Answer: 50 to 80mmHg
Explanation:
Retinopathy of prematurity (ROP) is described as an ocular condition associated with O2 administration to premature infants. Excessive blood O2 levels produce retinal vasoconstriction, which (if prolonged) can result in permanent blindness. This risk poses a serious management problem, for the premature infant is often in need of supplemental O2. The American Academy of Pediatrics has recommended that keeping PaO2s between 50 to 80mmHg is the best way to minimize the risk of Retinopathy of Prematurity.
Answer:
C
Explanation:
Well, we can see from the equation that hydrogen gas and ethane are in a 2:1 ratio based on their mole coefficients.
You need 2 mol H2 to make 1 mol C2H6. So that means you need 2*13.78 mol H2 to make 13.78 mol C2H6!
2*13.78 = 27.56 mol
C is correct.
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:
drops a glass cup of water outcome mad parents
Explanation:
Mad pagents mean death no surviva.