0.048 percent is the answer
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
Answer:
152 pm
Explanation:
According to the question, we can estimate the bond length from the given values of the atomic radii. This now is the upper limit of the bond length for the molecule.
Since we have that;
Atomic radius of H= 37.0 pm
Atomic radius of Br = 115.0 pm
Bond length = Atomic radius of H + Atomic radius of Br
Bond length = 37.0 pm + 115.0 pm
Bond length = 152 pm
<u>Answer:</u> The molarity of Iron (III) chloride is 0.622 M.
<u>Explanation:</u>
Molarity is defined as the number of moles present in one liter of solution. The equation used to calculate molarity of the solution is:
Or,
We are given:
Mass of iron (III) chloride = 1.01 g
Molar mass of iron (III) chloride = 162.2 g/mol
Volume of the solution = 10 mL
Putting values in above equation, we get:
Hence, the molarity of Iron (III) chloride is 0.622 M.
Answer:
c i guess, decrease temperature
