Answer:
n=N/NA
n= 3.754×10²³/6.02×10²³
n= 6.24 s
Explanation
since there is number of molecules, make use of Avogadro's constant to get number of moles.
You can cross multiply and then get k alone so it would look like this:
76 81
----- = ------ = 76k= 24624
304 k ----- ----------- = k = 324
76 76
so k is equal to 324
hope this helps have a nice day
Answer:
50.96g
Explanation:
Given parameters:
Number of moles of H₃PO₄ = 0.52moles
Unknown:
Mass of the compound = ?
Solution:
To find the mass of the compound:
Mass = number of moles x molar mass of H₃PO₄
Molar mass of H₃PO₄ = 3(1) + 31 + 4(16) = 98g/mol
Mass = 0.52 x 98 = 50.96g
Answer:
The concentration of cyclobutane after 875 seconds is approximately 0.000961 M
Explanation:
The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M
The final concentration of cyclobutane, [
] = 0.00119 M
The time for the reaction, t = 455 seconds
Therefore, the Rate Law for the first order reaction is presented as follows;
![\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }](https://tex.z-dn.net/?f=%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%20%3D%20%5Ctext%20%7B-k%7D%20%5Ccdot%20t%20%7D)
Therefore, we get;
![k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}} {-t }](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7B%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%7D%20%20%7B-t%20%7D)
Which gives;
k ≈ 5.088 × 10⁻⁴ s⁻¹
The concentration after 875 seconds is given as follows;
[] = [A₀]·
Therefore;
[] = 0.00150 × 
= 0.000961
The concentration of cyclobutane after 875 seconds, [] ≈ 0.000961 M
Atomic number of F = 9
Electron configuration of F:-
⁹F = 1s² 2s² 2p⁵
Incase of F2: Total electrons: 9 + 9 = 18
For F2 the ground state electron configuration would be:
F2 = (1σ)²(1σ*)²(2σ)²(2σ*)²(2pσ)²(2pπ)⁴(2pπ*)⁴
Incase of F2+: There is loss of one electron from neutral F2. Total electrons: 18-1 = 17
For F2+ the ground state electron configuration would be:
F2 = (1σ)²(1σ*)²(2σ)²(2σ*)²(2pσ)²(2pπ)⁴(2pπ*)³
