I believe that the answer is 1.8^24 of Ni atoms in 3.6 mol of Ni.
Hope this helps. :)
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
Answer:
So, you're dealing with a sample of cobalt-60. You know that cobalt-60 has a nuclear half-life of
5.30
years, and are interested in finding how many grams of the sample would remain after
1.00
year and
10.0
years, respectively.
A radioactive isotope's half-life tells you how much time is needed for an initial sample to be halved.
If you start with an initial sample
A
0
, then you can say that you will be left with
A
0
2
→
after one half-life passes;
A
0
2
⋅
1
2
=
A
0
4
→
after two half-lives pass;
A
0
4
⋅
1
2
=
A
0
8
→
after three half-lives pass;
A
0
8
⋅
1
2
=
A
0
16
→
after four half-lives pass;
⋮
Explanation:
now i know the answer
C. Temperature the average <span> kinetic energy of the particles in an object is directly proportional to its temperature </span>
Carbon discovered in Prehistoric times.. Discoverer will probably never be known
