<h3>Chemical reaction formula</h3>
Be2+ + S2- -> BeS
<h3>Product formula : BeS</h3>
Answer:
Heat is absorbed.
Explanation:
Endothermic reactions: Heat is absorbed.
It is true. Since it is very reactive with oxygen and water, it is kept in kerosene or some other some other oil
Answer:
a) The half life of the radioactive sample is 0.22 years.
b) It will take 0.16 years to sample to decay to 40% of its original amount.
Explanation:
a) Initial amount of radioactive substance = ![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
Final amount of radioactive substance after 1 year= ![[A]=(100\%-96\%)[A_o]=4\%[A_o]=0.04[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D%28100%5C%25-96%5C%25%29%5BA_o%5D%3D4%5C%25%5BA_o%5D%3D0.04%5BA_o%5D)
Decay constant = k
Decaying of radio active sample follows first order kinetics:
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![0.04[A_o]=[A_o]\times e^{-k\times 1 year}](https://tex.z-dn.net/?f=0.04%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-k%5Ctimes%201%20year%7D)
![k=3.2189 year^{-1}](https://tex.z-dn.net/?f=k%3D3.2189%20year%5E%7B-1%7D)
Half life of the sample = ![t_{1/2}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D)
![t_{1/2}=\frac{0.693}{k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B0.693%7D%7Bk%7D)
![=\frac{0.693}{3.2189 year^{-1}}=0.2152 year\approx 0.22 year](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.693%7D%7B3.2189%20year%5E%7B-1%7D%7D%3D0.2152%20year%5Capprox%200.22%20year)
The half life of the radioactive sample is 0.22 years.
b)
Initial amount of radioactive substance = ![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
Final amount of radioactive substance after t years= ![[A]=(100\%-40\%)[A_o]=60\%[A_o]=0.6[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D%28100%5C%25-40%5C%25%29%5BA_o%5D%3D60%5C%25%5BA_o%5D%3D0.6%5BA_o%5D)
Decay constant = k = ![3.2189 year^{-1}](https://tex.z-dn.net/?f=3.2189%20year%5E%7B-1%7D)
Decaying of radio active sample follows first order kinetics:
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![0.6[A_o]=[A_o]\times e^{-3.2189 year^{-1}\times t}](https://tex.z-dn.net/?f=0.6%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-3.2189%20year%5E%7B-1%7D%5Ctimes%20t%7D)
Solving for t:
t = 0.1587 years ≈ 0.16 years
It will take 0.16 years to sample to decay to 40% of its original amount.
Answer:
86.9 g
Explanation:
Use the conversion factor: 4.82 mol H2O x (18.02g H2O/1 mol H2O)
The moles cancel out so we have to multiply 4.82 x 18.02 = 86.9
The answer is 3 sig. figs.