Answer:
The rate of reaction changes by a factor 6.
Explanation:
Given the reaction is first order.
Let us assume the initial concentration of 
is 
and the initial concentration of 
is 
.
So, we can write the rate of the reaction as
![R=k[A][B]](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5BB%5D)
Where,
is the rate of reaction. And 
is the rate constant.
Also, the initial concentration of has been increased by 
, and the initial concentration of was increased by factor 
.
So, we have
![R'=k[1.5A][4B]\\R'=6k[A][B]\\R'=6R](https://tex.z-dn.net/?f=R%27%3Dk%5B1.5A%5D%5B4B%5D%5C%5CR%27%3D6k%5BA%5D%5BB%5D%5C%5CR%27%3D6R)
We can see the rate of reaction changes by a factor 6.
Answer: E=∆H*n= -40.6kj
Explanation:
V(CO) =15L=0.015M³
P=11200Pa
T=85C=358.15K
PV=nRT
n=(112000×0.015)/(8.314×358.15)
n(Co)= 0.564mol
V(Co)= 18.5L = 0.0185m³
P=744torr=98191.84Pa
T= 75C = 388.15k
PV=nRT
n= (99191.84×0.0185)/(8.314×348.15)
n(H2) = 0.634mol
n(CH30H) =1/2n(H2)=1/2×0.634mol
=0.317mol
∆H =∆Hf{CH3OH}-∆Hf(Co)
∆H=-238.6-(-110.5)
∆H = 128.1kj
E=∆H×n=-40.6kj.
#2
They are ordered by increasing number of protons (aka atomic number)
I think it is the first choice
Hope that helped!
The elements whose electron configurations end with f electrons are in the Lanthanide and Actinide series: those long series at the bottom of the Periodic Table (see image).
The atoms are adding electrons into f orbitals, but the f electrons are not always the last electrons in the electron configuration.
For example, the electron configuration of Eu is [Xe]6s²4f⁷, but that of Gd
is [Xe] 6s²4f⁷5d.
I have blocked off in <em>red</em> all the <em>exceptions</em> like Gd.
