×
Log in to StudySoup
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 18 - Problem 18.68
Join StudySoup for FREE
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 18 - Problem 18.68

Already have an account? Login here
×
Reset your password

The oxidation of 25.0 mL of a solution containing Fe21

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 18.68 Chapter 18

Chemistry: A Molecular Approach | 3rd Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

4 5 1 393 Reviews
31
0
Problem 18.68

The oxidation of 25.0 mL of a solution containing Fe21 requires 26.0 mL of 0.0250 M K2Cr2O7 in acidic solution. Balance the following equation and calculate the molar concentration of Fe21: Cr2O7 22 1 Fe21 1 H1 Cr31 1 Fe31

Step-by-Step Solution:
Step 1 of 3

Chapter 7: Quantum Mechanical Model of Atom  Quantum Mechanics Model: Explains strange behavior of electrons  Classical Psychics o Unable to define quantum mechanics due to deterministic views o Deterministic environment is defined as a present set of conditions completely determining the future  Quantum Mechanics o Not Deterministic o If quantum mechanics doesn’t confuse you, you haven’t understood quantum mechanics yet  Section 7.2: Light  Objective: Describe Characteristics of light o Wave: Disturbance in a medium o Electromagnetic Radiation: Type of energy embodied in oscillating electric and magnetic fields  Magnetic Field: Region of space where a magnetic particles experiences a force (i.e. When holding magnetic material close to a magnet, you can feel a pull between the two, that is a force)  Electric Field: Region of space where an electrically charged particle experiences a force (i.e. A proton has an electric field around it. If you bring another charged particle into that field, that particle will experience a force) o Amplitude: Height from the midline to the maximum/minimum of the wave  Different amplitudes, different brightness (intensity) o Wavelength: Measured distance from crest to crest  Different wavelengths, different colors in the spectrum o Frequency: Count of wavelengths passing a given point in a given period of time  Inversely proportional to wavelength ( λ )  Directly proportional to speed  Speed of Light (c) 8 o c= 3.0∙10 m/s o c=λυ Frequency (Hz) Wavelength (m) o Example: What is the frequency of 650nm of light −9 Convert 650 nm to meters by just adding (x10 ) **Prefixes!** 8m −9 14 o Ex(mple: Cs)culate the wavelength (in nm) of the red light emitted by a barcode scanner that has a frequency of If you are given the frequency you can easily figure out the wavelength using this equation: c=λυ , by rearranging the equation to fit what we need we get: λ= c υ Convert wavelength from meters to nanometers by using the conversion factor between the two: (1nm= 10 - 9 m) 4.62∙10 s −1  Electromagnetic Spectrum o Interference: Two waves are in the same space  Constructive: “In phase”; peak overlaps peak making the overall amplitude of the wave increase  Destructive: “Out of phase”; Amplitude of the wave decreases because peaks overlap toughs, canceling out the wave (as seen above)  Diffraction: Bend of wave passing through object  Diffraction Pattern: Series of bright and dark spots past a double slit  Photoelectric Effect: When a metal can emit electrons when light hits it o Example: Care head light turns on when its dim enough outside o Specific energy required to emit electron o Threshold Frequency: Minimum frequency to eject electrons  No electrons are emitted from metal, no matter how long the light shines on the metal  Low-frequency (long wavelength) light does not eject electrons from metal regardless of its intensity or its duration  High-frequency (short-wavelength) light does eject electrons, even if its intensity is low  Albert Einstein: “Light energy must come in packets”  Photon: Quantized packet of energy  E=hν o Example: A nitrogen gas lazer with a wavelength of 337 nm contains 3.83 mJ of energy. How many photons does it Frequency(Hz) Planck’s Constant: contain Given: Epulse.83mJ λ=337nm Find: Number of Photons 10 m −7 λ=337nm∙ =3.37∙10 m 1nm −34 8m hc (6.626∙10 J ∙s)(3.0∙10s ) E Photon = =5.8985∙10 −1J v 3.37∙10−7m −3 3.83mJ ∙10 J =3.83∙10−3J 1mJ E −3 Numberof Photons= Pul=e 3.83∙10 J =6.49∙10 photons EPhoton5.8985∙10−19J  Section 7.4: The Wave Nature of Matter: The de Broglie Wavelength, the Uncertainty Principle, and Indeterminacy  de Broglie Relation: h λ= mv o **Notice: The velocity of a moving electron is related to its wavelength: Knowing one is equivalent to knowing the other  Heinsenberg’s Uncertainty Principle o Conclusion: Best we can do is probability Section 7.5: Quantum Mechanical Model A constant Uncertainty in position Uncertainty in velocity  Instead of orbits. . . o Orbitals: Probability distribution map showing where elctrongs most likely are found o Schrodingers Equation:  Wave function( ψ ) describes wave behavior of electrons  ( ψ ) = orbital o Quantum Numbers  Objective: Describe what characteristic each quantum number represents  Objective: Know which Quantum Numbers are allowed  n  principle number  Overall size and energy that we actually have (Bohr put it as a distance away from the nucleus)  n= 1,2,3,. . . ∞ (positive integers for n- values)  l  angular quantum number  Shape of orbital “Subshell”  If n=1, then the only possible number for l is 0 according to equation for l= n-1 so 1-1=0  If n=3, the only possible number for 1 is (n-1  3-1=2) 0,1,2  l= o 0  s orbital o 1  p orbital o 2  d orbital o 3  f orbital  M lM sub l)  Magnetic Quantum Number  Orientation in space “orbital”  Integers of M = -l, . . . -2, -1, 0, 1, 2, . . . +l l  M sM sub s)  Spin Quantum Number (not required by Schrodingers equation)  Electron spin  M s +1/2 , -1/2 o Objective: Know relationships between n, l, M l  Subshells in a principal level (n) is always equal to the n-value  Number of orbitals in a subshell is always equal to 2(l)+1  Number of orbitals in an energy level is always n 2  Energy level in atom and spectra  Bohr’s equation  Hydrogen atom o En=−2.18∙10 −18J/n2 Example: What is the wavelength of light emitted when electron move from n=2 to n=1 in a hydrogen atom −18 1 1 −18 1 1 −18 Δ E=−2.18∙10 J ( )− 2 →=−2.18∙10 J ( )− 2→ΔE=−1.635∙10 J→ nF nI 1 2 −34 8m (6.626∙10 J∙s)(3.0∙10 ) −1.635∙10 −1= s →λ=1.216∙10 meters λ  Section 7.6: Shapes of Atomic Orbitals  Objective: Describe orbital shape given n and l o S orbital (n=1 & l=0) **For s orbitals, l is always 0**  Probability Density: Probability  (Volume @ shell=r) Unit volume  Radical Distribution  Probability o “Node”: Zero probability; at the nucleus, the volume is 0 therefore making the probability 0 of finding an electron in the nucleus  1s orbital  Spherical  2s orbital  Sphere inside a sphere with a node in the middle  3s orbital  Sphere inside a sphere inside a sphere o P Orbital (n=2 & l=1)  2p orbital  Above are all the orientations of a p orbital  3p orbitals are just like 2p but there is a lope inside each lope like such:  And so on with increasing numbers o D orbitals  Same with inside layers increasing with numbers  The shape of d orbitals below: o F orbitals  Same with inside layers increasing with numbers  The shape of F orbitals below:

Step 2 of 3

Chapter 18, Problem 18.68 is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

The full step-by-step solution to problem: 18.68 from chapter: 18 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. Since the solution to 18.68 from 18 chapter was answered, more than 241 students have viewed the full step-by-step answer. This full solution covers the following key subjects: solution, equation, calculate, concentration, containing. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. The answer to “The oxidation of 25.0 mL of a solution containing Fe21 requires 26.0 mL of 0.0250 M K2Cr2O7 in acidic solution. Balance the following equation and calculate the molar concentration of Fe21: Cr2O7 22 1 Fe21 1 H1 Cr31 1 Fe31” is broken down into a number of easy to follow steps, and 40 words. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247.

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

The oxidation of 25.0 mL of a solution containing Fe21