For the 1st order reactions,rate constant (k) is mathematically expressed as
k =

where, t = time
Co = initial conc. of reactant
Ct = conc. of reactant after time 't'
Given: k = <span>2.20 × 10^-5 s-1, t = 2 hours = 7200 s
Therefore, we have
</span>2.20 × 10^-5 =

∴

= 0.06877
∴,

= 1.1716
∴, Ct = 85.35%
Thus, <span>
85.35 % of the initial amount of SO2Cl2 will remain after 2.00 hours.</span>
Answer:
96g
Explanation:
Given parameters:
Number of moles of NaOH = 2.4moles
Unknown:
Mass of NaOH = ?
Solution:
The mass of a substance given the number of moles can be found using the expression below;
Mass of NaOH = Number of moles x molar mass;
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Now input the parameters and solve;
Mass of NaOH = 2.4 x 40 = 96g
Answer:
0.6749 M is the concentration of B after 50 minutes.
Explanation:
A → B
Half life of the reaction = 
Rate constant of the reaction = k
For first order reaction, half life and half life are related by:


Initial concentration of A = ![[A]_o=0.900 M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.900%20M)
Final concentration of A after 50 minutes = ![[A]=?](https://tex.z-dn.net/?f=%5BA%5D%3D%3F)
t = 50 minute
![[A]=[A]_o\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_o%5Ctimes%20e%5E%7B-kt%7D)
![[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}](https://tex.z-dn.net/?f=%5BA%5D%3D0.900%20M%5Ctimes%20e%5E%7B-0.02772%20min%5E%7B-1%7D%5Ctimes%2050%20minutes%7D)
[A] = 0.2251 M
The concentration of A after 50 minutes = 0.2251 M
The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M
0.6749 M is the concentration of B after 50 minutes.
Hydroxide ion is a strong and would react with H+ to form water
OH-+H+---->H2O