Chem 102 chapter 11

Which statement best describes the polarity of SCl4I2?

Depending on the arrangement of outer atoms, this molecule could be polar or nonpolar.

You are a researcher for a golf club manufacturer. You are given two identical looking cubes of a metal alloy. You are informed that they are made of the exact same material, but one is crystalline, whereas the other is amorphous. It is your job to determine which one is amorphous because this one is more stress-resistant and is useful in reinforcing golf clubs. Which of the following is the best way to determine which is which? Determine the density of each cube. The less dense cube is the amorphous solid. Melt both cubes and look for a broader range of melting temperatures. The one that melts over a broader range of temperatures is the amorphous solid. Melt both cubes and measure the range of melting temperatures. The one that melts over a narrower range of temperatures is the amorphous solid. Determine the density of each cube. The more dense one is the amorphous solid.

Determine the density of each cube. The less dense cube is the amorphous solid. It is also important to note that many crystals are made from more than one kind of atom. For example, proteins will form crystals under the right conditions, and proteins are more complex than a single metal atom. The unit cell concept applies to protein crystals as it does to single atoms. The unit cell may contain one or more proteins and weigh thousands of atomic mass units. Since it is in the form of a crystal, one of the unit cells will enable you to recreate the entire protein crystal. This fact is exploited when structural biologists perform X-ray crystallography to determine the atomic structures of proteins.

All three of the boron-fluorine single bonds in BF3 are polar. In which direction should the polarity arrows point?

point away from the central boron atom.

There are three main types of cubic unit cells:

primitive cubic (or simple cubic), body-centered cubic, and face-centered cubic.

The dipole moment (μ) of HBr (a polar covalent molecule) is 0.790D (debye), and its percent ionic character is 11.7 % . Estimate the bond length of the H−BrH−Br bond in picometers. Note that 1 D=3.34×10−30 C⋅m and in a bond with 100% ionic character, Q=1.6×10−19 C.

r = 140 pm The H−Br bond length is 140 pm, and its bond strength is 366 kJ/mol.

Polonium metal crystallizes in a simple cubic arrangement, with the edge of a unit cell having a length d = 334 pm. What is the radius in picometers of a polonium atom?

r = 167 pm

The distance between the atoms of H−Br is 1.41Å. What is the distance in meters?

see slide above

The dipole moment of HBr is 0.80D. What is the dipole moment of HBr in C⋅m?

μ = 2.67×10−30 C⋅m

Polonium metal crystallizes in a simple cubic arrangement, with the edge of a unit cell having a length d = 334 pm. What is the density of polonium in g/cm3?

ρ = 9.31 g/cm3

Which pair of molecules gives the species with the lowest viscosity first? CH3CH2CH2Cl and CH3CH2CH2CH3 CH3CH2OH and CCl4 CBr4 and CCl4 CH3—O—CH3 and CH3CH2OH

CH3—O—CH3 and CH3CH2OH

Tell which of the following compounds is likely to have a dipole moment, and show the direction of each. Which of the following compounds is likely to have a dipole moment? CHCl3 H2C=CH2 CH2Cl2 SF6

CHCl3 CH2Cl2

The vapor pressure of dichloromethane, CH2Cl2, at 0 ∘C is 134 mmHg. The normal boiling point of dichloromethane is 40. ∘C. Calculate its molar heat of vaporization.

Clausius-Claperyon equation: ln(P2/P1) = (-DHvap/R) x (1/T2 - 1/T1) Molar gas constant R = 8.314 J/mol.K T1 = 0 deg C = 273.15 K, P1 = 134 mmHg At normal boiling point T2 = 40 deg C = 313.15 K => P2 = atmospheric pressure = 760 mmHg Substituting values into equation: ln(760/134) = (-DHvap/8.314) x (1/313.15 - 1/273.15) Heat of vaporization = DHvap = 30855 J/mol = 30.9 kJ/mol

An unknown substance has a melting point of 735 ∘C, is soluble in water, does not conduct heat as a solid, and is hard. Given these properties which of the following are possible identities for the unknown substance? K Cdiamond CHCl3 CoCl2

CoCl2

The dipole moment of HF is μ = 1.83 D, and the bond length is 92 pm. Calculate the percent ionic character of the H−F bond. Is HF more ionic or less ionic than HCl (Worked Example 10.1 in the textbook)?

% ionic character = 41% more ionic than HCl

The dipole measured for HI is 0.380 D. The bond length is 161 pm. What is the percent ionic character of the HI bond?

%ionic character = 4.92 % The bond in an HI is 4.92% ionic in character. This means that the hydrogen atom carries a +0.0492 charge and the iodine atom carries a −0.0492 charge.

How much energy (in kilojoules) is absorbed when 10.0 g of liquid water at 75.0 °C is converted to water vapor at 150.0 °C? (The ΔH vap of water is 40.67 kJ/mol, and the molar heat capacity is 75.3 J/K·mol for the liquid and 33.6 J/K·mol for the vapor.)

(10.0 g H2O) / (18.01532 g H2O/mol) = 0.5551 mol H2O (75.3 J/K·mol) x (100.0 - 75.0)K x (0.5551 mol) = 1044.98 J to warm the liquid to 100°C ( 40.67 kJ/mol) x (0.5551 mol) = 22.576 kJ to vaporize the liquid (33.6 J/K·mol) x (150.0 - 100.0)K x (0.5551 mol) = 932.57 J to heat the steam to 150°C 1.04498 kJ + 22.576 kJ + 0.93257 kJ = 24.6 kJ total

What is the percent ionic character of the H−Br bond?

(see slide above) 11.8%

Consider the following two substances and their vapor pressures at 298 K. Substance Vapor pressure (mmHg) A 275 B 459 Based on this information, compare the characteristics of the two substances.

1)Substance A has the higher heat of vaporization. In other words, more heat has to be applied to get it to vaporize. We really can't tell which substance has the weaker intermolecular forces, because substance A may just be a much larger molecule with higher molecular weight than substance B. Substance A will have the higher boiling point, and therefore substance B will be a gas at 300mmHg pressure. 2)ln (P1/P2) = (?Hvap/R) (1/T2-1/T1) At the normal boiling point, the vapor pressure is 760 mm Hg. Convert your temperatures to Kelvin. Then, you have all the information you need to solve the question, so just plug in and solve for ?Hvap: ln (134/760) = ?Hvap/(8.3145 J/molK) ( 1/313K - 1/273K) ?Hvap = 3.08X10^4 J/mol = 30.8 kJ/mol

Calculate the number of atoms per unit cell in each type of cubic unit cell.

1,2,4 atoms per unit cell Simple cubic cell: 1/8 x 8 = 1 atom/cell Body centered cell: 1/8 x 8 + 1 = 2 atoms/cell Face cubic cell: 1/8 x 8 + 1/2 x 6 = 4 atoms/cell

How long would it take for 1.50 mol of water at 100.0 ∘C to be converted completely into steam if heat were added at a constant rate of 19.0 J/s ?

1.5mol water = 27g latent heat of vaporisation of water at 100°C to steam at 100°C is: Latent heat = 2258kJ/kg You have 27g, So heat required is 2258*27/1000 = 61kJ Heat is supplied at 19J/sec It will take 61000/19 = 3,210 seconds or 53.5 minutes.

Calculate the amount of energy in kilojoules needed to change 333 g of water ice at −10 ∘C to steam at 125 ∘C. The following constants may be useful: Cm (ice)=36.57 J/(mol⋅∘C) Cm (water)=75.40 J/(mol⋅∘C) Cm (steam)=36.04 J/(mol⋅∘C) ΔHfus=+6.01 kJ/mol ΔHvap=+40.67 kJ/mol

1st transition from -10 deg C 0 deg C q = m C delta T Delta T = final T - initial T , m is mass in g , C is specific heat =333 g x (1 mol/ 18.0148 g) x 36.57 J / mol deg C x ( 0 -(-10) deg C =6759.89 J 2nd Transition From 0 to 0 deg C q = Delta H(fus) x mol = 6.01 x 1000 J /mol x (333 / 18.0148) mol =111093.66 J 3rd from 0 to 100 deg q = 333 g x (1 mol/ 18.0148 g) x 75.40 J / mol deg C x ( 100 - 0) deg C =139375.4 J 4th from 100 to 100 deg C q = Delta Hvap x mol = 40.67 x 1000 J/mol x (333 /18.0148 ) mol =751776.87 J 5th from 100 to 125 q = 333 g x (1 mol/ 18.0148 g) x 36.04 J / mol deg C x ( 125-100 0 deg C = 16654.81 J Lets add q values from all transition Total q = 6759.89 J +111093.66 J +139375.4 J +751776.87 J +16654.81 J Total q = 1025660.63 J Answer in kJ = 1025.66 kJ

For future reference, you would want to draw the Lewis structure for each of these simple molecules, determine the geometry, and look to see if the molecule is symmetrical or not. A symmetrical molecule is nonpolar with a dipole moment of zero, while an asymmetrical molecule has a net dipole moment and is polar. You may have noticed that the compounds are listed in terms of increasing electron pair geometry complexity of the central atom, and with the molecular geometry that corresponds to the electron pair geometry.

2 electron pairs - linear electron pair geometry BeH2 - symmetrical - linear 3 electron pairs - trigonal planar electron pair geometry BF3 - symmetrical - trigonal planar SO2 - not symmetrical - bent 4 electron pairs - tetrahedral electron pair geometry CH4 - symmetrical - tetrahedral NH3 - not symmetrical - trigonal pyramidal H2O - not symmetrical - bent 5 electron pairs - trigonal bipyramidal electron pair geometry PCl5 - symmetrical - trigonal bipyramidal SCl4 - not symmetrical - see-saw ClF3 - not symmetrical - T-shaped XeF2 - symmetrical - linear 6 electron pairs - octahedral electron pair geometry SF6 - symmetrical - octahedral IF5 - not symmetrical - square pyramidal XeF4 - symmetrical - square planar

X-rays with a wavelength of 1.70 Å scatter at an angle of 36.5 ∘ from a crystal. If n=1, what is the distance between planes of atoms in the crystal that give rise to this scattering?

According to Bragg's law nλ = 2d sin θ Where n = 1 λ = wavelength of ligth used =1.70 Ao = 1.70 * 10^-10 m θ = glancing angle = 36.5 degrees d = distance between the palnes = ? Plug the values we get d = nλ / 2 sin θ = ( 1 * 1.70*10^-10 ) / ( 2*sin 36.5 ) = 1.43*10^-10 m d = 1.43 Å

± Vapor Pressure and Phase Changes

All liquids evaporate to a certain extent. The pressure exerted by the gas phase in equilibrium with the liquid is called vapor pressure, Pvap. The vapor pressure of a particular substance is determined by the strength of the intermolecular forces. But for any given substance, the vapor pressure only changes with temperature. The Clausius-Clapeyron equation expresses the relationship between vapor pressure and temperature: lnP2=lnP1+(ΔHvapR)(1T1−1T2) where P2 and P1 are the vapor pressures that correspond to temperatures T2 and T1, respectively, ΔHvap is the molar heat of vaporization, and R=8.3145J/(mol⋅K) is the gas constant.

Molecules at the surface of a liquid, if they possess enough kinetic energy, can escape to the gas state. As a result, a liquid will exert a vapor pressure. If the liquid is confined to a closed container, eventually the pressure reaches a constant value as a result of a dynamic equilibrium between molecules forming the gas state and those returning to the liquid state. The magnitude of the vapor pressure is determined by the strength of the intermolecular forces in the liquid and the temperature of the sample.

Atmospheric pressure decreases as altitude increases. In other words, there is more air pushing down on you at sea level, and there is less air pressure pushing down on you when you are on a mountain.

An unknown substance has a melting point of 1064 ∘C, is insoluble in water, conducts electricity as a solid, and is hard. Given these properties, which of the following are possible identities for the unknown substance? Au O2 SiO2 NaCl

Au

The unit cell of a crystal can be described in term of its edge length, l, and the number of atoms it contains. The edge length defines the volume of a cubic unit cell, V, as V=l3 The number of atoms per unit cell depends on the type of unit cell. For cubic unit cells, the position (body, face, or corner) of the atoms determines the fraction of the atoms that are completely contained within the unit cell.

Based on the number of atoms per unit cell and the mass of the atom, the mass m of the unit cell can be calculated. The density of the unit cell and the material as a whole can be determined from the mass mand the volume V of the unit cell as. density=mV The usual units of density are grams per cubic centimeter or g/cm3.

X-ray diffraction can be used to obtain structural information of crystalline compounds. X-ray wavelengths are about the same size as the space between atoms in solids. X-rays aimed at a crystal are diffracted by the atoms in the crystalline lattice. This results in an X-ray diffraction pattern, which can be seen on a detector placed behind the crystal. X-ray diffraction is caused by constructive and destructive interference of the X-ray waves interacting with the electron clouds of the atoms in the crystal. Waves that are in phase interfere constructively, reinforcing each other. Waves that are out of phase interfere destructively with each other, canceling each other out. The wave interference can be described mathematically by Bragg's equation: 2d×sin(θ) = nλ where : d s the distance between parallel planes of atoms in the crystal, θ is the angle at which the x-ray is hitting the material relative to the plane, n is a nonzero integer, and λ is the wavelength of the X-ray. note: The units of d and λ do not matter so long as they are the same.

Example: X-rays with a wavelength of 1.54062Å scatter at an angle of 12.251∘ from a copper crystal. If n=1, what is the distance between planes of atoms in the copper crystal that give rise to this scattering? Solution: We are given n, λ , and θ, and asked to solve for d. For an answer in units of Å, we do can calculate it directly. d = nλ2sin(θ) = (1×1.54062Å )/(2sin(12.251∘)) =3.63Å This is the distance between the planes in the bcc crystal structure of copper.

Gallium crystallizes in a primitive cubic unit cell. The length of an edge of this cube is 362 pm. What is the radius of a gallium atom?

For a cubic unit cell edge length = 2r 2r = a here a = edge of cell = 362 pm Hence r = a/2 = 181 pm

Which molecule has a dipole moment of 0 D? ClF3 NF3 BF3 OF2

For future reference, you would want to draw the Lewis structure for each of these simple molecules, determine the geometry, and look to see if the molecule is symmetrical or not. A symmetrical molecule is nonpolar with a dipole moment of zero, while an asymmetrical molecule has a net dipole moment and is polar. BF3

Which allotrope of carbon is correctly matched to the forces that hold the structural units together? Graphite : dispersion forces Graphite : hydrogen bonding Diamond : dipole-dipole interactions Fullerene : dipole-dipole interactions

Graphite : dispersion forces

A dipole moment tends to stabilize the liquid state of the compound as molecules align to form attractive molecular interactions. A liquid state that is more stable, that is one that is held together by stronger dipole forces, will have a higher boiling point since it takes more energy to break these intermolecular forces. Part B Rank the following compounds in order of decreasing boiling point: sodium chloride (NaCl), carbon tetrafluoride (CF4), and iodomethane (CH3I)

Highest boiling point sodium chloride ( NaCl ) iodomethane ( CH3I ) carbon tetrafluoride ( CF4 ) Lowest boiling point The ionic compound NaCl has the highest boiling point of the group, at 1413 ∘C. Iodomethane, CH3I, is asymmetric and polar, with a boiling point of 42 ∘C. The symmetric molecule CF4 has no net dipole moment. It is thus nonpolar and has a boiling point of -127 ∘C.

A molecule can be polar or nonpolar depending upon the nature of the bonds and the shape of the molecule. For a molecule that has different outer atoms the molecular symmetry will decide the polarity. If the molecular geometry is such that the dipole moments of each polar bond cancel each other then the molecule is nonpolar.

However, if the the molecular geometry is such that the dipole moments of each polar bond don't cancel each other then the molecule is polar.

In the molecule HCl, which atom is the negative pole?

Hydrochloric acid (HCl) is a strong acid pKa= -3. It ionizes into hydrogen ion (H+) and chloride ion (Cl-). Thus hydrogen is a positive pole since it carries positive charge and chlorine is a negative pole as it carries negative charge.

Identify which of the following molecules can exhibit hydrogen bonding as a pure liquid.

Hydrogen bonding has an important effect on the physical properties of compounds. The melting and boiling points of compounds are affected by hydrogen bonds.

To recognize what intermolecular forces are present in a given compound and which of those forces is predominant. Chemists use the term intermolecular forces to describe the attractions between two or more molecules. Dipole-dipole forces result from the attraction of the positive end of one polar molecule to the negative end of another polar molecule. Compounds consisting of atoms with different electronegativities may have a dipole moment, or partial charge, caused by an asymmetry of electrons.

Hydrogen bonding is a particularly strong type of dipole-dipole force that occurs when hydrogen is attached to nitrogen, oxygen, or fluorine. Water is an example of a substance in which hydrogen bonding occurs. Because of oxygen's high electronegativity and the electron deficiency of the hydrogen atom, the hydrogen atoms are attracted to the lone pairs of electrons on the oxygen of another water molecule. All substances have dispersion forces, also known as London forces. These forces are very weak and are only important in the absence of any other intermolecular force. Nonpolar covalent molecules and single-atom molecules are examples of substances that lack all other intermolecular forces except for dispersion. Dispersion forces result from shifting electron clouds, which can cause a weak, temporary dipole.

Watch the animation and identify the correct conditions for forming a hydrogen bond. A hydrogen bond is equivalent to a covalent bond. Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an N, O, or F atom. A hydrogen atom acquires a partial positive charge when it is covalently bonded to an F atom. A hydrogen bond is possible with only certain hydrogen-containing compounds. The CH4 molecule exhibits hydrogen bonding.

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an N, O, or F atom. A hydrogen atom acquires a partial positive charge when it is covalently bonded to an F atom. A hydrogen bond is possible with only certain hydrogen-containing compounds. Hydrogen bonding is the intermolecular force that exists between an H atom of one molecule and an electronegative atom (N, O, or F) of another molecule. If a molecule has N−H, O−H, or F−H bonds, that molecule exhibits hydrogen bonding.

To understand basic information about unit cells and how it relates to crystal packing. Atoms and molecules generally form one of two solids: a crystal or an amorphous solid. In an amorphous solid, the spacing of the individual units (atoms or molecules) are not ordered. Most materials can pack in both forms, depending on the rate at which they solidify. A rapid liquid-to-solid transition does not allow the atoms time to arrange themselves, and an amorphous solid results. A slow liquid-to-solid transition generally results in the crystalline form. For example, glass is the amorphous form of silicon dioxide (SiO2), whereas quartz is a crystalline form. Quartz is more dense than glass because the atoms are packed more efficiently. Amorphous solids have a much broader melting temperature range than crystalline metals. In the crystalline solid, the atoms are all arranged in the same way. Therefore, they all require the same amount of energy to undergo the transition from solid to liquid. In contrast, the disordered atoms in an amorphous solid will require different amounts of energy to undergo the same transition because each atom is a different distance from its neighbor.

In a crystal, the atoms are arranged in a pattern that repeats throughout the structure. Since the spatial arrangement of the atoms repeats in a stepwise fashion in a crystal, one can define the smallest nonrepeating unit of the crystal. This is called the unit cell. The entire structure of a crystal can be recreated by translating copies of the unit cell in three dimensions. In a simple-cubic (sc) crystal lattice, the unit cell is composed of eight atoms that each take a corner of a cube. (Figure 1) Because this is such an inefficient way to pack atoms, it is a rare arrangement for metals. Only one metal is known to pack in this form: polonium, Po. The body-centered cubic (bcc) crystal lattice consists of a simple cubic unit cell and an additional atom at the center of the unit cell. (Figure 2) Metals such as iron, chromium, tungsten, and sodium pack in the bcc crystal lattice form. The face-centered cubic (fcc) unit cell consists of a simple cubic unit cell with an extra atom at the center of each of the six sides of the cube. (Figure 3) Some examples of metals that pack in this form are nickel, silver, gold, copper, and aluminum. Another crystal lattice is the hexagonal close packing (hcp) crystal lattice. (Figure 4) Some metals that pack in the hcp crystal lattice are zinc, titanium, and cobalt.

What is the radius of a copper atom (in picometers) that crystallizes in a face-centered cubic unit cell with an edge length of 3.62 × 10−8 cm?

In face-centered cubic packing, the spheres touch along the face diagonal thus the face diagonal = 4r and the edge = 2*sqrt(2)*r. r = 362 pm/(2*sqrt(2)) = 127.98 pm

Viscosity, Surface Tension, and Intermolecular Forces

Intermolecular forces determine the physical properties of substances. The viscosity and surface tension of liquids are two such properties. Viscosity is the resistance of a liquid to flow. Surface tension is the energy required to increase the surface area of a liquid by a certain amount. The stronger the forces of attraction, the more difficult it is for the molecules to move past one another, thus increasing the viscosity and surface tension of the liquid.

The melting point of a solid is related to the way in which that substance is held together. Qualitatively, the relative melting points for a series of solids may be estimated using the following trends:

It is important to keep in mind that a high or low melting point is relative, and that while this trend can be used for a quick estimation, if exact values are needed there are sources to look them up.

Bromine has Pvap = 400 mm at 41.0 ∘C and a normal boiling point of 331.9 K. What is the heat of vaporization, ΔHvap, of bromine in kJ/mol?

Let us use the Clausius-Clapeyron Equation: ln (P1 / P2) = (ΔH / R) (1/T2 - 1/T1) P1 =400mm T1= 314 K P2= 760 mm natural boiling point pressure T2 =331.9 K ln (400 / 760) = (ΔH / 8.314 ) (1/331.9 - 1/314) = 3.11*104 J = 31.1 KJ

Enthalpy and Entropy for Phase Changes

Matter can change from one physical state (phase) to another without any change in chemical identity. Each change is characterized by a specific name, a ΔH (enthalpy) value, and a ΔS (entropy) value. The energy involved in phase changes and temperature changes can be calculated by using the following equations. For a phase change (with no change in temperature) heat=ΔH×n where ΔH is the enthalpy change for that transition and n is the number of moles of the substance. For a temperature change (with no change in phase) heat=Cm×ΔT×n where Cm is the molar heat capacity, n is the number of moles, and ΔT is the temperature change in degrees Celsius.

Each set of solids is grouped according to a particular type of crystalline solid. In which set of solids is there one compound that is a mismatch in that set? Na, Cr, and LiCl I2, H2S, and CH3CH3 O2, CH3COOH, and CH3Cl C(diamond), C(graphite), and SiO2

Na, Cr, and LiCl

How many atoms are in the following: One body-centered cubic unit cell of a metal One face-centered cubic unit cell of a metal

Number of atoms per unit cell in primitive cubic = 1 Number of atoms present in body centered cubic cell = 2 Number of atoms present in face centered cubic unit cell = 4

In the liquid and solid states, molecules are held together by attractions called intermolecular forces. There are several types of intermolecular forces. London dispersion forces, found in all substances, result from the motion of electrons. These work to attract both polar and nonpolar molecules to one another via instantaneous dipole moments. Dipole-dipole forces arise from molecular dipole moments. Ion-dipole forces result from the interaction of an ion and a molecular dipole. Hydrogen-bond forces result from the attraction of a hydrogen atom bonded to a small highly electronegative atom (N, O, and F) and the unshared electron pairs of another electronegative atom

Physical properties such as boiling point, melting point, vapor pressure, viscosity, and surface tension are all affected by the strength of the intermolecular forces within a substance.

What is the charge associated with each side of the HBr molecule?

Q = 1.89×10−20 C

How much heat energy, in kilojoules, is required to convert 55.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ?

Q1 = m*Cpice(0 --18) Q2 = m*Hmelt Q3 = m*Cpwate(25-0) Substitute all Q1 = 55*2.01(0 --18) = 1989.9 Q2 = 55*333 = 18315 Q3 = 55*4.184(25-0) = 5753 QT = 1989.9 + 18315 + 5753 = 26057.9 J QT = 26.05 kJ

In ionic solids, the anions form the unit cell and the smaller cations fill the "holes" within that unit cell. Complete each statement based on the images below. Notice that the anions are represented by large red spheres and the cations are represented by smaller blue spheres.

See next slide

The dipole moment of methanol is μ=1.70 D. Indicate the direction in which electrons are displaced.

See picture

Of the molecules SiCl4 and SiI4, which has bonds that are more polar?

SiCl4

Nickel, Ni, has a face-centered cubic structure(Figure 1) with an edge length of 352 pm. What is the density of this metal? Use the periodic table as needed.

Step 1 : To calculate volume of the unit cell Edge length, a = 352pm = 352X10-12 m Volume, V = a3 = (352X10-12 m)3 V = = 4.36X10-29 m3 Step 2: To calculate mass of an atom of Nickel Molar mass of Nickel = 58.69 g/mol 1 mole = 6.023X1023 atoms So, mass of 6.023X1023 atoms of Nickel = 58.69g mass of 1 atom of Nickel = 58.69g / (6.023X1023) = 9.744X10-23 g Step 3 : To calculate density of the unit cell / Nickel metal No. of atoms per fcc unit cell , n = 4 Mass of 4 atoms of Nickel,m = mass of 1 atom of Ni X 4 = (9.744X10-23 g) X 4 m = 38.977 X 10-23 g Density = m/V = (38.977 X 10-23 g ) / ( 4.36X10-29 m3) = 8.94 X 106 g/m3 Density = = 8.94 g/cm3 ( 1m3 = 106 cm3)

Rank these liquids by their expected surface tension.

Surface tension helps to determine whether a liquid will "wet" or spread out on a given surface. Fabrics are often treated with compounds that prevent water from spreading out on the surface and penetrating the fabric. This is a result of the cohesive forces among the water molecules being stronger than the adhesive forces between the water and the fabric.

Learning Goal: To understand the relation between intermolecular forces and the observable properties of liquids. Viscosity, surface tension, boiling point, and vapor pressure are properties of liquids that are affected by the intermolecular forces within them. Viscosity is a liquid's resistance to flowing. Examples of viscous liquids are molasses, olive oil, and maple syrup.

Surface tension is a phenomenon that allows insects to walk on water and causes water to "bead" on windows. Surface tension is defined as the energy required to increase the surface area of a given amount of liquid by a given unit of area. Boiling point is the temperature at which a liquid becomes a gas. It is also defined as the temperature at which the vapor pressure reaches atmospheric pressure. Vapor pressure is the pressure exerted by the vapor above a liquid. Liquids that readily evaporate have relatively high vapor pressures.

What is the boiling point of water at an elevation of 9500 ft ?

T = 90.6 ∘C The boiling point of water is said to decrease by 0.05∘C for every 1 mmHg drop in atmospheric pressure the answer should be reported to one significant figure in literature or a paper: 90 ∘C or 9×10¹∘C.

The normal boiling point of benzene is 80.1 ∘C, and the heat of vaporization is ΔHvap = 30.7 kJ/mol . What is the boiling point of benzene in ∘C on top of Mt. Everest, where P = 260 mmHg?

Tboil = 47.2 ∘C

Methylamine, CH3NH2, is responsible for the odor of rotting fish. Look at the following electrostatic potential map of methylamine, and explain the observed polarity.

The N atom is electron rich because of its high electronegativity. The C and H atoms are electron poor because they are less electronegative.

Compounds that experience hydrogen bonding between molecules generally have higher boiling points. Because of hydrogen bonding, the intermolecular forces of attraction between molecules within the compound are high. Consequently, more energy is required to separate these molecules from one another through boiling. Watch the animation in the introduction to examine the effect of hydrogen bonding on the boiling points of hydrides belonging to groups 4A and 6A. The hydrides of group 5A are NH3, PH3, AsH3, and SbH3. Arrange them from highest to lowest boiling point.

The boiling point of NH3 is higher than that predicted by periodic trends alone because of hydrogen bonding. The molecule SbH3 actually has a higher boiling point than NH3. Both intermolecular forces, such as hydrogen bonding, and molecular mass contribute to the observed boiling point of a molecule. This means that you cannot always qualitatively predict the relative boiling points of two compounds that differ greatly. You could predict that methanol (CH3OH) has a higher boiling point than methane (CH4), but decane (C10H22) has an even higher boiling point without hydrogen bonding partly due to its significantly higher molar mass.

The vapor pressure of a substance describes how readily molecules at the surface of the substance enter the gaseous phase. At the boiling point of a liquid, the liquid's vapor pressure is equal to or greater than the atmospheric pressure exerted on the surface of the liquid. Since the atmospheric pressure at higher elevations is lower than at sea level, the boiling point of water decreases as the elevation increases. The atmospheric pressure at sea level is 760 mmHg. This pressure decreases by 19.8 mmHg for every 1000-ft increase in elevation.

The boiling point of water decreases 0.05∘C for every 1 mmHg drop in atmospheric pressure.

Complete the sentences describing the steps needed to calculate the energy change associated with the conversion of 333 g of water ice at −10 ∘C to steam at 125 ∘C.

The change from ice at −10 ∘C to steam at 125 ∘C involves five steps: Raising the temperature of ice from −10 ∘C to the melting point, 0 ∘C. Melting the ice to liquid water. Raising the temperature of water from 0 ∘C to the boiling point, 100 ∘C. Boiling the liquid water to steam. Raising the temperature of the steam from 100 ∘C to 125 ∘C. To calculate the energy required for the process, you need to calculate the energy component of each step. Steps 1, 3, and 5 involve a temperature change and therefore you should use the following equation: heat=Cm×ΔT×n Steps 2 and 4 involve a phase change and therefore you should use the following equation: heat=ΔH×n

PI3Br2 is a nonpolar molecule. Based on this information, determine the I−P−I bond angle, the Br−P−Br bond angle, and the I−P−Br bond angle. Enter the number of degrees of the I−P−I, Br−P−Br, and I−P−Br bond angles, separated by commas (e.g., 30,45,90)

The geometry of PI3Br2 is trigonal bipyramidal. Given that the molecule is non-polar that means three I atoms occupy equitorial positions and two Br atoms occupy axial positions in trigonal bipyramidal geometry. Then I-P-I bond angle is 1200 , Br-P-Br bond angle is 1800 and I-P-Br bond angle is 900.

Consider heating solid water (ice) until it becomes liquid and then gas (steam). (Figure 1) Alternatively, consider the reverse process, cooling steam until it becomes water and, finally, ice. (Figure 2) In each case, two types of transitions occur, those involving a temperature change with no change in phase (shown by the diagonal line segments on the graphs) and those at constant temperature with a change in phase (shown by horizontal line segments on the graphs).

The heat energy associated with a change in temperature that does not involve a change in phase is given by q=msΔT where q is heat in joules, m is mass in grams, s is specific heat in joules per gram-degree Celsius, J/(g⋅∘C), and ΔT is the temperature change in degrees Celsius. The heat energy associated with a change in phase at constant temperature is given by q=mΔH where q is heat in joules, m is mass in grams, and ΔH is the enthalpy in joules per gram.

Describe the molecular dipole of OCl2.

The molecular dipole is measurable and directed toward oxygen.

Which statement best describes the polarity of CF2I2?

The molecule is always polar. It will always be tetrahedral, and there is no possible way to cancel out the dipoles.

Molecules can be polar because of the unsymmetrical distribution of electrons. The dipole moment, μ, is a measure of the net polarity and is defined as the charge 1.87×10−20 C , in coulombs, times the distance between the charges, which is estimated by the bond length, r, in meters: μ=Q×r The SI unit of dipole moment is the coulomb-meter (C⋅m), but another common unit is the debye (D). The two are related as 3.336×10−30 C⋅m=1 D

The percent ionic character is a comparison of the measured dipole to the expected dipole: %ionic character=measured dipoleexpected dipole×100% The expected dipole is one for which a full unit of charge (1.60×10−19 C) exists on each end of the bond.

The atoms of crystalline solid pack together into a three-dimensional array of many small repeating units called unit cells. The simplest of the unit cells are those that have cubic symmetry, with atoms positioned at the corners of a cube. Atoms can also be found in the sides (faces) of the cube, or centered within the body of the cube. It is important to realize that a unit cell is surrounded by other unit cells in every direction. Therefore, face and corner atoms are shared with neighboring unit cells. The fraction varies with the type of atom as shown in the following table.

The size of a unit cell in any given solid can be calculated by using its density. This and the reverse calculation are common test questions in general chemistry courses.

The Society of Automotive Engineers has established an accepted numerical scale to measure the viscosity of motor oil. For example, SAE 40 motor oil has a higher viscosity than an SAE 10 oil. Rank the following hydrocarbons by their expected viscosity.

The viscosity of a given liquid varies with temperature. Different grades of motor oil are often used in summer and winter because of this variation with temperature.

Chloroform (CHCl3) has ΔHvap = 29.2 kJ/mol and ΔSvap = 87.5 J/ (K⋅mol). What is the boiling point of chloroform in kelvin?

Use the following equation to solve the problem ΔSvap = ΔHvap / T T = ΔHvap / ΔSvap ΔHvap = 29.2KJ/mol = 29200J/mol, ΔSvap = 87.5 J/ (K⋅mol) T = 29200 / 87.5 = 333.71K

The face-centered gold crystal has an edge length of 407 pm. Based on the unit cell, calculate the density of gold.

We know that density , d = (Z x atomic mass of gold ) / ( No x a 3 x 10 -30 cm 3 ) ---(1) Where Z = No . of atoms in a unit cell = 4 Since it is a fcc structure atomic mass of gold = 197 g No = avagadro number = 6.023x10 23 a = edge length = 407 pm Plug these values in (1) we get d = (4x197 ) / ( 6.023x10 23 x 4073x 10 -30 cm 3 ) = 19.4 g / cm 3

Qualitatively estimate the relative melting points for each of the solids, and rank them in decreasing order.

While the general trend for melting points is shown in the introduction, there are deviations from this rule. Mercury is a well known exception, and is a liquid at room temperature. However, the trend is a good way to estimate how materials will behave quickly. If quantitative measures are needed, it is always good to look up the information.

Arrange the liquids pentane (CH3CH2CH2CH2CH3), pentanol (CH3CH2CH2CH2CH2OH), and pentanal (CH3CH2CH2CH2CHO) in order of decreasing viscosity, surface tension, and boiling point.

iscosity and surface tension are both related to intermolecular forces (IMFs) pentane is nonpolar and cannot H-bond, so only LDF is possible (London Dispersion Force) pentanal is polar, but cannot H-bond => (Dipole-Dipole) pentanol has an OH group, so it can H-bond => (Hydrogen Bonding) With higher IMFs, you get greater surface tension and greater viscosity (since the liquid molecules are more attracted to each other) In order of decreasing IMF: (and thus decreasing viscosity/surface tension) pentanol pentanal pentane