Answer:
c. −1.22 × 10^4 K
Explanation:
Generally, the slope of an Arrhenius plot can be determined either from the graph of In k against 1/T or the available experimental data of the rate constant (k) and the absolute temperature. Mathematically,
![slope = \frac{ln k_{2} -ln k_{1} }{\frac{1}{T_{2} }-\frac{1}{T_{1} } }](https://tex.z-dn.net/?f=slope%20%3D%20%5Cfrac%7Bln%20k_%7B2%7D%20-ln%20k_%7B1%7D%20%7D%7B%5Cfrac%7B1%7D%7BT_%7B2%7D%20%7D-%5Cfrac%7B1%7D%7BT_%7B1%7D%20%7D%20%20%20%7D)
Where:
L/(mol.s)
L/(mol.s)
K
K
slope = (ln 2.9*10^-1 - ln 6.6*10^-4)/(1/500 - 1/400)
slope = (-1.238+7.323)/(0.002-0.0025)
slope = 6.085/-0.0005 = -1.22*10^4 K
<span>The reaction rate increases.
Why </span><span>Well a catalyst usually lower the activation barrier in an energy diagram. The lower and smaller that gap means the reaction is taking place rapidly compared to when that activation barrier gap is higher. </span>
Answer:
for
system would be -1.53 V
Explanation:
Determination of standard reduction potential of any system is done by placing the system in cathode and a reference half cell in anode and then evaluate the cell potential. The cell potential is the standard reduction potential of the system.
So ![E_{cell}^{0}=E_{system}^{0}-E_{reference}^{0}](https://tex.z-dn.net/?f=E_%7Bcell%7D%5E%7B0%7D%3DE_%7Bsystem%7D%5E%7B0%7D-E_%7Breference%7D%5E%7B0%7D)
As
is equal to 0 therefore cell potential is equal to reduction potential of any system by taking hydrogen electrode as a reference.
equal to -0.76 V with respect to hydrogen
equal to 0.77 V with respect to hydrogen
Therefore standard reduction potential of
system when
system is taken as reference is-
![E_{cell}^{0}=E_{Zn^{2+}\mid Zn}^{0}-E_{Fe^{3+}\mid Fe^{2+}}^{0}](https://tex.z-dn.net/?f=E_%7Bcell%7D%5E%7B0%7D%3DE_%7BZn%5E%7B2%2B%7D%5Cmid%20Zn%7D%5E%7B0%7D-E_%7BFe%5E%7B3%2B%7D%5Cmid%20Fe%5E%7B2%2B%7D%7D%5E%7B0%7D)
= -0.76 V - 0.77 V
= -1.53 V
Nitrogen fixation. most nitrogen fixation is performed by certain kinds of bacteria . Some of this bacteria live in bumps called nadules (NAHJ oolz) on the roots of certain plants.
An element can be represented symbolically in terms of its atomic number and mass number as:
\[_{Z}^{A}\textrm{X}\]
where X = symbol of the element
A = atomic mass
Z = atomic number
In this case we have the representation as:
\[_{0}^{1}\textrm{n}\]
where n = symbol for neutron
mass number of neutron = 1
charge/atomic number of neutron = 0
Ans: D) A neutron of zero charge and one unit mass