![EBK CHEMICAL PRINCIPLES](https://www.bartleby.com/isbn_cover_images/9781305856745/9781305856745_largeCoverImage.jpg)
(a)
Interpretation: The formula of the four superconductors needs to be determined.
Concept Introduction: The superconductor is a substance which is capable of superconducting at very low temperature values.
(a)
![Check Mark](/static/check-mark.png)
Explanation of Solution
The given structures are as follows:
By considering the color code for spheres, the number of atoms of each element can be counted.
It can be observed that in molecule (a) there is 1 Tl, 2 Ba, 1 Cu and 5 O atoms in 1unit. Thus, the formula of the molecule will be:
Similarly, the atoms in molecule (b) are 1 Tl, 2 Ba, 1 Ca, 1 Cu and 7 O in 1unit. Thus, the formula will be:
The structure (c) contains 1 Tl, 2 Ba, 2 Ca, 3 Cu and 9 O atoms in 1unit. The formula will be:
Similarly, the structure (d) has 1 Tl, 2 Ba, 2 Ca, 4 Cu and 11 O atoms in 1unit thus, the molecular formula will be:
(b)
Interpretation: The four structures needs to be ordered from lowest to highest superconducting temperature.
Concept introduction: The number of sheets in the unit cell is directly proportional to the temperature for the superconductivity.
(b)
![Check Mark](/static/check-mark.png)
Explanation of Solution
If the number of sheets in each unit cell increases, the temperature for the superconductivity increases. The number of sheets in the structure (a) is less then (b) which is further less than (c) and (d). Thus, the increasing order of the superconductivity for 4 structures will be as follows:
(c)
Interpretation: The oxidation state needs to be assigned to Cu in each mixture. It is assumed that the oxidation state of Tl is +3. Also, the oxidation state of Ca, Ba, O is +2, +2 and -2 respectively.
Concept Introduction: The oxidation state of any atom in the molecule is equal to the charge on it. It can be calculated if oxidation state of all the other atoms is given.
For example, the molecule H2O2 is neutral. The oxidation state of H is +1 (general) thus, the oxidation state of O can be calculated as follows:
Thus, the oxidation state of O in H2O2 is -1.
(c)
![Check Mark](/static/check-mark.png)
Explanation of Solution
The oxidation state of Cu in structure can be calculated by taking the overall charge on the molecule equal to zero.
The oxidation state of Tl, Ca, Ba and O is assumed +3, +2, +2 and -2 respectively.
For
Thus, the oxidation state of Cu in molecule (a) is +3.
For
There are 2 Cu atoms, thus, there is a mixed oxidation state one is
For
Thus, copper has mixed oxidation state that is 1
For
There will be one
(d)
Interpretation: The reason for the copper to display a mixture of oxidation states in the superconductor needs to be explained.
Concept Introduction: The superconductor is a substance which is capable of superconducting at very low temperature values.
(d)
![Check Mark](/static/check-mark.png)
Explanation of Solution
There are variable oxidation states of copper due to varying the numbers of Ca, Cu and O atoms in each unit cell. The different oxidation states of Cu are calculated for molecules (b), (c) and (d). In the other superconductor in exercise 79, in YBa2Cu3O7, there is variable oxidation state of copper by omitting oxygen atom at various sites in the lattice.
Want to see more full solutions like this?
Chapter 16 Solutions
EBK CHEMICAL PRINCIPLES
- 3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forward
- in which spectral range of EMR, atomic and ionic lines of metal liesarrow_forwardQ2: Label the following molecules as chiral or achiral, and label each stereocenter as R or S. CI CH3 CH3 NH2 C CH3 CH3 Br CH3 X &p Bra 'CH 3 "CH3 X Br CH3 Me - N OMe O DuckDuckarrow_forward1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward
- 1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward6. How many peaks would be observed for each of the circled protons in the compounds below? 8 pts CH3 CH3 ΤΙ A. H3C-C-C-CH3 I (₁₁ +1)= 7 H CI B. H3C-C-CI H (3+1)=4 H LIH)=2 C. (CH3CH2-C-OH H D. CH3arrow_forward
- Physical ChemistryChemistryISBN:9781133958437Author:Ball, David W. (david Warren), BAER, TomasPublisher:Wadsworth Cengage Learning,Chemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStaxChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
- Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781133958437/9781133958437_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781938168390/9781938168390_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305957404/9781305957404_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305079243/9781305079243_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781133611097/9781133611097_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781285199047/9781285199047_smallCoverImage.gif)