15. Choose correct IUPAC name for KIV(OH2)2Brs A. potassium dihydroxytetrabromov
ID: 698764 • Letter: 1
Question
15. Choose correct IUPAC name for KIV(OH2)2Brs A. potassium dihydroxytetrabromovanadate (III) B. potassium d D. vanadium d e(IV) 16. Which particle has the highest penetrating power? A) alpha particle D) positron emission B) beta particle Celectron c E) gamma particle 17. Describe what changes occur during electron capture. A) The mass number and atomic number decrease. B) The mass number and atomic number increase C) The mass number is unchanged and the atomic number decreases D) The mass number is unchanged and the atomic number increases. E) The mass number and atomic number do not change. 18. Determine the identity of the daughter nuclide from the alpha decay of s, Rn 218 A) Po 226 224 223 230 19. Determine the identity of the daughter nuclide from the electron capture by O'Pa. 226 230 230 20. The following reaction represents what nuclear process? 137 55 Cs +-1 e 54 Xe A) beta emission D) electron capture E alpha capture B) positron emissionC) gamma emission 21. Identify the missing particle in the following nuclear equation: 141 144 143 143Explanation / Answer
15)
IUPAC name of K[V(OH)22Br4] is PotassiumdiaquatetrabromoVandate(III)
Answer : C
16)
Gamma particle has highest penetration power due to chargeless and very less mass.
Answer : E
17)
During electron capture there is decrease in atomic number by one unit and no change in the mass number.
Answer : C
18)
222 Rn 86 -------> 218 pO 84 + 4 He 2 (alpha particle)
Answer : A
19)
230 Pa 91 + 0 e -1 -------> 230 Th 90
Answer : E
20)
The given reaction is electron capture process since in the reactant side there is absorption electron takes place
Answer : D
21)
235 U 92 -----> 90 Sr 38 + 2 1 n 0 + 4 0 gamma 0 + 143 Xe 54
Answer : C
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