Answer: -
When bismuth-212 undergoes alpha decay, it becomes ²⁰⁸Tl
Explanation: -
Mass number of ²¹²Bi = 212
Atomic number of ²¹²Bi = 83
When alpha decay occurs the mass number decreases by 4 and the atomic number decreases by 2.
Mass number of daughter = 212 - 4 = 208
Atomic number of daughter = 83 - 2 = 81
It is the atomic number of Thallium Tl.
Thus the daughter nucleide is ²⁰⁸Tl.
KE=.5mv^2
M=mass
v=velocity
.5(4)(100)=200
That should be the answer.
Answer:
972.3 Torr
Explanation:
P2=P1V1/V2
You can check this by knowing that P and V at constant T have an an inverse relationship. Hence, this is correct.
Answer:
A = Metallic Bond
B = Strong bonding, strong conductor, high melting and boiling points
Explanation:
Since the bond is between two metals (located in groups 11 and 12), they would experience metallic bonding. Metallically bonded molecules have high melting and boiling points due to the strength of the metallic bond. They also experience strong electrical current due to the there delocalized electrons.
To get the value of ΔG we need to get first the value of ΔG°:
when ΔG° = - R*T*㏑K
when R is constant in KJ = 0.00831 KJ
T is the temperature in Kelvin = 25+273 = 298 K
and K is the equilibrium constant = 4.5 x 10^-4
so by substitution:
∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4
= -19 KJ
then, we can now get the value of ΔG when:
ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]
when ΔG° = -19 KJ
and R is constant in KJ = 0.00831
and T is the temperature in Kelvin = 298 K
and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m
so, by substitution:
ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )
= -40
