Question

need help on number 12,14 16,18,20,22,24 please help me pass chemistry please

C412 Write balanced nuclear equations for the following: (a) Formation ofTi through positron emission (b) Formation of silver-107 through electron capture (c) Formation of polonium-206 through a decay 24.13 Write balanced nuclear equations for the following: (a) Formation of Am through B decay 228 (b) Formation of 9Ac through B decay 203. (c) Formation of Bi through a decay 241 2414 Write balanced nuclear equations for the following: aFormation of 18 Ir through electron capture (b) Formation of francium-221 through a decay (c) Formation of iodine-129 through B decay 24.15 Write balanced nuclear equations for the following: (a) Formation of Mn through positron emission (b) Formation of polonium-215 through a decay (c) Formation of Kr through electron capture C C4.16 Which nuclide (s) would you predict to be stable? Why? (a) 20 24.17 Which nuclide(s) would you predict to be stable? Why? (a) Nd (b)2Co (c) Li 146y 114 Cd (c) Mo (b) 88- 48 2A18 Which nuclide(s) would you predict to be stable? Why? (a) 1271 (b) tin-106 24.19 Which nuclide(s) would you predict to be stable? Why? (b) 7 Br 68 (c) As (a) 48K (c) argon-32 2420 What is the most likely mode of decay for each nuclide? (a) U 24.21 What is the most likely mode of decay for each nuclide? (a) 47Ag 238. 92 (b) Cr (c) Mn 111 (b) CI 110p (c) 44RU (24.22 What is the most likely mode of decay for each nuclide! (a) 1C (b) 120Xe (c) 224TH 24.23 What is the most likely mode of decay for each nuclide? fa) 1051n (h) 141Fu (c) 241 Am

Answer 1

24.12 Balanced nuclear equations for:

a) Formation of through positron emmision

22

In nuclear balance, atomic and masic number must be equal at both sides of equation. Positron is an electron with positive charge, we will refers to it as:
1
. Then, formation of
22
through positron emission must be written as:

22

Identity of X atom will be given by its atomic numer (Z). Because atomic and masic (A) number are balanced, for X, we have:

Z22 123

It is to say: vanadium. The balanced nuclear equation is

48 48 V Ti e 23

b) Formation of silver-107 through electron capture

In this case, the initial atom catches an electron ($\tiny ^{0}_{-1}e$) to form silver-107 (Z= 47):

107 X +e Ag

Balancing atomic and masic number:

TO Ag 47 LOCd e 48

c) Formation of polonium-206 through alpha decay

An alpha particle is an helius nucleus, and we will refers to it as: $^{4}_{2}\alpha$

We have for polonium-206 (Z = 84)

X26Po+a 84

Balancing atomic and mass number:

206 Po a 84 210 Rn 86