Ni = Nickel
SO4 = Sulphate
2 and 3 in the sub-script are the charges.
So , its name is NICKEL SULPHATE
Answer:

Explanation:
Let A₀ = the original amount of ⁵⁵Co
.
The amount remaining after one half-life is ½A₀.
After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.
After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.
The general formula for the amount remaining is:
A =A₀(½)ⁿ
where n is the number of half-lives
n = t/t_½
Data:
A = 1.90 ng
t = 45 h
t_½ = 18.0 h
Calculation:
(a) Calculate n
n = 45/18.0 = 2.5
(b) Calculate A
1.90 = A₀ × (½)^2.5
1.90 = A₀ × 0.178
A₀ = 1.90/0.178 = 10.7 ng
The original mass of ⁵⁵Co was
.
Dinitrogen tetroxide decomposes to give nitrogen dioxide according to equation below
N2O4--->2NO2
According to avogadros law
1 moles =22.4 l
what about 2.45 l
=2.45l x1 mole/22.4=0.1094 moles
by use of reacting ratio that 1 :2
the moles of No2 = 0.1094 x2=0.2188 moles
1 moles= 22.4 l
what about 0.2188moles
=(0.2188moles x22.4 l) /1moles=4.90 L
Explanation:
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