For plato users
the answer is a. O2(l) O2(g)
hope this helps!
Answer:
![\boxed{\text{8 da}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7B8%20da%7D%7D)
Explanation:
The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"
The half-life of the element is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
![\begin{array}{cccl}\textbf{No. of} & & \textbf{Fraction} & \\\textbf{half-lives} & \textbf{t/da} & \textbf{remaining} & \\1 & 4 & \dfrac{1}{2} & \\\\2 & 8 & \dfrac{1}{4}& \\\\3 & 12 & \dfrac{1}{8}& \\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bcccl%7D%5Ctextbf%7BNo.%20of%7D%20%26%20%26%20%5Ctextbf%7BFraction%7D%20%26%20%5C%5C%5Ctextbf%7Bhalf-lives%7D%20%26%20%5Ctextbf%7Bt%2Fda%7D%20%26%20%5Ctextbf%7Bremaining%7D%20%26%20%5C%5C1%20%26%204%20%26%20%5Cdfrac%7B1%7D%7B2%7D%20%26%20%5C%5C%5C%5C2%20%26%208%20%26%20%5Cdfrac%7B1%7D%7B4%7D%26%20%5C%5C%5C%5C3%20%26%2012%20%26%20%5Cdfrac%7B1%7D%7B8%7D%26%20%5C%5C%5Cend%7Barray%7D)
![\text{We see that 8 da is two half-lives, and the fraction of the element remaining is $\frac{1}{4}$.}\\\text{It takes $\boxed{\textbf{8 da}}$ for three-fourths of the element to decay}](https://tex.z-dn.net/?f=%5Ctext%7BWe%20see%20that%208%20da%20is%20two%20half-lives%2C%20and%20the%20fraction%20of%20the%20element%20remaining%20is%20%24%5Cfrac%7B1%7D%7B4%7D%24.%7D%5C%5C%5Ctext%7BIt%20takes%20%24%5Cboxed%7B%5Ctextbf%7B8%20da%7D%7D%24%20for%20three-fourths%20of%20the%20element%20to%20decay%7D)
Activity series of metals: K,Na,Mg,Al,Zn,Fe,Cu,Ag. Metals on the left are more reactive than metals on the right. For example Zn is more reactive than Fe and can displace him.
Reaction than can occur is: <span>CuSO4(aq) + Fe(s) → FeSO4(aq) + Cu(s).</span>
Primary producers i believe