A. And so we take the quotient, 169 g 40.1 g mol1, and multiply this by N A,Avogadro's number of molecules, where N A = 6.022 1023 mol1. D. N2O4 How many calcium atoms can fit between the Earth and the Moon? There are two atoms in a body-centered cubic. Figure 12.5 The Three Kinds of Cubic Unit Cell. A single layer of close-packed spheres is shown in part (a) in Figure 12.6. What is the atomic radius of platinum? (a) What is the atomic radius of Ag in this structure? First Law of Thermodynamics and Work (M6Q3), 30. Ionic Crystals and Unit Cell Stoichiometry (M11Q6), Appendix E: Specific Heat Capacities for Common Substances (M6Q5), Appendix F: Standard Thermodynamic Properties (M6), Appendix G: Bond Enthalpy, Bond Length, Atomic Radii, and Ionic Radii. a. C. 126 What is the mass in grams of NaCN in 120.0 mL of a 2.40 x 10^ -5 M solution? D. FeBr3 And thus we can find the number of calcium atoms in a lump of metal, simply by measuring the mass of the lump and doing a simple calculation. For example, platinum has a density of 21.45 g/cm3 and a unit cell side length a of 3.93 . Why is the mole an important unit to chemists? Why is it valid to represent the structure of a crystalline solid by the structure of its unit cell? Explaining Solubility and Surface Tension through IMFs (M10Q4), 58. B) CHN B The molar mass of iron is 55.85 g/mol. atomic mass Ca = 40.08 g/mol Find mols of Ca that you have: 149 g Ca x 1 mol Ca / 40.08 g = 3.718 mols Ca Find the number of atoms in 3718 mols of Ca. C. 25 We get an answer in #"moles"#, because dimensionally #1/(mol^-1)=1/(1/(mol))=mol# as required. By calculating the molar mass to four significant figures, you can determine Avogadro's number. Thus the unit cell in part (d) in Figure 12.2 is not a valid choice because repeating it in space does not produce the desired lattice (there are triangular holes). Determine the mass in grams of NaCl that are in 23.4 moles of NaCl? Because density is mass per unit volume, we need to calculate the mass of the iron atoms in the unit cell from the molar mass and Avogadros number and then divide the mass by the volume of the cell (making sure to use suitable units to get density in g/cm3): \[ mass \; of \; Fe=\left ( 2 \; \cancel{atoms} \; Fe \right )\left ( \dfrac{ 1 \; \cancel{mol}}{6.022\times 10^{23} \; \cancel{atoms}} \right )\left ( \dfrac{55.85 \; g}{\cancel{mol}} \right ) =1.855\times 10^{-22} \; g \], \[ volume=\left [ \left ( 286.6 \; pm \right )\left ( \dfrac{10^{-12 }\; \cancel{m}}{\cancel{pm}} \right )\left ( \dfrac{10^{2} \; cm}{\cancel{m}} \right ) \right ] =2.345\times 10^{-23} \; cm^{3} \], \[ density = \dfrac{1.855\times 10^{-22} \; g}{2.345\times 10^{-23} \; cm^{3}} = 7.880 g/cm^{3} \]. Protons, Neutrons, and Electrons (M2Q1), 6. To do so, I will use the Pythagorean Theorem. DeBroglie, Intro to Quantum Mechanics, Quantum Numbers 1-3 (M7Q5), 39. This is the calculation in Example \(\PageIndex{2}\) performed in reverse.
Therefore, 127 g of What are the answers to studies weekly week 26 social studies? This is called a body-centered cubic (BCC) solid. A. P4H10 A. (Assume the volume does not change after the addition of the solid.). In this example, multiply the mass of K by the conversion factor: \[\dfrac{1\; mol\; K}{39.10\; grams\; K} \nonumber \]. 2. Converting the mass, in grams, of a substance to moles requires a conversion factor of (one mole of substance/molar mass of substance). Find the number of atoms in 3718 mols of Ca. Similarly, an atom that lies on the edge of a unit cell is shared by four adjacent unit cells, so it contributes 14 atom to each. What are the Physical devices used to construct memories?
Grams To Atoms Calculator Electron Configurations, Orbital Box Notation (M7Q7), 41.
Solid Crystal Lecture Flashcards | Quizlet Amounts may vary, according to . 35,000 worksheets, games, and lesson plans, Spanish-English dictionary, translator, and learning, a Question .75 Explain how the intensive properties of a material are reflected in the unit cell. 9. Who is Katy mixon body double eastbound and down season 1 finale? So #"Moles of calcium"# #=# #(197*cancelg)/(40.1*cancelg*mol^-1)#. The hcp and ccp arrangements fill 74% of the available space and have a coordination number of 12 for each atom in the lattice, the number of nearest neighbors. B. A FCC unit cell contains four atoms: one-eighth of an atom at each of the eight corners (8 [latex]\frac{1}{8}[/latex] = 1 atom from the corners)) and one-half of an atom on each of the six faces (6 [latex]\frac{1}{2}[/latex] = 3 atoms from the corners) atoms from the faces). Calculate the density of gold, which has a face-centered cubic unit cell (part (c) in Figure 12.5) with an edge length of 407.8 pm. 2. Legal. First we calculate the
How many atoms are in 195 grams of calcium? - Answers Actually, however, these six sites can be divided into two sets, labeled B and C in part (a) in Figure 12.6. What is are the functions of diverse organisms? 1 point How many chlorine atoms are there in 20.65 moles of aluminum chloride? C) CH Atoms in the corners of a BCC unit cell do not contact each other but contact the atom in the center. Metallic gold has a face-centered cubic unit cell (part (c) in Figure 12.5). Each carbon-12 atom weighs about \(1.99265 \times 10^{-23}\; g\); therefore, \[(1.99265 \times 10^{-23}\; g) \times (6.02214179 \times 10^{23}\; atoms) = 12\; g\; \text{ of carbon-12} \nonumber \]. Direction of Heat Flow and System vs. Surroundings (M6Q2), 28. (CC BY-NC-SA; anonymous by request). If the metallic radius of nickel is 125 pm, what is the structure of metallic nickel? Why is the mole an important unit to chemists? B) CH Figure 12.2 Unit Cells in Two Dimensions. Measurements, Units, Conversions, Density (M1Q1), 4. After we have found the moles of Ca, we can use the relationship between moles and Avogadro's number: 1 mole of atoms = 6.022 1023 atoms. E. S2O, What is the mass percent of oxygen in HNO3? cubic close packed (identical to face-centered cubic).
How many calcium atoms can fit between the Earth and the Moon? The edge length of its unit cell is 409 pm. If we place the second layer of spheres at the B positions in part (a) in Figure 12.6, we obtain the two-layered structure shown in part (b) in Figure 12.6. The structures of many metals depend on pressure and temperature. Avogadro's Number of atoms. Each packing has its own characteristics with respect to the volume occupied by the atoms and the closeness of the packing. Orbitals and the 4th Quantum Number, (M7Q6), 40. Calorimetry continued: Phase Changes and Heating Curves (M6Q6), 33. Check Your Learning What value do you obtain? 147 grams calcium (1 mole Ca/40.08 grams)(6.022 X 1023/1 mole D) CH.N, A compound that contains only carbon, hydrogen, and oxygen is 58.8% C and 9.87% H by mass. 4.45 x 10 ^26 atoms. (The mass of one mole of calcium is 40.08 g.).00498 mol. (b) Because atoms are spherical, they cannot occupy all of the space of the cube. A link to the app was sent to your phone. B. B. In the United States, 112 people were killed, and 23 are still missing0. D) CHO 7) Let's do the bcc calculation (which we know will give us the wrong answer). Based on your answer for the number of formula units of TlCl(s) in a unit cell, (b) how is the unit cell of TlCl(s) likely to be structured? 14.7 Charge of Ca=+2. Determine the number of atoms of O in 10.0 grams of CHO, What is the empirical formula of acetic acid, HCHO? The third layer of spheres occupies the square holes formed by the second layer, so that each lies directly above a sphere in the first layer, and so forth. 22% Calorimetry continued: Types of Calorimeters and Analyzing Heat Flow (M6Q5), 31. What is the total number of atoms contained in 2.00 moles of iron? Note that an answer that uses #N_A# to represent the given number would be quite acceptable; of course you could multiply it out. Because the atoms are on identical lattice points, they have identical environments. (ac) Three two-dimensional lattices illustrate the possible choices of the unit cell. Lithium crystallizes in a bcc structure with an edge length of 3.509 . Silver crystallizes in an FCC structure. For example, the unit cell of a sheet of identical postage stamps is a single stamp, and the unit cell of a stack of bricks is a single brick. 1. Is the structure of this metal simple cubic, bcc, fcc, or hcp? D. 2.0x10^23 Cl gains 1 electron each. Finally, if you are asked to find the number of atoms in one mole, for example, the number of H atoms in one mole of H2O, you multiply the number of atoms by. D. 3.6 x 10 ^24 Answer (1 of 5): It's not fix like no. Report. What are the 4 major sources of law in Zimbabwe. definition of Avogadro's Number, each gram atomic mass contains When the metal reacts with excess water, the reaction produces 539.29 mL of hydrogen gas at 0.980 atm and 23C.
The arrangement of the atoms in a solid that has a simple cubic unit cell was shown in part (a) in Figure 12.5. Ionic Bond. .5 So: The only choice to fit the above criteria is answer choice b, Na3N. C. C4H14O Table 12.1: Properties of the Common Structures of Metals. E. none, A compound is 50% S and 50% O. B. 0.316 mols (6.022x1023 atoms/ 1mol) = 1.904x1023 atoms of O, 0.055 mols (6.022x1023 atoms/ 1mol) = 3.312x1022 atoms of K, 4. What are the Physical devices used to construct memories? Step 1 of 4. Cubic closest packed structure which means the unit cell is face - centered cubic. See the answer. 6 In contrast, atoms that lie entirely within a unit cell, such as the atom in the center of a body-centered cubic unit cell, belong to only that one unit cell. To think about what a mole means, one should relate it to quantities such as dozen or pair. Using a periodic table, give the molar mass of the following: Convert to moles and find the total number of atoms. Problem #7: Tungsten has an atomic radius of 137 pm and crystallizes in a cubic unit cell having an edge length d = 316 pm. To calculate the number of atoms in the unit cell, multiply the number of atoms on vertices times the fraction of each atom that is within the unit cell. The hexagonal close-packed (hcp) structure has an ABABAB repeating arrangement, and the cubic close-packed (ccp) structure has an ABCABC repeating pattern; the latter is identical to an fcc lattice. The total number of atoms in a substance can also be determined by using the relationship between grams, moles, and atoms. So Moles of calcium = 197 g 40.1 g mol1 =? 8 d. Determine the packing efficiency for this structure. Assuming that the rest of the sample is water, how many moles of H2O are there in the sample? A) C.HO 12% #5xxN_A#, where #N_A# is #"Avogadro's number"#. Solution: 1) Calculate the average mass of one atom of Fe: 55.845 g mol1 6.022 x 1023atoms mol1= 9.2735 x 1023g/atom 2) Determine atoms in 1 cm3: 7.87 g / 9.2735 x 1023g/atom = 8.4866 x 1022atoms in 1 cm3 3) Determine volume of the unit cell: 287 pm x (1 cm / 1010pm) = 2.87 x 108cm D. 4.5g amount in moles of calcium in a 98.5g pure sample.Amount of Ca = Because the atoms are on identical lattice points, they have identical environments. 100% (27 ratings) for this solution. 10. Dec 8, 2015 0.650 g Au contain 1.99 1021atoms. \[3.0\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.130\; mol\; Na \nonumber \], \[0.130548\; \cancel{ mol\; Na} \left(\dfrac{6.02214179 \times 10^{23}\; atoms \;Na}{1\; \cancel{ mol\; Na}}\right) = 7.8 \times 10^{22} \; atoms\; of\; \; Na \nonumber \]. Who were the models in Van Halen's finish what you started video? Any intensive property of the bulk material, such as its density, must therefore also be related to its unit cell. If we choose the first arrangement and repeat the pattern in succeeding layers, the positions of the atoms alternate from layer to layer in the pattern ABABAB, resulting in a hexagonal close-packed (hcp) structure (part (a) in Figure 12.7). Then, we need to convert moles to atoms which we do with Avogadro's constant which is 6.022*10^23atoms/mol. This page titled 12.2: The Arrangement of Atoms in Crystalline Solids is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Anonymous. Atoms on a corner are shared by eight unit cells and hence contribute only \({1 \over 8}\) atom per unit cell, giving 8\({1 \over 8}\) =1 Au atom per unit cell. Explain your answer. Unit cells are easiest to visualize in two dimensions. 4.366mol * 6.022*10^23atoms/mol = 2.629*10^24 atoms of Ca in 175g of Ca. A 21.64 g sample of a nonreactive metal is placed in a flask containing 12.00 mL of water; the final volume is 13.81 mL. A. Gold does not crystallize bcc because bcc does not reproduce the known density of gold. D. 4.5 x 10^23 A. The concept of unit cells is extended to a three-dimensional lattice in the schematic drawing in Figure 12.3. How many grams of calcium chloride do you need? 197 g Actiu Go to This problem has been solved! How many atoms are in a 3.5 g sample of sodium (Na)? Most questions answered within 4 hours. Complete reaction with chlorine gas requires 848.3 mL of chlorine gas at 1.050 atm and 25C. 1) I will assume the unit cell is face-centered cubic. We take the quotient \text{moles of carbon atoms}=\dfrac{\text{mass of carbon}}{\text{molar mass of carbon}}=\dfrac{1.70g}{12.01gmol^{-1}}=0.1415mol And I simply got the molar mass of carbon from a handy Per. 11. 35,000 worksheets, games, and lesson plans, Spanish-English dictionary, translator, and learning, a Question There is only one Ca atom. A We know from Example 1 that each unit cell of metallic iron contains two Fe atoms. = 2.21 X 1024 atoms of calcium How many moles of water is this? around the world. Scientists who study ancient marine life forms usually obtain fossils not from the sea floor, but from areas that were once undersea and have been uplifted onto the continents. 10. Acids, Bases, Neutralization, and Gas-Forming Reactions (M3Q3-4), 13. The transition temperature, the temperature at which one phase is converted to the other, is 95C at 1 atm and 135C at 1000 atm.
The mole concept is also applicable to the composition of chemical compounds. B. So, 3.17 mols 6.022 1023 atoms/1 mol = 1.91 1024 atoms. (CC BY-NC-SA; anonymous by request). A simple cubic cell contains one metal atom with a metallic radius of 100 pm. From there, we take the 77.4 grams in the original question, divide by 40.078 grams and we get moles of Calcium which is 1.93 moles. The density of tungsten is 19.3 g/cm3. B) CHO One mole is equal to \(6.02214179 \times 10^{23}\) atoms, or other elementary units such as molecules. .25 Verifying that the units cancel properly is a good way to make sure the correct method is used. Problem #5: A metal nitride has a nitrogen atom at each corner and a metal atom at each edge. The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. Multiply moles of Ca by the conversion factor 40.08 g Ca/ 1 mol Ca, with 40.08 g being the molar mass of one mole of Ca. That's because of the density. How many moles of C3H6 are in 25.0 grams of the substance (propylene)? A.
- You want to make a molecule that has Calcium Ca and Chloride Cl. 7. sodium, unit cell edge = 428 pm, r = 185 pm. Join Yahoo Answers and get 100 points today. E. 7.2 x 10^23 g, How many moles are in a 45g sample of C6H12O6? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. D. 340 g How many Au atoms are in each unit cell? What is the mass in grams of 6.022 1023 molecules of CO2?
Types of Unit Cells: Body-Centered Cubic and Face-Centered - Unizin 1 Ca unit cell [latex]\frac{4\;\text{Ca atoms}}{1\;\text{Ca unit cell}}[/latex] [latex]\frac{1\;\text{mol Ca}}{6.022\;\times\;10^{23}\;\text{Ca atoms}}[/latex] [latex]\frac{40.078\;\text{g}}{1\;\text{mol Ca}}[/latex] = 2.662 10. . Melting and Boiling Point Comparisons (M10Q2), 55. How many grams are 10.78 moles of Calcium (\(\ce{Ca}\))? You need to prepare 825. g of a 7.95% by mass calcium chloride solution. For instance, consider methane, CH4. 32g Which of the following is this compound? ?mol. Solutions and Solubility (part 1) (M3Q1), 11. #6.022xx10^23# individual calcium atoms have a mass of #40.1*g#. Ca) Shockingly facts about atoms. Isomorphous metals with a BCC structure include K, Ba, Cr, Mo, W, and Fe at room temperature. (Hint: there is no empty space between atoms.). Then the number of moles of the substance must be converted to atoms. 9. 100.0 mL of a 0.500 M solution of KBr is diluted to 500.0 mL. Explanation: We're asked to calculate the number of atoms of Ca in 153 g Ca. Then, we need to convert moles to atoms which we do with Avogadro's constant which is 6.022*10^23atoms/mol. 1. What are the most important constraints in selecting a unit cell? B. C6H6 The simple cubic and bcc lattices have coordination numbers of 6 and 8, respectively. Determine the mass, in grams, of 0.400 moles of Pb (1 mol of Pb has a mass of 207.2 g). 1 atom. D. C4H4 How many grams of water Atoms in an FCC arrangement are packed as closely together as possible, with atoms occupying 74% of the volume. How many sodium atoms (approx.) A metal has two crystalline phases. The metal crystallizes in a bcc lattice.
Problem #1: Many metals pack in cubic unit cells. One simply needs to follow the same method but in the opposite direction. Problem #4: Many metals pack in cubic unit cells. 28.5 mol of P4O10 contains how many moles of P. Q. Add the contributions of all the Au atoms to obtain the total number of Au atoms in a unit cell. E. 6.0 x 10^24, How many oxygen atoms are in 1.5 moles of N2O4? C. 2 Calcium sulfate, CaSO4, is a white, crystalline powder. 44 Use Avogadro's number 6.02x1023 atoms/mol: 3.718 mols Ca x 6.02x1023 atoms/mol = 2.24x1024 atoms (3 sig. 3) Calculate the mass of NaCl inside the cube: 4) The molar mass divided by the mass inside the cube equals Avogadro's Number. C. 51% UALR 1402: General Chemistry I Therefore, the answer is 3.69 X Solution. With the reference cube having 4 vertices of Na and 4 vertices of Cl, this means there is a total of 1/2 of a Na atom and 1/2 of a Cl atom inside the reference cube. C. 2.25 The cubic hole in the middle of the cell is empty. If your sample is made of one element, like copper, locate the atomic mass on the periodic table. Report your answer with the correct significant figures using scientific notation. See the answer Show transcribed image text Expert Answer 100% (1 rating) This arrangement is called a face-centered cubic (FCC) solid. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Each sphere is surrounded by six others in the same plane to produce a hexagonal arrangement. As shown in part (b) in Figure 12.5, the body-centered cubic structure consists of a single layer of spheres in contact with each other and aligned so that their centers are at the corners of a square; a second layer of spheres occupies the square-shaped holes above the spheres in the first layer. How many 5 letter words can you make from Cat in the Hat? Any atom in this structure touches four atoms in the layer above it and four atoms in the layer below it. Oxidation-Reduction Reactions (M3Q5-6), 19. Emission Spectra and H Atom Levels (M7Q3), 37. In a cubic unit cell, corners are 1/8 of an atom, edges are 1/4 of an atom, and faces are 1/2 of an atom. If the cubic unit cell consists of eight component atoms, molecules, or ions located at the corners of the cube, then it is called simple cubic (part (a) in Figure 12.5). Above any set of seven spheres are six depressions arranged in a hexagon. (b) Density is given by density = [latex]\frac{\text{mass}}{\text{volume}}[/latex]. Follow. The mass of a mole of substance is called the molar mass of that substance. Valence Bond Theory and Hybridization (M9Q3), 51. 0.650g Au 196.966569 g molAu = 0.00330 mol Au atoms 1mol atoms = 6.022 1023atoms Multiply the calculated mol Au times 6.022 1023atoms 1mol. Then, multiply the number of moles of Na by the conversion factor 6.022141791023 atoms Na/ 1 mol Na, with 6.022141791023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na. In order to find the number of atoms in a given mass of a substance, you need to first find the molar mass of the substance in question. (b) Placing an atom at a B position prohibits placing an atom at any of the adjacent C positions and results in all the atoms in the second layer occupying the B positions. A face-centered cubic (fcc) unit cell contains a component in the center of each face in addition to those at the corners of the cube. What is the atomic radius of barium in this structure? What is the length of the edge of the unit cell? How many atoms are in 10.0 g of gold? In this arrangement, each atom touches 12 near neighbors, and therefore has a coordination number of 12. Molarity, Solutions, and Dilutions (M4Q6), 23. Using Figure 12.5, identify the positions of the Au atoms in a face-centered cubic unit cell and then determine how much each Au atom contributes to the unit cell. The edge length of its unit cell is 558.8 pm. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum. Why is polonium the only example of an element with this structure? My avg. B. A BCC unit cell contains two atoms: one-eighth of an atom at each of the eight corners (8 [latex]\frac{1}{8}[/latex] = 1 atom from the corners) plus one atom from the center. Explanation: By definition, 40.1 g of calcium atoms contains Avogadro's number of molecules. E. N4O, LA P&C Insurance Licensing - Bob Brooks Quest. (Elements or compounds that crystallize with the same structure are said to be isomorphous.). Most questions answered within 4 hours. We can find the number of moles of this substance by dividing our given mass in grams by our molar mass. .85 g Now that we know how to count atoms in unit cells, we can use unit cells to calculate the densities of simple compounds. 175 g Ca (1 mol / 40.078 g) (6.022x10^23 atoms / 1 mol) = 2.63x10^24 Ca atoms There are 2.63x10^24 c.alcium atoms in 175 grams of. UW-Madison Chemistry 103/104 Resource Book by crlandis is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. A. C5H18 We're asked to calculate the number of atoms of #"Ca"# in #153# #"g Ca"#. A sample of an alkali metal that has a bcc unit cell is found to have a mass of 1.000 g and a volume of 1.0298 cm3. (197 g/mol divided by 6.022 x 1023 atoms/mol) times 2 atoms = 6.5427 x 10-22 g, 6.5427 x 10-22 g / 3.6776 x 10-23 cm^3 = 17.79 g/cm^3. Explain your reasoning. By multiplying the number of moles by Avogadro's constant, the mol units cancel out, leaving the number of atoms. Also, one mole of nitrogen atoms contain, Example \(\PageIndex{1}\): Converting Mass to Moles, Example \(\PageIndex{2}\): Converting Moles to mass, constant, it is also easy to calculate the number of atoms or molecules present in a substance (Table. D. CH3CH2OH A. SO2 Since each vertex is in a total of 8 cells, we have 1 F atom in the unit cell. 1:07. D. 76% 10 Each atom in the lattice has only six nearest neighbors in an octahedral arrangement. Calculate the edge length of the face-centered cubic unit cell and the density of platinum. To calculate the density of a solid given its unit cell. All the alkali metals, barium, radium, and several of the transition metals have body-centered cubic structures. Calculation of Atomic Radius and Density for Metals, Part 2 What is meant by the term coordination number in the structure of a solid? C. SO3 A link to the app was sent to your phone. In this case, the mole is used as a common unit that can be applied to a ratio as shown below: \[2 \text{ mol H } + 1 \text{ mol O }= 1 \text{ mol } \ce{H2O} \nonumber\].
CHEM 1411 - chapter 3 quiz Flashcards | Quizlet Get off Wiki Answers Mrs. Z's chemistry class, Quite a few! B. NO3 Upvote 0 Downvote Add comment Report Still looking for help? 4. Isotopes, Atomic Mass, and Mass Spectrometry (M2Q3), 10. Simple cubic and bcc arrangements fill only 52% and 68% of the available space with atoms, respectively. 5. Cell 1: 8 F atoms at the 8 vertices.
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