Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after 
is 0.00345 mol/L.
Explanation:
Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t =
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after is 0.00345 mol/L.
The answer for the following questions is explained below.
Explanation:
The two variables that affect kinetic energy are:
- mass and
- velocity
- velocity - The faster an object moves,the more the kinetic energy it has.
- mass - Kinetic energy increases as mass increases
The kinetic energy of an object depends on both its mass and its velocity
Kinetic energy increases as mass increases
For example,think about rolling a bowling ball and a golf ball down a bowling lane at same velocity
Here,the bowling ball has more mass than the golf ball
Therefore you use more energy to roll the bowling ball than to roll the golf ball
The bowling ball is more likely to knock down the pins because it has more kinetic energy than the golf ball
Answer:
We need 3910.5 joules of energy
Explanation:
Step 1: Data given
Mass of aluminium = 110 grams
Initial temperature = 52.0 °C
Final temperature = 91.5 °C
Specific heat of aluminium = 0.900 J/g°C
Step 2: Calculate energy required
Q = m*c*ΔT
⇒with Q = the energy required = TO BE DETERMINED
⇒with m = the mass of aluminium = 110 grams
⇒with c = the specific heat of aluminium = 0.900 J/g°C
⇒with ΔT = the change in temperature = T2 - T1 = 91.5 °C - 52.0 °C = 39.5 °C
Q = 110 grams * 0.900 J/g°C * 39.5
Q = 3910.5 J
We need 3910.5 joules of energy
