Answer:the answer is 18.01528
Explanation:
Answer:
A) 0.065 M is its molarity after a reaction time of 19.0 hour.
B) In 52 hours ![[Co(NH_3)5Br]^{2+}](https://tex.z-dn.net/?f=%5BCo%28NH_3%295Br%5D%5E%7B2%2B%7D)
will react 69% of its initial concentration.
Explanation:
![Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)](https://tex.z-dn.net/?f=Co%28NH_3%29_5%28H_2O%29_3%2B%5BCo%28NH_3%295Br%5D%5E%7B2%2B%7D%28Purple%29%28aq%29%2BH_2O%28l%29%5Crightarrow%20%5BCo%28NH_3%29_5%28H_2O%29%5D%5E%7B3%2B%7D%28Pinkish-orange%29%28aq%29%2BBr%5E-%28aq%29)
The reaction is first order in :
Initial concentration of = ![[A_o]=0.100 M](https://tex.z-dn.net/?f=%5BA_o%5D%3D0.100%20M)
a) Final concentration of after 19.0 hours= ![[A]](https://tex.z-dn.net/?f=%5BA%5D)
t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)
Rate constant of the reaction = k = 
The integrated law of first order kinetic is given as:
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}](https://tex.z-dn.net/?f=%5BA%5D%3D0.100%20M%5Ctimes%20e%5E%7B-6.3%5Ctimes%2010%5E%7B-6%7D%20s%5E%7B-1%7D%5Ctimes%2019.0%5Ctimes%203600%20s%7D)
![[A]=0.065 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.065%20M)
0.065 M is its molarity after a reaction time of 19.0 h.
b)
Initial concentration of = ![[A_o]=x](https://tex.z-dn.net/?f=%5BA_o%5D%3Dx)
Final concentration of after t = ![[A]=(100\%-69\%) x=31\%x=0.31x](https://tex.z-dn.net/?f=%5BA%5D%3D%28100%5C%25-69%5C%25%29%20x%3D31%5C%25x%3D0.31x)
Rate constant of the reaction = k =
The integrated law of first order kinetic is given as:
t = 185,902.06 s = 
≈ 52 hours
In 52 hours will react 69% of its initial concentration.
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
