Answer : The partial pressure of
is, 222.93 torr
Explanation :
Half-life = 2.81 hr = 168.6 min
First we have to calculate the rate constant, we use the formula :



Now we have to calculate the partial pressure of 
The balanced chemical reaction is:

Initial pressure 760 0 0
At eqm. (760-2x) 4x x
Expression for rate law for first order kinetics is given by:

where,
k = rate constant
t = time passed by the sample = 215 min
a = initial pressure of
= 760 torr
a - x = pressure of
at equilibrium = (760-2x) torr
Now put all the given values in above equation, we get:


The partial pressure of
= x = 222.93 torr
C not sure 100 percent but my best guess
crop rotation, green manure, and bone meal
Explanation:
I just looked it up. hope it helps
Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M