<span>The answer is hypertonic. In osmosis, water
molecules move from a hypotonic solution to the hypertonic solution, through a
semipermeable membrane. This occurs until
both solutions become isotonic relative to each other. In osmosis, only
the movement of water molecules occurs since the ions are large enough to pass
through the pores of the semipermeable membrane,
in this case, the cell membrane. Due to
loss of water in the process of osmosis, the cells in the fingers of the swimmers
shrunk hence looked shriveled.</span>
N(Ca)/2 = n(O)/1 = n(CaO)/2
The calcium and the Calcium Oxide are divided by 2 because of their coefficients
there is no number in front of the oxygen so it is over one.
Hope this helped!!
The average atomic weight is calculated by adding up the products of the percentage abundance and atomic weight. In this item, we have the equation,
A = (0.412)(21.016 amu) + (0.5012)(21.942 amu) + (0.0868)(23.974 amu)
Simplifying the operation will give us the answer of 21.74 amu.
<em>Answer: 21.74 amu</em>
<u>Answer:</u> The amount of Iodine-131 remain after 39 days is 0.278 grams
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:

where,
= half life of the reaction = 8.04 days
Putting values in above equation, we get:

Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 39 days
= initial amount of the sample = 8.0 grams
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![0.0862=\frac{2.303}{39}\log\frac{8.0}{[A]}](https://tex.z-dn.net/?f=0.0862%3D%5Cfrac%7B2.303%7D%7B39%7D%5Clog%5Cfrac%7B8.0%7D%7B%5BA%5D%7D)
![[A]=0.278g](https://tex.z-dn.net/?f=%5BA%5D%3D0.278g)
Hence, the amount of Iodine-131 remain after 39 days is 0.278 grams