Answer:A chlorine atom in its ground state has a total of seven electrons in orbitals related to the atoms third energy level.
Explanation:
Answer:
1.64g
Explanation:
The reaction scheme is given as;
2-bromocyclohexanol --> 1,2-epoxycyclohexane + HBr
From the reaction above,
1 mol of 2-bromocyclohexanol produces 1 mol of 1,2-epoxycyclohexane
3.0 grams of trans-2-bromocyclohexanol.
Molar mass = 179.05 g/mol
Number of moles = mass / molar mass = 3 / 179.05 = 0.016755 mol
This means 0.016755 mol of 1,2-epoxycyclohexane would be produced.
Molar mass = 98.143 g/mol
Theoretical yield = Number of moles * Molar mass
Theoretical yield = 0.016755 * 98.143 ≈ 1.64g
<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
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= 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
Answer:
We have to take 37.5 mL of a 0.400 M solution
Explanation:
Step 1: Data given
Stock volume = 100 mL = 0.100L
Stock concentration 0.400 M
Volume of solution he wants to make = 100 mL = 0.100L
Concentration of solution he wants to make = 0.150 M
Step 2: Calculate the volume of 0.400 M CuSO4 needed
C1*V1 = C2*V2
⇒with C1 = the stock concentration = 0.400M
⇒with V1 = the volume of the stock = TO BE DETERMINED
⇒with C2 = the concentration of the solution he wants to make = 0.150 M
⇒with V2 = the volume of the solution made = 0.100 L
0.400 M * V1 = 0.150M * 0.100L
V1 = (0.150M*0.100L) / 0.400 M
V1 = 0.0375 L = 37.5 mL
We have to take 37.5 mL of a 0.400 M solution
