Explanation:
In a magnetic field, the radius of the charged particle is as follows.
r = 
where, m = mass, v = velocity
q = charge, B = magnetic field
Therefore, q will be calculated as follows.
q = 
= 
= 
= 
= +2e
Thus, we can conclude that the charge of the ionized atom is +2e.
Answer:
All are having different valent cation and anion like mono,di and trivalent polyatomic ions .
A. RbNO3
B. K2S
C. NaHS
D. Mg3(PO4)2 formed by divalent Mg+2 and trivalent PO43-
E. CaHPO4
F. PbCO3 , lead is in Pb+2 form
G. SnF2
H. (NH4)2SO4
I. AgClO4
J. BCl3
Answer:
7. 3–ethyl–6 –methyldecane
8. 5–ethyl–2,2–dimethyl–4–propyl–4 –heptene
Explanation:
It is important to note that when naming organic compounds having two or more different substituent groups, we simply name them alphabetically.
The name of the compound given in the question above can be written as follow:
7. Obtaining the name of the compound.
Compound contains:
I. Decane.
II. 3–ethyl.
III. 6 –methyl.
Naming alphabetically, we have
3–ethyl–6 –methyldecane
8. Obtaining the name of the compound.
Compound contains:
I. 2,2–dimethyl.
II. 4–propyl.
III. 4 –heptene.
IV. 5–ethyl.
Naming alphabetically, we have
5–ethyl–2,2–dimethyl–4–propyl–4 –heptene
