(a)
Interpretation:
The balanced nuclear reaction of the alpha decay of
Concept introduction:
The nuclear reactions are a type of chemical process which leads to the formation of some new nuclei with the emission of certain particles. Usually, alpha or beta particles or gamma rays are emitted as a side product with some new daughter nuclei. The nuclear reactions follow the conservation of
(b)
Interpretation:
Themass change in g/atom and energy change in kJ/mol for the given nuclear reaction of the alpha decay of
Concept introduction:
The nuclear reactions are a type of chemical process which lead to the formation of some new nuclei with the emission of certain particles. Usually, alpha or beta particles or gamma rays are emitted as a side product with some new daughter nuclei.
The force between protons and neutrons holds these particles together. It can be measured by measuring the amount of energy necessary to break these forces.
It can be calculated with the help of Einstein’s mass − energy equation:
The mass defect can be calculated as the difference in the
(c)
Interpretation:
The energy released or absorbed during the radioactive decay of
Concept introduction:
The nuclear reactions are a type of chemical process which leads to the formation of some new nuclei with the emission of certain particles. Usually, alpha or beta particles or gamma rays are emitted as a side product with some new daughter nuclei.
The force between protons and neutrons holds these particles together. It can be measured by measuring the amount of energy necessary to break these forces.
It can be calculated with the help of Einstein’s mass-energy equation:
The mass defect can be calculated as the difference in the atomic mass and the sum of masses of all atomic particles.
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Chapter 20 Solutions
LCPO CHEMISTRY W/MODIFIED MASTERING
- Which of the following would be the expected major product and why? HCI CI II CI Product I because of resonance stabilization of the carbocation intermediate Product II because of resonance stabilization of the carbocation intermediate Product I because of inductive stabilization of the carbocation intermediate Product II because of inductive stabilization of the carbocation intermediatearrow_forwardNonearrow_forward9.20 How many ¹H NMR signals (not peaks) would you predict for each of the following compounds? (Consider all protons that would be chemical shift nonequivalent.) OH Br OHarrow_forward
- 9.22 Propose a structure for an alcohol with molecular formula C5H12O that has the 1H NMR spectrum given in Figurearrow_forwardFor each set of carbonyl additions, circle the carbonyl addition that occurs at the faster rate (assuming everything is the same except that the reagent/substrate differs - i.e., same temperature, and ratios/concentrations of reagent and substrate). Electrostatic attraction has a greater impact on the relative rates than steric hindrance. (a) CH3OH HO OCH3 H H CH3 i CH₂OH HO OCH 3 H F3C CH3 (b) F3C NaOCH3 HO OCH3 H3C CH3 H3C CH3 CH3OH HO OCH3 H3C CH3 H3C CH3 (c) NaSCH3 OSCH 3 H3C CH3 H3C CH3 NaOCH3 O OCH 3 H3C CH3 H3C CH3arrow_forward9.34. Assign the chemical shifts and splitting patterns to specific aspects of the structure you propose. C5H12O 1H 2H 2 6H ille H(ppm) 1 3H и 0arrow_forward
- HO (c) (1 pt) Both of the following are hydride donors. Circle the harder nucleophile of -P-Cu-H Н H-AI-H HINIH Н (d) (4 pts) The following reaction involves two steps. Draw the anionic intermediate that forms after sodium hydride reacts and the final organic product. Hints: what type of nucleophile is NaH and where does that mean it will react? Also, the second step is not a proton transfer. What's the most likely reaction for that intermediate to undergo? NaH anionic intermediate final productarrow_forwardPredict the product(s) for the reaction shown. O excess HBr heatarrow_forwardPlease help graph these plotts belowarrow_forward
- Please graph the image below:arrow_forward7. Our textbook says that the fragmentation that occurs in the mass spectrometry of alkanes can be understood by realizing that "the differences in energy among ... tertiary, secondary, primary and methyl carbocations in the gas phase are much greater than the differences among comparable radicals. Therefore, where alternative modes of fragmentation are possible, the more stable carbocation tends to form in preference to the more stable radical." Given this information, which one of the following hexane isomers (all C6H14) is most likely to have a strong M-15 peak (that is, a peak at m/z 71)? HINT: You're looking for a compound that forms a 3° carbocation after loss of an electron and a CH³· radical. A) n-hexane D) 2-methylpentane B) 2,2-dimethylbutane E) 3-methylpentane C) 2,3-dimethylbutanearrow_forwardPlease help graph these plots below:arrow_forward
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