Answer:
When a reversible reaction happens in a closed container, it reaches a dynamic equilibrium . At equilibrium: the forward and backward reactions are still happening. the forward and backward reactions have the same rate of reaction
Explanation:
Explanation:
The given data is:
The half-life of gentamicin is 1.5 hrs.
The reaction follows first-order kinetics.
The initial concentration of the reactants is 8.4 x 10-5 M.
The concentration of reactant after 8 hrs can be calculated as shown below:
The formula of the half-life of the first-order reaction is:

Where k = rate constant
t1/2=half-life
So, the rate constant k value is:

The expression for the rate constant is :

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.
3! You have to ensure balance of all the different elements.
Answer:
you can solve the rest of the equation. I only reduced it to that much to show you how to derive it
Answer:
Density of the He atom = 12.69 g/cm³
Explanation:
From the information given:
Since 1 mole of an atom = 6.022x 10²³ atoms)
1 atom of He = 

The volume can be determined as folows:
since the diameter of the He atom is approximately 0.10 nm
the radius of the He =
= 0.05 nm
Converting it into cm, we have:


Assuming that it is a sphere, the volume of a sphere is
= 
= 
= 
Finally, the density can be calcuated by using the formula :


D = 12.69 g/cm³
Density of the He atom = 12.69 g/cm³