<u>Answer:</u> The amount remained after 151 seconds are 0.041 moles
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
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 = 151 sec
= initial amount of the reactant = 0.085 moles
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}](https://tex.z-dn.net/?f=4.82%5Ctimes%2010%5E%7B-3%7D%3D%5Cfrac%7B2.303%7D%7B151%7D%5Clog%5Cfrac%7B0.085%7D%7B%5BA%5D%7D)
![[A]=0.041moles](https://tex.z-dn.net/?f=%5BA%5D%3D0.041moles)
Hence, the amount remained after 151 seconds are 0.041 moles
<span>rutherfordium element # 104</span>
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
The answer is A hope this help