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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 1/33 VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) Due: 12:00am on Monday, December 12, 2022 You will receive no credit for items you complete after the assignment is due. Grading Policy Exercise 11.32 - Enhanced - with Feedback MISSED THIS? Read Section 11.2 (Page) ; Watch KCV 11.2 , IWE 11.1 . A molecule with the formula has a trigonal planar geometry. Part A How many electron groups are on the central atom? Express your answer as an integer. ANSWER: Exercise 11.34 - Enhanced - with Feedback MISSED THIS? Read Section 11.2 (Page) , 11.3 (Page) ; Watch KCV 11.2 , 11.3 , IWE 11.1 , 11.2 . The following figures show several molecular geometries. Part A Give the number of total electron groups, the number of bonding groups, and the number of lone pairs for the geometry depicted in (a). Express your answers as integers separated by commas. ANSWER: Part B Give the number of total electron groups, the number of bonding groups, and the number of lone pairs for the geometry depicted in (b). Express your answers as integers separated by commas. ANSWER: Part C Give the number of total electron groups, the number of bonding groups, and the number of lone pairs for the geometry depicted in (c). Express your answers as integers separated by commas. ANSWER: electron groups , , = , , =
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 2/33 Exercise 11.36 - Enhanced - with Feedback MISSED THIS? Read section 11.3 (Page) , 11.4 (Page) ; Watch KCV 11.3 , IWE 11.2 . Determine the electron geometry, molecular geometry, and idealized bond angles for each of the following molecules. a. b. c. d. In which cases do you expect deviations from the idealized bond angle? Part A Determine the electron geometry for each molecule. Drag the appropriate items to their respective bins. ANSWER: Part B Determine the molecular geometry for each molecule. Drag the appropriate items to their respective bins. ANSWER: , , = Reset Help Linear Tetrahedral Trigonal planar Trigonal bipyramidal
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 3/33 Part C Determine the idealized bond angle for each molecule. Drag the appropriate items to their respective bins. ANSWER: Part D In which cases do you expect deviations from the idealized bond angle? Check all that apply. ANSWER: Reset Help Reset Help Tetrahedral Trigonal pyramidal Bent Linear 90 180 109.5 120
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 4/33 Exercise 11.39 - Enhanced - with Feedback MISSED THIS? Read Section 11.3 (Page) , 11.4 (Page) ; Watch KCV 11.3 , IWE 11.2 . Part A Determine the molecular geometry for . ANSWER: Correct has 34 valence electrons (6 on sulfur + (4 7) on the fluorine). There are five electron groups around the sulfur atom: four bonding groups and one lone pair. This means that the atoms has a seesaw molecular geometry. Part B Determine the molecular geometry for . ANSWER: Correct The Lewis structure for has 28 valence electrons. There are five electrons groups around the atom: three bonding groups and two lone pairs. The molecular geometry for is T-shaped. Part C Determine the molecular geometry . ANSWER: trigonal bipyramidal square planar T-shape seesaw square pyramidal seesaw square planar T-shape square pyramidal trigonal bipyramidal
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 5/33 Correct The Lewis structure for has 22 valence electrons. There are five electrons groups around the central atom: two bonding groups and three lone pairs. The molecular geometry for is linear. Part D Determine the molecular geometry . ANSWER: Correct The ion has 36 valence electrons. There are six electron groups around the central atom: four bonding groups and two lone pairs. The molecular geometry for is square planar. Part E For each molecule or ion choose the appropriate sketch. Drag the appropriate labels to their respective targets. ANSWER: bent seesaw trigonal bipyramidal linear T-shape square pyramidal trigonal bipyramidal seesaw square planar octahedral
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 6/33 Exercise 11.48 - Enhanced - with Feedback MISSED THIS? Read Section 11.5 (Page) ; Watch KCV 11.5 , IWE 11.5 . is a polar molecule, even though the tetrahedral geometry often leads to nonpolar molecules. Part A Draw the Lewis structure of . Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons and all hydrogen atoms. ANSWER: Help Reset
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 7/33 Part B Complete previous part(s) Exercise 11.50 - Enhanced - with Feedback and Hints MISSED THIS? Read Section 11.5 (Page) ; Watch KCV 11.5 , IWE 11.5 . Part A Determine whether each molecule given below is polar or nonpolar. Drag the appropriate items to their respective bins. You did not open hints for this part. ANSWER: + - C H O N S P F Br Cl I X More
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 8/33 Exercise 11.52 - Enhanced - with Feedback and Hints MISSED THIS? Read Section 11.5 (Page) ; Watch KCV 11.5 , IWE 11.5 . Part A Determine whether each molecule is polar or nonpolar. Drag the appropriate items to their respective bins. You did not open hints for this part. ANSWER: Reset Help Reset Help Polar Nonpolar Nonpolar Polar
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 9/33 Exercise 11.53 - Enhanced - with Feedback MISSED THIS? Read Section 11.6 (Page) ; Watch KCV 11.6 . The valence electron configurations of several atoms are shown. How many bonds can each atom make without hybridization? Part A Express your answer as an integer. ANSWER: Part B Express your answer as an integer. ANSWER: Part C Express your answer as an integer. ANSWER: Exercise 11.59 - Enhanced - with Feedback MISSED THIS? Read Section 11.7 (Page) ; Watch KCV 11.7 . Part A Which of the following hybridization schemes allows the formation of at least one bond? Check all that apply. ANSWER: Exercise 11.60 - Enhanced - with Feedback MISSED THIS? Read Section 11.7 (Page) ; Watch KCV 11.7 .
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 10/33 Part A Which of the following hybridization schemes allows the central atom to form more than four bonds? ANSWER: Exercise 11.62 - Enhanced - with Feedback MISSED THIS? Read Section 11.7 (Page) ; Watch KCV 11.7 , IWE 11.8 . Write a hybridization and bonding scheme for each molecule. Part A Identify the hybridization of the atom in . ANSWER: Correct To allow bonding with four other atoms, the carbon atom must hybridize its four valence orbitals to equally accommodate each bond. Part B In the sketch of the structure of label all bonds. Drag the appropriate labels to their respective targets. Labels can be used once, more than once, or not at all. ANSWER:
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 11/33 Part C Identify the hybridization of the atom in . ANSWER: Correct The standard atomic orbitals of sulfur has one filled orbital, one filled orbital, and two partially filled orbitals. To accommodate a lone pair and bonding with two oxygen atoms (i.e., three electron groups), the valence orbitals of sulfur are hybridized to , which leaves an unhybridized orbital. Part D In the sketch of the structure of label all bonds. Drag the appropriate labels to their respective targets. Labels can be used once, more than once, or not at all. ANSWER: Help Reset
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 12/33 Part E Identify the hybridization of the atom in . ANSWER: Correct Part F In the sketch of the structure of label all bonds. Drag the appropriate labels to their respective targets. Labels can be used once, more than once, or not at all. ANSWER: Help Reset Lone pair in orbital Lone pair in orbital
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 13/33 Part G Identify the hybridization of the atom in . ANSWER: Correct For each atom to have a formal charge of 0, the valence shell of boron is only partially filled. The valence orbital and two of the valence orbitals of boron hybridize to form three equivalent orbitals. Part H In the sketch of the structure of label all bonds. Drag the appropriate labels to their respective targets. Labels can be used once, more than once, or not at all. ANSWER: Help Reset Lone pair in orbital Lone pair in orbital Lone pair in orbital
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 14/33 Exercise 11.67 - Enhanced - with Feedback and Hints MISSED THIS? Read Section 11.7 (Page) ; Watch KCV 11.7 , IWE 11.8 . Part A Consider the structure of the amino acid alanine. Indicate the hybridization about each interior atom. Drag the labels to the appropriate targets. You did not open hints for this part. ANSWER: Help Reset Empty orbital Empty orbital Lone pair in orbital Lone pair in orbital
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 15/33 Exercise 11.68 - Enhanced - with Feedback MISSED THIS? Read Section 11.7 (Page) ; Watch KCV 11.7 , IWE 11.8 . Part A Consider the structure of the amino acid aspartic acid. Indicate the hybridization about each interior atom. Drag the appropriate labels to their respective targets. ANSWER: Help Reset Help Reset sp sp 2 sp 3
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 16/33 Exercise 11.72 - Enhanced - with Feedback MISSED THIS? Read Section 11.8 (Page) ; Watch IWE 11.10 . Part A Complete the atomic orbital (AO) and molecular orbital (MO) energy diagram for . Drag the appropriate labels to their respective targets. ANSWER: Part B Complete the AO and MO energy diagram for . Assume the left AO comes from and the right AO comes from . Drag the appropriate labels to their respective targets. ANSWER: Help Reset
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 17/33 Part C Complete previous part(s) Part D Complete previous part(s) PhET Simulation - Molecule Shapes The shapes of molecules depend on the number of electron groups that surround a central atom. For molecules in which all electrons around the central atom are participating in bonding, the molecular geometry is the same as the electron geometry, and the molecular shapes are linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. However, nonbonded electrons, which wouldn't be observed in the molecular geometry, affect the overall distribution of electron groups; therefore, the molecular and electron geometries will be different when nonbonding electrons are present. This can be exemplified in the case of ammonia, . Ammonia has three bonding groups, but it will not exhibit trigonal planar geometry because the lone pair of electrons exerts a repulsive force. There are four electron groups (three bonding groups and one nonbonding group) in ammonia, which means that the electron geometry will be tetrahedral . When only the molecular structure is examined, the lone pair is not seen, and the molecular geometry will adopt a more pyramidal structure that can be seen in the image below ( trigonal pyramidal ). Electron geometry of is tetrahedral Molecular geometry of is trigonal pyramidal The valence-shell electron-pair repulsion (VSEPR) model encompasses the geometries that result from the various interactions that occur between electron groups (also called electron domains ) and the relative repulsive forces exerted by each type of electron group (lone pair, single bond, double bond, and triple bond). VSEPR models also predict both the electron and molecular geometries, but not all reference charts may indicate bond angles. Click on the image to explore this simulation , which allows you to build molecules with lone pairs and various types of bonds while showing the resulting geometries. When you click the simulation link, you may be asked whether to run, open, or save the file. Choose to run or open it. Help Reset
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 18/33 When the simulation is opened, you will the option for Model or Real Molecules modes. In the Model mode, you can add single, double, and triple bonds as well as remove them using the Bonding box. Lone pairs can be added or removed using the Lone Pair box. If you want to easily reset the molecule, click the Remove All button to have just the central atom displayed. There are also checkbox options for showing lone pairs and bonding angles (when they exist) and the applicable Molecule Geometry and Electron Geometry . In the Real Molecules mode, you have the same options that existed in the Model mode, but the displayed molecule is selected from a dropdown menu. The radio buttons labeled Real and Model that are centered at the top alters the display in the following manner: The color of each atom is generic in the model view, but colors are specific to each element in the real view; the predicted bond angles based on geometry are shown in the model view, but the experimentally observed bond angles are shown in the real view. Part A The simulation shows the electron and molecular geometries for a variety of compounds. For example, click the Real Molecules mode and select as the molecule. Check the boxes marked Show Lone Pairs (under Options to the right) as well as Molecule Geometry and Electron Geometry (under Name in the bottom left), and the respective geometries should display, which are Square Pyramidal and Octahedral , respectively. The structure of bromine pentafluoride consists of one lone pair and five single bonds. You can recreate this molecule in the Model mode by adding a single lone pair and five single bonds to the central atom (click the Remove All button to start over). You will have successfully recreated the molecule if the displayed molecular and electron geometries are square pyramidal and octahedral, respectively. Use the Model mode in the simulation to identify the geometries for ozone, phosphate ion, and argon fluorohydride (for which the Lewis structures are depicted, and the resonance forms can be ignored). Drag the labels to the respective targets You did not open hints for this part. ANSWER:
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 19/33 Bond angles The types of electron groups not only affect the electron and molecular geometries but have differing effects on bond angles. Lone pairs of electrons repulse other electron groups more strongly than single bonds. For instance, four single bonds in a tetrahedral formation will yield 109.5 bond angles because all bonds equally repulse each other. In the case of three single bonds and one lone pair (e.g., in , the electron geometry is tetrahedral, and the molecular geometry is trigonal pyramidal), the bond angle between the three single bonds will actually be more acute than the predicted 109.5 (instead, they are 107.8 ) because the repulsion from the lone pair forces the bonds closer together. This behavior can also be seen in the simulation. Click the Real Molecules mode, and select as the molecule. Check the box marked Show Bond Angles , and switch between the Real and Model radio buttons centered at the top of the simulation. What is being demonstrated is the increased repulsion of lone pairs, and that is why the model predicts 109.5 , and the actual bond angle is 104.5 . These differences in the bond angles can be seen for the following compounds in the simulation: , , , , , and . There are no changes to predicted bond angles when all the bonds are in the same plane because the lone pairs are equally repulsing on each side of the plane, which occurs in the case of and . Part B Consider the following pairs of molecules that have the same electron geometries. Based on your observations between the Model and Real bond angles in the PhET simulation, complete the sentences to accurately compare the bond angles in each molecule. Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. You did not open hints for this part. ANSWER: Help Reset seesaw linear trigonal planar T-shaped bent trigonal bipyramidal square planar octahedral square pyramidal tetrahedral trigonal pyramidal
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 20/33 Effects of electronegativity in molecular bonds When atoms participate in a covalent bond, the electrons they share are not always evenly distributed within that bond. Electronegativity describes the tendency for an atom to draw electrons toward its nucleus. When there is a significant enough difference between the electronegativities of two atoms, the bond is considered polar covalent since the electrons spend more time near the more electronegative nucleus. In a polar covalent bond, the less electronegative atom will have a partial positive charge ( ), and the more electronegative atom will have a partial negative charge ( ). If there is no significant difference between the electronegativities of the atoms, then the bond is considered nonpolar covalent (and the nonpolar aspect can be characterized by the symbol). Finally, when the electronegativity difference is very large, ionic bonds tend to form instead of covalent bonds. The table below summarizes how to classify bonds based on the numeric difference between the Pauling electronegativity values of atoms. Pauling electronegativity difference Type of bond <0.4 Nonpolar covalent 0.4 to 1.7 Polar covalent >1.7 Ionic Part C The structures of , , and are provided below along with the Pauling electronegativity values of , , and . Use the figure and values to complete the following statements regarding the polarity of bonds in water, carbon dioxide, and methane. Element Pauling electronegativity 2.55 2.20 3.44 Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. You did not open hints for this part. ANSWER: Reset Help less than equal to greater than The bond angle in is the bond angle in . The bond angle in is the bond angle in . The T-junction (one axial and one equatorial T atom) bond angle in is the T-junction bond angle in . The bond angle in is the bond angle in . The bond angle in is the bond angle in .
11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 21/33 Polar molecules and their dependence on geometry Molecular geometry impacts whether a molecule is polar. Polarity is important because polar molecules interact more strongly with each other (intermolecular interaction) than nonpolar molecules. This explains why nonpolar molecules tend to be gases at room temperature and why polar molecules tend to be liquids at room temperature. Stronger intermolecular forces require more energy to separate them from each other. The molecular geometry and bond polarities both play critical roles in determining whether or not a molecule is polar. A polar molecule has a net dipole moment, which means that there is an uneven distribution of electrons across the entire molecular structure. This dipole moment is sometimes indicated by an arrow, which has a cross at the end of the lowest electron density (positive end) and points in the direction of the highest electron density (negative end). This dipole moment is achieved when the molecule's polar covalent bonds have an asymmetric orientation. Examine the structures of carbon dioxide and carbonyl sulfide below. Both molecules contain at least one polar covalent bond, but only is symmetric. The asymmetry in is due to oxygen being more electronegative than sulfur. As you can see, the symmetry in effectively cancels out all bond dipole moments, but a net dipole moment can exist in because of its asymmetry. Part D Examine the three-dimensional structures of each of the following molecules in the simulation, which can be found in the Real Molecules mode. Then, identify which molecules are polar and which are nonpolar. Assume that every bond in each molecule is polar covalent. It may be easier to visualize if you uncheck the box labeled Show Lone Pairs . Drag the appropriate items to their respective bins. You did not open hints for this part. ANSWER: Reset Help ionic more polar nonpolar less 1. For the water molecule, oxygen is electronegative than hydrogen. 2. The difference in electronegativity signifies that the hydrogen-oxygen bond is . 3. According to the electronegativity difference between the atoms in water, it would be appropriate to label hydrogen with the symbol and oxygen with the symbol . 4. Based on the electronegativity difference between carbon and oxygen, the bonds in carbon dioxide are . 5. According to the electronegativity difference between the atoms in carbon dioxide, it would be appropriate to label oxygen with the symbol and carbon with the symbol . 6. Based on the electronegativity difference between carbon and hydrogen, the bonds in methane are . 7. According to the electronegativity difference between the atoms in methane, it would be appropriate to label hydrogen with the symbol and carbon with the symbol .
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 22/33 PhET Simulation – Molecule Polarity Electronegativity is a term used to define an atom’s ability to attract electrons to itself when it is bonded to another atom. A larger electronegativity value indicates a greater ability for an atom to attract electrons to itself. Electronegativity values follow trends (with exceptions) across the periodic table. They increase from left to right across a period and increase from the bottom of a group to the top. Thus, fluorine is the most electronegative element with a value of 4.0. Notice that since the noble gases have completely filled electron shells and are inert, they don’t have assigned electronegativity values. In any given molecule comprised of two or more nonidentical atoms, the electrons contained in the bond are not shared equally. Bond polarity is a measure of how unevenly the electrons in a bond are shared and is indicated by the difference between the electronegativities of the atoms involved. A large difference in electronegativities indicates that a bond is polar and contains a dipole (i.e., the bond character is more ionic), which is created when two equal but opposite charges are separated. A smaller difference in electronegativity makes the bond character more covalent. Click on the image below to explore this simulation , which demonstrates the cause and effect of molecular polarity. When you click this simulation link, you may be asked whether to run, open, or save the file. Choose to run or open it. When the simulation is opened you should select the option for Two Atoms . You will then see two atoms (A and B) with a dipole moment drawn directly over the bond. As the electronegativity for each atom is adjusted using the sliders above, the dipole moment arrow will grow, shrink, or change directions. There are check boxes to the right of the molecule to control whether bond dipoles, partial charges, or bond character are displayed. There are also switches to control the display of electrostatic potential, the display of electron density, and the ability to enable an electric field. Clicking the Three Atoms tab brings up a similar display but with an additional atom. Clicking the Real Molecules tab enables you to view the properties of various real compounds. Part A In the PhET simulation window, click the Two Atoms option. Click on the partial charges check box to see the partial charges of the atoms. Vary the electronegativity of atom A and atom B and observe the effects on the partial charge of each accordingly. Use the PhET to complete the sentences describing how electronegativity is related to bond dipole experienced by the molecule. Reset Help Polar Nonpolar
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 23/33 Match the words to the appropriate blanks in the sentences below. Make certain each sentence is complete before submitting your answer. You did not open hints for this part. ANSWER: Bond polarity Bond polarity results from two bonded atoms having differing electronegativity values. Covalent bonds exist when the two atoms involved in the bond share the electrons. These electrons are shared either evenly or unevenly to a degree depending on the electronegativity differences between the two atoms. Ionic bonds exist when the two atoms transfer electrons. These electrons are so strongly attracted to the more electronegative atom that there is virtually no sharing of electrons and the atoms are simply attracted to another by having opposite charges. The difference in electronegativity values can be used to predict the type of bond formed between two atoms. Nonpolar covalent : Bonds between atoms with electronegativity differences of 0.4 and under are generally considered nonpolar covalent bonds since the electrons are shared relatively evenly. This is true even if there are two different atoms in the bond, such as a bond. Polar covalent : Bonds between atoms with electronegativity differences between 0.4 and 2.0 are generally considered polar covalent since the electrons are shared unevenly. Ionic : Bonds between atoms with electronegativity differences greater than 2.0 are generally considered ionic since the electrons are localized on the more electronegative atom. Part B Using the given table of electronegativity values, determine whether each bond is nonpolar covalent, polar covalent, or ionic. Element Electronegativity Element Electronegativity 4.0 2.5 3.5 2.5 3.0 2.1 2.9 1.6 2.7 0.9 2.6 0.8 Drag the appropriate items to their respective bins. You did not open hints for this part. ANSWER: Reset Help more less increase negative positive lowest ionic highest decrease covalent dipole 1. When two atoms with different electronegativities are bonded together, a bond exists. These are displayed with an arrow originating at the atom with the electronegativity and pointing toward the atom with the electronegativity. 2. Partial charges will exist on each atom when a bond exists between two atoms with different electronegativities. Since the atom with a higher electronegativity will have a stronger ability to attract electrons, it will have a partial charge. The atom with a lower electronegativity will have an equal and opposite charge. As the difference in electronegativity increases, the resulting magnitudes of the partial charges will . 3. Bond character is used to describe the severity of the electronegativity differences between the two atoms involved in a bond. Bonds that have character correspond to a larger difference in electronegativities, and bond character corresponds to a smaller difference. 4. For example, if the electronegativity of both atom "A" and "B" in the PhET are set to be equivalent, then increasing the electronegativity of atom "A" in the PhET simulation and decreasing the electronegativity of atom "B" will result in a bond that is ionic in character.
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 24/33 Molecular dipoles and shape Molecular dipoles are overall dipoles for an entire molecule that result from individual bond dipoles and molecular shape. For molecules where the shape is uncertain, Lewis structures can be used to predict molecular geometry. If the only bond dipoles that exist are equal in magnitude and opposite in direction, then there is no molecular dipole and the molecule is considered nonpolar. If bond dipoles exist and this is not the case, then the molecule has a molecular dipole and it is considered polar. For example, since the molecule below is linear and the dipoles are equal and opposite, they cancel out, leaving the molecule nonpolar with no net molecule dipole. However, if instead the molecule contained atoms , , and , where the bond dipoles for and were not equal, we would expect a molecule dipole and a polar molecule. Keep in mind there are instances in which a molecule can be made of a single element and still have a molecular dipole. For example, ozone contains a molecular dipole because the electrons involved in the bonds between the oxygen atoms are unequally shared, and its molecular shape doesn't place the resulting dipoles in equally opposite directions. The electrons involved in the bonds of ozone are shared unequally because the central oxygen atom has to share electrons with the two oxygen atoms on the ends, while the others only have to share electrons with the central atom. This gives the central oxygen a slight positive partial charge. Part C NOTE: The simulation required for this part uses Java, and is therefore not screen-reader accessible and may not work on a mobile device. If the browser you’re using no longer supports Java, try a different browser and download the Java plugin for this content. In the PhET simulation window, click the Real Molecules option. Select the corresponding molecule and click on the molecular dipole check box in the View menu on the right. Use the Molecule dropdown box to choose a different molecule. Indicate whether each molecule is polar or nonpolar. Drag the appropriate items to their appropriate bins. You did not open hints for this part. ANSWER: Reset Help bond between and bond between and bond between and bond between and bond between and bond between and bond between and bond between and bond between and bond between and Nonpolar covalent Polar covalent Ionic
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 25/33 Electric field Electric fields can orient molecules based upon their molecular dipoles and partial charges. Positive partial charges will align toward the negative end of the electric field, and negative charges will align toward the positive end of the field. Part D In the PhET simulation window, click the Three Atoms option. Drag the switch to "on" under Electric Field . Vary the shape and electronegativity of each atom to observe the effect of the electric field. After experimenting with the PhET, draw a molecule of dibromomethane ( ). Use this drawing to assign partial charges and determine whether or not the molecule is polar. Color Element Electronegativity Dark gray (large) Carbon 2.5 Light gray (small Hydrogen 2.1 Brown Bromine 2.9 The positive plate is on the right, and the negative plate is on the left. An electric field is set up between the plates. The molecule will orient according to the electric field to minimize charge distribution. Complete the diagram showing how dibromomethane ( ) will orient itself in an electric field according to partial positive and negative charges. Drag the appropriate labels to their respective targets. You did not open hints for this part. ANSWER: Reset Help Polar molecules Nonpolar molecules
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 26/33 Exercise 11.31 - Enhanced - with Feedback MISSED THIS? Read Section 11.2 (Page) ; Watch KCV 11.2 , IWE 11.1 . A molecule with the formula has a trigonal pyramidal geometry. Part A How many electron groups are on the central atom ( )? Express your answer as an integer. ANSWER: Chapter 11 Algorithmic Question 9 Part A Determine the electron geometry (eg) and molecular geometry (mg) of SiF 4 . ANSWER: Help Reset Group 2 Group 1 Group 1 Group 3 Group 3 Group 3 Group 3 Group 1 Group 1 polar nonpolar
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 27/33 Correct Chapter 11 Algorithmic Question 10 Part A Determine the electron geometry (eg) and molecular geometry (mg) of PF 5 . ANSWER: Correct Chapter 11 Algorithmic Question 14 Part A Identify which of the following statements best describes ClBr 2 - ? ANSWER: Correct Chapter 11 Algorithmic Question 19 eg = tetrahedral, mg = tetrahedral eg = octahedral, mg = trigonal bipyramidal eg = tetrahedral, mg = bent eg = trigonal bipyramidal, mg = trigonal bipyramidal eg = tetrahedral, mg = trigonal pyramidal eg = trigonal planar, mg = trigonal pyramidal eg = tetrahedral, mg = trigonal pyramidal eg = trigonal bipyramidal, mg = trigonal bipyramidal eg = octahedral, mg = square pyramidal eg = trigonal bipyramidal, mg = trigonal pyramidal It has a molecular geometry that is linear with lone pairs on the Cl atom. linear with no lone pairs on the Cl atom. trigonal pyramidal with lone pairs on the Cl atom. square planar with no lone pairs on the Cl atom.
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 28/33 Part A Determine the electron geometry (eg) and molecular geometry (mg) of XeF 2 . ANSWER: Correct Chapter 11 Algorithmic Question 20 Part A Determine the electron geometry (eg) and molecular geometry (mg) of KrF 4 . ANSWER: Correct Chapter 11 Algorithmic Question 24 Part A What is the molecular geometry of IF 4 - ? ANSWER: eg = tetrahedral, mg = bent eg = trigonal bipyramidal, mg = tetrahedral eg = tetrahedral, mg = tetrahedral eg = trigonal bipyramidal, mg = linear eg = linear, mg = trigonal pyramidal eg = trigonal bipyramidal, mg = square planar eg = tetrahedral, mg = trigonal pyramidal eg = octahedral, mg = square planar eg = tetrahedral, mg = square planar eg = linear, eg = square planar
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 29/33 Correct Chapter 11 Algorithmic Question 38 Part A What is the O-S-O bond angle in SO 3 ? ANSWER: Correct Chapter 11 Algorithmic Question 48 Part A Determine the electron geometry (eg), molecular geometry (mg), and polarity of N 2 O (N central). ANSWER: Correct Chapter 11 Algorithmic Question 45 trigonal bipyramidal tetrahedral square pyramidal square planar seesaw 109.5° 107° 60° 120° 240° eg = linear, mg = linear, nonpolar eg = trigonal planar, mg = linear, nonpolar eg = tetrahedral, mg = bent, polar eg = linear, mg = linear, polar eg = tetrahedral, mg = linear, nonpolar
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 30/33 Part A Choose the compound below that contains at least one polar covalent bond, but is nonpolar. ANSWER: Correct Chapter 11 Algorithmic Question 55 Part A The orbital hybridization on the carbon atoms in SCN - is ANSWER: Correct Chapter 11 Algorithmic Question 57 Part A The hybrid orbital set used by the central atom in BrO 3 - is ________. ANSWER: AsBr 5 GeH 2 Br 2 SCl 2 CH 2 Cl 2 All of the above are nonpolar and contain a polar covalent bond. sp. sp 3 d. sp 2 . sp 3 . sp 2 d.
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 31/33 Correct Chapter 11 Algorithmic Question 61 Part A A molecule containing a central atom with sp hybridization has a ________ electron geometry. ANSWER: Correct Chapter 11 Algorithmic Question 76 Part A Give the electron geometry (eg), molecular geometry (mg), and hybridization for H 2 O. ANSWER: Chapter 11 Multiple Choice Question 36 sp sp 3 d 2 sp 2 sp 3 d sp 3 T-shaped linear square pyramidal tetrahedral trigonal bipyramidal eg = trigonal pyramidal, mg = bent, sp 3 eg = tetrahedral, mg = bent, sp 3 eg = bent, mg = trigonal bipyramidal, sp 2 eg = trigonal planar, mg = tetrahedral, sp 2 eg = tetrahedral, mg = trigonal pyramidal, sp 3
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 32/33 Part A Give the hybridization for the Br in BrCl 3 . ANSWER: Chapter 11 Multiple Choice Question 38 Part A Give the hybridization for the C in H 2 CCH 2 . ANSWER: Chapter 11 Multiple Choice Question 43 Part A Use the molecular orbital diagram shown to determine which of the following is most stable. sp 3 sp sp 3 d 2 sp 2 sp 3 d sp 3 sp 3 d 2 sp sp 3 d sp 2
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11/12/22, 11:32 PM VSEPR and MO Diagrams Workshop Activity - Nov 7th/8th (Workshop 10) https://session.masteringchemistry.com/myct/assignmentPrintView?assignmentID=11274219 33/33 ANSWER: Chapter 11 Multiple Choice Question 46 Part A Draw the molecular orbital diagram shown to determine which of the following is paramagnetic. ANSWER: Score Summary: Your score on this assignment is 16.5%. You received 5.63 out of a possible total of 34 points. N 2 2 + C 2 2 - B 2 2 + C 2 2 + B 2 O 2 2+ F 2 2+ Ne 2 2+ O 2 2- None of the above are paramagnetic.
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Week 8-1 X + " https://app.101edu.co 79°F Mostly cloudy Substance NH:(9) HCI (g) NH.CI (S) Chapter 5 x 00 5.3 The Fu X 00 Quiz 5.1 to x 0o DESU Calc X 2 W " # 3 * E $ 4 27 O R 12 Calculate AS for NH3(g) + HCI(g) → NH4Cl(s). % 5 " Search 40 6 الت " S* (J/mol-K) 192.5 186.7 94.6 4- & 7 YIL M Log In | Fe X Aktiv Chen X U Question 8 of 23 4+ 8 DA hp 9 ► 11 M ) IL O O Flashcards X O A4 P C { Photograp X A + L A = O Welcome X Ⓒ P } 1 4 7 +/- G prt sc 。 2 5 8 delete backspace B Log In | Fe X + J/mol K 3 6 9 0 home . num lock x 1 end G 7 home
MasteringChemistry: Chapter X P Introduction: Chemical Quantit X + C https://session.masteringchemistry.com/myct/itemView?assignmentProblemID=114908580 Spring 2019
Solve fast and accurate,
General Chemistry I Laboratory Manual, 2021 Revision 3) Reaction of Hydrogen with Oxygen: AHreaction per mole of H₂O Formed: Give a reference for this value, making sure you cite author, year, text name and edtion, publisher, and page number. pu Reaction of Magnesium with Oxygen: Summarize Data from the first three parts. odl Reaction 1) MgO(s) + 2HCl(aq) → MgCl2(aq) + H₂O(1) 2) Mg(s) + 2HCl(aq) → MgCl2(aq) + H₂(g) 3) H₂(g) + ¹/2O2(g) → H₂O(1) RI 4) Use your data and Hess' law to calculate AHreaction for: Mg(s) + 1/2O2(g) → MgO(s) Contex M 20 to tagio SPP ΔΗ S 117 T A Joost hard Maan HA
The heat of formation at 25°C of liquid acetone, CH,C(0)CH,(1) is -247.5 kJ/mol. Which of the following is the correct heat of formation reaction for CH,C(O)CH,() )?
ps//testnavclient.psonsvc.net/#/question/37ae15ac-69b1-43c8-97b4-937c4ad3606e/d2839865-2121-4791859c 14226e61458c Review ABookmark CE CHEMISTRY HONORS MP2 2021-22 / 19 of 21 Identify which of the following oxide of nitrogen will have 1.513 mol in 114.95 grams of its compound. O A NO2 O B. N203 O C NO O D. N20 P Type here to search 35°F IOI ho hor #3 3. $ 4 @ & 5. 7 8. W R Y A D F J K
9:54 AM Mon Aug 9 * 92% AA saplinglearning.com + Sapling Learning SCC 110 Final Exam Audrea Formantes - macmillan learning Sapling Learning > LaGuardia Community College - SCC 110 - Summer21 - TSAI > Activities and Due Dates > sCC 110 Final Exam Check Answer Assignment Score: 74% Resources Hint 01 49 -- Consider the following reaction. MgCl, (aq) + 2 NaOH(aq) → Mg(OH), (s) + 2 NaCI(aq) Calculate the volume of 0.275 M MgCl, that is needed to react completely with 26.2 mL of 0.432 M NaOH. O 41.2 mL O 8.34 mL O 20.6 mL O 82.3 mL !!!
Complete the following reactions ?
You may want to reference (Pages 245-246) Section 7.8 while completing this problem. 3 E Course Home. Calcium cyanamide reacts with water to form calcium carbonate and ammonia via the following reaction: CaCN2 (s) + 3H₂O(1)→ CaCO3(s) + 2NH3(g) 80 F3 $ 4 000 000 F4 % 5 M MasteringChemistry: Homework Chapter 7 F5 openvellum.ecollege.com ^ 6 ▼ Part A F6 How many grams of water are needed to react with 71.0 g CaCN₂? mass H₂O = Submit Previous Answers Request Answer Part B X Incorrect; Try Again; 5 attempts remaining mass NH3 = IVE ΑΣΦ How many grams of NH3 are produced from 5.22 g CaCN₂? MacBook Air b Answered: Attempt 1 ed with Feedback w.. Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining & 7 [ΨΕ ΑΣΦ Copyright © 2022 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy | Permissions Contact Us | P Pearson Ad F7 * 8 ∞ DII F8 Swed ( 9 F9 ? ? g ) 0 g 21 Shopping Ba Review | Constants F10 () F11
Document1 - Word Search (Alt+Q) References Mailings Review View Help Write the corresponding letters to the following descriptions. Each proposal can be associated with 0 or 1 match and each letter is not necessarily associated with a proposal. a. [Ar] 4s? b. 1s 2s2p63s23p3 c. 1s2s2p2 d. [Ne] 3s23ps e. [Ar] 4s23d10 f. 1s22s! g. None of those answers 1. I am a halogen 2. I am a transition metal 3. I am alkaline 4. I have exactly 2 single electrons 5. I am the smallest of the atoms presented here 6. I have exactly 3 valence electrons 7.I am the most paramagnetic of the atoms presented here 8. I am the copper 9.I am isoelectronic with Ti2+ IF cessibility: Good to go IA 16 SIS
2. A reaction was found to proceed mechanistically in two steps. The following data was reported: Step 1: Ez = 35 kJ/mol; AHº = +25 kl/mol Step 2: Eact = 7 kl/mol; AH° = -45 kJ/mol (a) Draw a potential energy diagram for the complete reaction. Identify the x and y axes, and label the diagram with the following: the transition states; the rate - determining step; the locations of the reactants, intermediates, and products of the reaction; the two Ega values; and the two AH° values given. (b) Calculate the overall AH° for the reaction from the graph, and label your diagram with it. (c) Are the first and second steps endothermic or exothermic? Justify your choices. (d) Are the transition states each for the first like? Justify your answers! second steps more product – like or more reactant -
7. You can see an MSDS below. Please answer the following questions related to the MSDS. a) What is the name of this chemical? b) What should you do if someone drinks the chemical? c) Would this chemical catch on fire if it was exposed to flames? d) If this chemical gets in your eye what should you do? e) What color is this chemical? f) What should you do if someone spills a small amount of the chemical?
rsity o... ingsbo... kboard... ARED4 ethod? 7... F2 Remaining Time: 1 hour, 22 minutes, 23 seconds. Question Completion Status: A Moving to the next question prevents changes to this answer. Question 1 # The angle of the water molecule (H₂O) is 180 degrees 120 degrees 100 degrees 90 degrees 109 degree A Moving to the next question prevents changes to this answer. MAR 14 80 F3 SA $ 000 F4 tv % F5 NA MacBook Air 22 F6 ∞r F7 45.113
Pls help ASAP ON ALL ASKED QUESTIONS PLS PLS
Chrome File Edit View History Bookmarks People Tab Window Help A ALEKS - Jacqueline Hoppenrey x Mcc Chapter 4: Reactions in Aqueo x G halogens - Google Search A www-awn.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNslkr7j8P3jH-IQİHqRdYV_6Ux63SypJXz0Coxvwqgg4JkWI7UFmRjQiHo9r7UFmxvG2Y0ofvn27IAf2ti O SIMPLE REACTIONS Predicting the products of a combustion reaction Predict the products of the following reaction. If no reaction will occur, use the NO REACTION button. Be sure your chemical equation is balanced! CH,CH,CH,(g) + 0,(8) -0 NO ローロ REACTION Explanation Check ©2021 McGraw-Hill Education II
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Calculus Preface (?) Brytwave Blackboard Quizlet Clutch E Knowledge Check Question 9 Black Forest Biologicals, a biotech startup, has a promising Alzheimer's drug candidate Compound SLT-88 entering Phase I trials this year. SLT-88 is the only product formed by the reaction of two precursor compounds A and B, both of which are quite expensive. The chief medicinal chemist of Black Forest is trying out different reaction conditions to minimize the cost of manufacturing SLT-88. In the table below are listed the initial and final amounts of A and B used under two different trial conditions, and also the actual amount of SLT-88 recovered in each case. Complete the table by calculating the theoretical yield of SLT-88 and the percent yield of SLT-88. Round your amounts to the nearest milligram and your percentages to the nearest whole percent. amount of A amount of B yield of SLT-88 Trial initial final initial final theoretical actual 1000. mg 0 mg 550. mg 286. mg 999. mg 150. mg 0 mg 1000. mg…
Calculate deltaSrxn for the following reaction. The S for each species is shown below the reaction. C2H2(g)+H2(g)-->C2H4(g) 200.9 130.7 219.3
10 View History Bookmarks People Tab Window Help G convert c to O GroupMes E Buy Essay On x A ALEKS - Jacq x Hoc Chapter 5: Th x Hue Dashboard -> A www-awn.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNslkr7j8P3jH-IQİHQRDYV_6Ux63SypJXz0Coxvwqgg4JkWI74Meq7fCi1Pjl. O GASES Jacqu Interconverting molar mass and density of ideal gases A sample of an unknown compound is vaporized at 200. °C. The gas produced has a volume of 1760. mL at a pressure of 1.00 atm, and it weighs 1.54 g. Assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. Round your answer to 3 significant digits. mol Explanation Check O 2021 McGraw-Hill Education, All Rights Reserved. Terms of Use | Privacy MacBook Air
Construct the expression for Kc for the following reaction. 4 HCI(aq) + O:(g) = 2 H:O(I) + 2 Cl:(g) 1 Drag the tiles into the numerator or denominator to form the expression. Each reaction participant must be represented by one tile. Do not combine terms. Kc 5 RESET [HCI] 2[HCI] 4[HCI] [HCIP [HCIJ* [0:] 2[0:] 4[O:] [H:O] 2[HŁO] 4[H.O] [H:OJ [CL] 2[CL] 4[CL] [CL*
SCH4U-Unit 3 Assignment-2023 1. Calculate AH for the reaction: HCl(g) + NaNO2(s)→ HNO2(1) + NaCl(s) using the following data: Reaction 2NaCl(s) + H₂O(1)→ 2HCI(g) + Na₂O(s) NO(g) + NO₂(g) + Na₂O(s)→ 2NaNO₂(s) NO(g) + NO₂(g) →N₂O(g) + O2(g) 2HNO2(1)→→ N₂O(g) + O2(g) + H₂O(1) ΔΗ(kJ) 382.0 -601.0 -40.0 Page 2 of 6 29.0 2. A 5.81g sample of quinone, C6H4O2, was burned in excess oxygen in a calorimeter. The heat capacity of the calorimeter is calibrated to 4.24 kJ/°C. Calculate the molar enthalpy of combustion if the temperature of the calorimeter increased from 17.0 °C to 38.0 °C.
Read and interpret the following information using the following abstract from a scientific journal article. In 2-3 sentences, summarize what you understand from this research
A&P II PSY Connect Part A Course Home H₂C pt Draw the major organic product formed when the compound shown below undergoes a reaction with an excess of EtOH under acidic conditions. Interactive 3D display mode chem in career H: 2D EXP 4 Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. I UZO 38 o Review | Constants | Periodic
I'm not sure how to solve 115. I saw one solution that used the ka to solve it, but I don't understand how to find the ka to use or where they were able to get the ka from with the given information in the problem. So if you could please make this clear in the answer. Thank you for your help.
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☐ 5:41 PM Sat Jun 29 kennesaw.view.usg.edu Name: Lab Section (Day & Time): 3361L Molecular Modeling and Stereochemistry Worksheet CHEM Ground Rules: 1. You are required to build models for various tasks presented in this exercise. Molecular modeling programs are not a substitute for 3-D hand-held molecular models. 2. You will be prompted to include a snapshot of your model with each task at the end of this worksheet. Read the question carefully to construct and orient the models appropriately. Use your smartphone to take a picture of the model you build, and insert it at the end of the handout with corresponding prompts. You will not get any credit for this assignment without correct snapshots of the models you build. 3. Hand-write your responses for each task and insert the snapshots for various models at the end of this docuent. Save the entire assignment as a single PDF. A consolidated PDF of the completed assignment (including relevant snapshots of your models should be submitted…
Dangerous Paint Stripper Jessica has a summer job working for the city parks program. She has been using a cleaner called “Graffiti Gone” to remove graffiti from the bathrooms. She has to take a lot of breaks, because the chemical makes her throat burn. It also makes her feel dizzy sometimes, especially when the bathrooms don’t have very many windows. On the label, she sees that the cleaner has methylene chloride in it. She feels like she’s managing to get the work done, but she is worried about feeling dizzy. She wants to find out more about this chemical, what harm it can cause, and whether there are safer ways to do this work. Questions for following story. 1. What went right in this situation? 2. What went wrong in this situation? 3. What steps should be taken in this workplace to make sure employees are better protected and prepared the next time?
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