Answer:
12 átomos de oxígeno hay presentes
Explanation:
Basados en la reacción:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
<em>6 moles de agua producen 1 mol de glucosa</em>
<em />
Si reaccionan 12 moleculas de agua, se producirán:
12 moleculas H₂O * (1 mol C₆H₁₂O₆ / 6 mol H₂O) =
2 moléculas de glucosa se producen.
Como cada molécula de glucosa tiene 6 átomos de oxígeno:
2 moléculas C₆H₁₂O₆ * (6 átomos Oxígeno / 1 molécula C₆H₁₂O₆) =
<h3>12 átomos de oxígeno hay presentes</h3>
<u>Answer:</u> The concentration of radon after the given time is 
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:

We are given:

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 = 3.00 days
= initial amount of the reactant = 
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}](https://tex.z-dn.net/?f=0.181days%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B3.00days%7D%5Clog%5Cfrac%7B1.45%5Ctimes%2010%5E%7B-6%7D%7D%7B%5BA%5D%7D)
![[A]=3.83\times 10^{-30}mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D3.83%5Ctimes%2010%5E%7B-30%7Dmol%2FL)
Hence, the concentration of radon after the given time is 
I think its b or a
i hope i help
Explanation:
The molecules in water is more tightly packed than in ice, so water has greater density than ice.