Answer:
The enthalpy of the solution is -35.9 kJ/mol
Explanation:
<u>Step 1:</u> Data given
Mass of lithiumchloride = 3.00 grams
Volume of water = 100 mL
Change in temperature = 6.09 °C
<u>Step 2:</u> Calculate mass of water
Mass of water = 1g/mL * 100 mL = 100 grams
<u>Step 3:</u> Calculate heat
q = m*c*ΔT
with m = the mass of water = 100 grams
with c = the heat capacity = 4.184 J/g°C
with ΔT = the chgange in temperature = 6.09 °C
q = 100 grams * 4.184 J/g°C * 6.09 °C
q =2548.1 J
<u>Step 4:</u> Calculate moles lithiumchloride
Moles LiCl = mass LiCl / Molar mass LiCl
Moles LiCl = 3 grams / 42.394 g/mol
Moles LiCl = 0.071 moles
<u>Step 5:</u> Calculate enthalpy of solution
ΔH = 2548.1 J /0.071 moles
ΔH = 35888.7 J/mol = 35.9 kJ/mol (negative because it's exothermic)
The enthalpy of the solution is -35.9 kJ/mol
4 carbon electrons. Hope this helps have a nice day
I think a shovel is a wedge simple machine.
Answer:
A ballon
Explanation:
Sorry it took me so long i just seen this , but gases are most compressible .
Answer:
a) 79.66 seconds is the half-life of the reaction.
b) It will take 
seconds for 34% of a sample of an acetone to decompose.
c) It will take 
seconds for 89% of a sample of an acetone to decompose.
Explanation:
The decomposition of acetone follows first order kinetics
The rate constant of the reaction = k = 
a)
Half life of the reaction = 
For the first order kinetic half life is related to k by :
79.66 seconds is the half-life of the reaction.
b)
Let the initial concentration of acetone be = ![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
Final concentration of acetone left after t time = [A]
![A=(100\%-34\%)[A_o]=66\%[A_o]=0.66[A_o]](https://tex.z-dn.net/?f=A%3D%28100%5C%25-34%5C%25%29%5BA_o%5D%3D66%5C%25%5BA_o%5D%3D0.66%5BA_o%5D)
For the first order kinetic :
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![0.66[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}](https://tex.z-dn.net/?f=0.66%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-8.7%5Ctimes%2010%5E%7B-3%7D%20s%5E%7B-1%7D%5Ctimes%20t%7D)
Solving for t;
It will take 47.76 seconds for 34% of a sample of an acetone to decompose.
c)
Let the initial concentration of acetone be =
Final concentration of acetone left after t time = [A]
![A=(100\%-89\%)[A-o]=11\%[A_o]=0.11[A_o]](https://tex.z-dn.net/?f=A%3D%28100%5C%25-89%5C%25%29%5BA-o%5D%3D11%5C%25%5BA_o%5D%3D0.11%5BA_o%5D)
For the first order kinetic :
![0.11[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}](https://tex.z-dn.net/?f=0.11%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-8.7%5Ctimes%2010%5E%7B-3%7D%20s%5E%7B-1%7D%5Ctimes%20t%7D)
Solving for t;
It will take seconds for 89% of a sample of an acetone to decompose.
