Explanation:
its answer is displacement reaction
The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
![rate=k[H_2]^2[I_2]^2](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BI_2%5D%5E2)
At a certain concentration of
and
, the initial rate of reaction is 4.0 × 10⁴ M/s. What would the initial rate of the reaction be if the concentration of
Answer : The initial rate of the reaction will be,
Explanation :
Rate law expression for the reaction:
![rate=k[H_2]^2[I_2]^2](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BI_2%5D%5E2)
As we are given that:
Initial rate = 4.0 × 10⁴ M/s
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Dividing 2 by 1, we get:
![\frac{R}{4.0\times 10^4}=\frac{k(\frac{[H_2]}{2})^2[I_2]^2}{k[H_2]^2[I_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B4.0%5Ctimes%2010%5E4%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BI_2%5D%5E2%7D%7Bk%5BH_2%5D%5E2%5BI_2%5D%5E2%7D)


Therefore, the initial rate of the reaction will be, 
That you have to do better
Answer:
4.D. 2KJ
5.B. close to 0 degree celsius
Explanation:
heat absorbed= heat given out
100g * 4.2 * (x-0)= 400g * 4.2* ( 20-x) where x is the new temp reached
then x= 20
i.e. the temperature of ice increases to 5 degree celsius and the temperature of water decreases to 5 degree celsius .
hope it helps
The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)