4 In the open chain, 5 in the cyclic. Just like glucose.
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:
0.00000363618
could be wrong.
double check me someone or just trust me
(don't blame me if you get it wrong)
Answer:
I think it's the second answer --If you increase the acidity..
I hope answer I can answer your question!
Answer:
2.42L
Explanation:
Given parameters:
V₁ = 1.8L
T₁ = 293K
P₁ = 101.3kPa
P₂ = 67.6kPa
T₂ = 263K
Unknown:
V₂ = ?
Solution:
To solve this problem, we are going to use the combined gas law to find the final volume of the gas. The combined gas law expression combines the equation of Boyle's law, Charles's law and Avogadro's law;
All the units are in the appropriate form. We just substitute and solve for the unknown;
101.3 x 1.8 / 293 = 67.6 x V₂ / 263
V₂ = 2.42L
