Concept explainers
(a)
Interpretation:
The missing particle in the
Concept Introduction:
Radioactive nuclides undergo disintegration by emission of radiation. This is a natural transmutation reaction where the nuclide of one element is converted into nuclide of another element. Radioactive decay happens naturally. This can also be done artificially in the laboratory by means of bombardment reaction. Bombardment reaction is the one where the target nuclei is hit by a small fast moving high-energy particle to give a daughter nuclide and a small particle such as proton or neutron. This can be represented in form of a nuclear equation. A balanced nuclear equation is the one in which the sum of subscripts on both sides are equal and sum of superscripts on both sides are equal.
(a)
Answer to Problem 15QRT
The missing particle in the given equation is
Explanation of Solution
Given nuclear equation is,
In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal. By looking into the above equation, the sum of superscript in the product side is 22 and the sum of subscript in the product side is 11. This has to be balanced in the reactant side. There is a short of 22 in the superscript and 11 in the subscript when comparing with the reactant side. The particle with mass number 22 and
Missing particle in the given nuclear equation is identified as
(b)
Interpretation:
The missing particle in the nuclear equation has to be filled.
Concept Introduction:
Refer part (a).
(b)
Answer to Problem 15QRT
The missing particle in the given equation is
Explanation of Solution
Given nuclear equation is,
In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal. By looking into the above equation, the sum of superscript in the product side is 122 and the sum of subscript in the product side is 54. This has to be balanced in the reactant side. Sum of superscript in the reactant side is 122. Sum of subscript is 53. There is an excess of 1 in the subscript when comparing with the reactant side. The particle with mass number 0 and atomic number -1 is electron. Therefore, the missing particle is
Missing particle in the given nuclear equation is identified as
(c)
Interpretation:
The missing particle in the nuclear equation has to be filled.
Concept Introduction:
Refer part (a).
(c)
Answer to Problem 15QRT
The missing particle in the given equation is
Explanation of Solution
Given nuclear equation is,
In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal. By looking into the above equation, the sum of superscript in the product side is 4 and the sum of subscript in the product side is 2. This has to be balanced in the reactant side. Sum of superscript in the reactant side is 210. Sum of subscript is 84. There is a short of 206 in the superscript and 82 in the subscript when comparing with the product side. The particle with mass number 206 and atomic number 82 is Lead. Therefore, the missing particle is
Missing particle in the given nuclear equation is identified as
(d)
Interpretation:
The missing particle in the nuclear equation has to be filled.
Concept Introduction:
Refer part (a).
(d)
Answer to Problem 15QRT
The missing particle in the given equation is
Explanation of Solution
Given nuclear equation is,
In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal. By looking into the above equation, the sum of superscript in the product side is 195 and the sum of subscript in the product side is 78. This has to be balanced in the reactant side. Sum of superscript in the reactant side is 195. Sum of subscript is 79. There is an excess of 1 in subscript on the reactant side when comparing with the product side. The particle with mass number 0 and atomic number -1 is an electron. Therefore, the missing particle is
Missing particle in the given nuclear equation is identified as
(e)
Interpretation:
The missing particle in the nuclear equation has to be filled.
Concept Introduction:
Refer part (a).
(e)
Answer to Problem 15QRT
The missing particle in the given equation is
Explanation of Solution
Given nuclear equation is,
In a nuclear equation the sum of subscript on both sides has to be equal and the sum of superscripts on both sides has to be equal. By looking into the above equation, the sum of superscript in the product side is 5 and the sum of subscript in the product side is 0. This has to be balanced in the reactant side. Sum of superscript in the reactant side is 257. Sum of subscript is 102. There is a short of 252 in the superscript and 102 in the subscript when comparing with the reactant side. The particle with mass number 252 and atomic number 102 is Nobelium. Therefore, the missing particle is
Missing particle in the given nuclear equation is identified as
Want to see more full solutions like this?
Chapter 18 Solutions
OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:
- Please correct answer and don't used hand raitingarrow_forwardPlease correct answer and don't used hand raitingarrow_forward(11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B Bond A Bond C a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest Bond Strongest Bond b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)!arrow_forward
- . 3°C with TH 12. (10pts total) Provide the major product for each reaction depicted below. If no reaction occurs write NR. Assume heat dissipation is carefully controlled in the fluorine reaction. 3H 24 total (30) 24 21 2h • 6H total ● 8H total 34 래 Br2 hv major product will be most Substituted 12 hv Br NR I too weak of a participate in P-1 F₂ hv Statistically most favored product will be major = most subst = thermo favored hydrogen atom abstractor to LL Farrow_forwardFive chemistry project topic that does not involve practicalarrow_forwardPlease correct answer and don't used hand raitingarrow_forward
- Q2. Consider the hydrogenation of ethylene C2H4 + H2 = C2H6 The heats of combustion and molar entropies for the three gases at 298 K are given by: C2H4 C2H6 H2 AH comb/kJ mol¹ -1395 -1550 -243 Sº / J K¹ mol-1 220.7 230.4 131.1 The average heat capacity change, ACP, for the reaction over the temperature range 298-1000 K is 10.9 J K¹ mol¹. Using these data, determine: (a) the standard enthalpy change at 800 K (b) the standard entropy change at 800 K (c) the equilibrium constant at 800 K.arrow_forward13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B Bond A Bond C a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest Bond Strongest Bond b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)! Googlearrow_forwardPrint Last Name, First Name Initial Statifically more chances to abstract one of these 6H 11. (10pts total) Consider the radical chlorination of 1,3-diethylcyclohexane depicted below. 4 4th total • 6H total 래 • 4H total 21 total ZH 2H Statistical H < 3° C-H weakest - product abstraction here bund leads to thermo favored a) (6pts) How many unique mono-chlorinated products can be formed and what are the structures for the thermodynamically and statistically favored products? Product 6 Number of Unique Mono-Chlorinated Products Thermodynamically Favored Product Statistically Favored Product b) (4pts) Draw the arrow pushing mechanism for the FIRST propagation step (p-1) for the formation of the thermodynamically favored product. Only draw the p-1 step. You do not need to include lone pairs of electrons. No enthalpy calculation necessary H H-Cl Waterfoxarrow_forward
- 10. (5pts) Provide the complete arrow pushing mechanism for the chemical transformation → depicted below Use proper curved arrow notation that explicitly illustrates all bonds being broken, and all bonds formed in the transformation. Also, be sure to include all lone pairs and formal charges on all atoms involved in the flow of electrons. CH3O II HA H CH3O-H H ①arrow_forwardDo the Lone Pairs get added bc its valence e's are a total of 6 for oxygen and that completes it or due to other reasons. How do we know the particular indication of such.arrow_forwardNGLISH b) Identify the bonds present in the molecule drawn (s) above. (break) State the function of the following equipments found in laboratory. Omka) a) Gas mask b) Fire extinguisher c) Safety glasses 4. 60cm³ of oxygen gas diffused through a porous hole in 50 seconds. How long w 80cm³ of sulphur(IV) oxide to diffuse through the same hole under the same conditions (S-32.0.0-16.0) (3 m 5. In an experiment, a piece of magnesium ribbon was cleaned with steel w clean magnesium ribbon was placed in a crucible and completely burnt in oxy cooling the product weighed 4.0g a) Explain why it is necessary to clean magnesium ribbon. Masterclass Holiday assignmen PB 2arrow_forward
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub CoIntroduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning