member of each group from Group IA to Group 7A to show that the number of valence electrons on an atom of the element is the same as its group number. Under group IA  Na, when ionized, loses 1 valence e- to yield +1 charge (Na+) Under group IIA  Mg, when ionized, loses 2 valence e- to yield +2 charge (Mg+2) Under group IIIA  Al, when ionized, loses 3 valence e- to yield +3 charge (Al+3) Under group IVA  Si, no net charge Under group VA  P, has 5 valence e- since as an ion P-3, the value 3 completes the octet with 5 (upon sharing) Under group VIA  S, has 6 valence e- since as an ion S-2, the value 2 completes the octet with 6 (upon sharing) Under group VIIA  Cl, has 7 valence e- since as an ion C-1, the value 1 completes the octet with 7 (upon sharing) (2) Explain the following: (a) ionic bond -- a bond formed by a cation and an anion from metallic and non-metallic elements, respectively, upon electron transfer. (b) nonpolar covalent bond -- a bond formed between non-metallic elements upon electron sharing (with no overall dipole moment). (c) polar covalent bond -- a bond formed between non-metallic elements upon electron sharing (with net dipole moment) (3) Specify which compound in the following pairs of ionic compounds has the higher lattice energy. Explain your choice. (a) KCl or MgO  MgO since each ion has +2 charge (b) LiF or LiBr  LiF since Br is greater in size compared to F (c) or NaCl  NaCl since this salt makes a stronger ionic bond (4) Compare the properties of ionic compounds and covalent compounds. Ionic compounds are crystalline solids made of ions with high melting and boiling points. These compounds are able to conduct electricity when melted and many of them are soluble in water but not in nonpolar liquid. On the other hand, covalent compounds are normally gases, liquids, solids that are molecular by nature. Covalent compounds have low melting and boiling points as well as poor conductivity. Most of them dissolve in nonpolar liquids but not in water. (5) Four atoms are arbitrarily labelled D, E, F, and G. Their electronegativities are as follows: D = 3.8, E = 3.3, F = 2.8, and G = 1.3. If the atoms of these elements form the molecules DE, DG, EG, and DF, how would you arrange these molecules in order of increasing covalent bond character? DG < EG < DF < DE (6) Classify the following bonds as ionic, polar covalent, or nonpolar covalent, and give your reasons. (a) the CC bond in  nonpolar covalent since both ends of the C-C bond are exactly the same and would therefore create no net dipole moment. (b) the KI bond in KI  ionic bond as it dissociates into ions K+ and I- (c) the NB bond in H3NBCl3  polar covalent bond since there would be a net dipole moment whose overall direction is towards the right (from the least electronegative H to the most electronegative Cl atoms) (d) the CF bond in CF4  nonpolar covalent bond since CF4 forms a tetrahedral geometry with no net dipole moment. (7) Write the correct Lewis structures for the following compounds (relative positions of atoms are shown correctly). (a) (b) H C C H (c) O Sn O (d) (e) (f) H C F O (g) (8) Classify the following substances as ionic compounds or covalent compounds containing discrete molecules: CH4, KF, CO, SiCl4, BaCl2. CH4  covalent compound ; KF  ionic compound CO  covalent compound ; SiCl4  covalent compound BaCl2  ionic compound (9) Predict the geometries of the following species using the VSEPR method. (a) PCl3 --- trigonal pyramidal (b) CHCl3 --- tetrahedral (c) SiH4 --- tetrahedral (d) TeCl4 --- see-saw (10) Predict the geometry of the following molecules using the VSEPR method: (a) HgBr2 --- linear (b) N2O --- linear (c) SCN- --- linear (11) Describe the geometry around each of the three central atoms in the CH3COOH molecule. The first carbon (leftmost) --- tetrahedral The second carbon (middle) --- trigonal planar The O connected to H (rightmost) --- bent (12) What is the relationship between the dipole moment and bond moment? How is it possible to have bond moments and yet be nonpolar? A bond moment is a measure of the asymmetry of charge distribution in the bond and its immediate vicinity and is affected by the environment of the molecular portion of interest. In effect, bond moment determines which direction a dipole moment is taking. If the bond moments add up to zero such that dipoles cancel each other out, then the 0 overall charge results causing the molecule to be nonpolar. (13) Specify which hybrid orbitals are used by carbon atoms in the following species: (a) CO : sp – hybrid orbital (b) CO2: sp – hybrid orbital (c) CN-: sp – hybrid orbital (14) How many sigma bonds and pi bonds are there in each of the following molecules? (a) 4 sigma bonds (b) 5 sigma bonds and 1 pi bond (c) 7 sigma bonds and 2 pi bonds (15) Which of the following species is not likely to have a tetrahedral shape? SF4 is not likely to have a tetrahedral shape as it forms a trigonal bi-pyramidal. ----------------------- ENCODING ASSIGNMENT 10 ---------------- Read More