Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae
------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e![^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}](https://tex.z-dn.net/?f=%5E%7B%5B%28-93.1%2A10%5E3%29%28J%2Fmol%29%5D%2F%5B%288.314%29%28J%2Fmol.K%29%28332K%29%7D)
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
Answer:
i don't know if this help tell me if i am wrong
Explanation:
Gravity is the force that pulls all elements of matter together. Matter refers to things you can physically touch. The more matter there is, the greater the amount of gravity or force. This means that the Earth or other planets have a great deal of pull and that everything on Earth is pulled back to Earth.
Some examples of the force of gravity include:
The force that holds the gases in the sun.
The force that causes a ball you throw in the air to come down again.
The force that causes a car to coast downhill even when you aren't stepping on the gas.
The force that causes a glass you drop to fall to the floor.
Answer:
a)
, b)
,
,
, c)
,
,
, 
Explanation:
a) The total number of users that can be accomodated in the system is:


b) The length of the side of each cell is:


Minimum time for traversing a cell is:



The maximum time for traversing a cell is:


The approximate time is giving by the average of minimum and maximum times:


c) The total number of users that can be accomodated in the system is:


The length of each side of the cell is:


Minimum time for traversing a cell is:



The maximum time for traversing a cell is:


The approximate time is giving by the average of minimum and maximum times:


Answer:
Define Variables and Use List methods to do the following
Explanation:
#<em>Conjoins two lists together</em>
all_names = male_names.union(female_names)
#<em>Finds the names that appear in both lists, just returns those</em>
neutral_names = male_names.intersection(female_names)
#<em>Returns names that are NOT in both lists</em>
specific_names = male_names.symmetric_difference(female_names)