The average atomic mass of her sample is 114.54 amu
Let the 1st isotope be A
Let the 2nd isotope be B
From the question given above, the following data were obtained:
- Abundance of isotope A (A%) = 59.34%
- Mass of isotope A = 113.6459 amu
- Mass of isotope B = 115.8488 amu
- Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
- Average atomic mass =?
The average atomic mass of the sample can be obtained as follow:

Thus, the average atomic mass of the sample is 114.54 amu
Learn more about isotope: brainly.com/question/25868336
Answer:
the same is what is this question like what did u exame
Through manipulation of equations, we are able to obtain the equation:
![-pOH= log [ OH^{-}]](https://tex.z-dn.net/?f=-pOH%3D%20log%20%5B%20OH%5E%7B-%7D%5D%20)
Then we can transform the equation into:
![[ OH^{-}]= 10^{-pOH}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D%2010%5E%7B-pOH%7D%20%20)
Then we are able to plug in the pOH and directly get [OH-]:
![[ OH^{-}] = 10^{-6.48}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%20%3D%2010%5E%7B-6.48%7D%20)
You have to use the equation F=ma and solve for m to get m=F/a.
m=mass in kg
F=force (in this case 350N)
a=acceleration (in this case 10m/s²)
when you plug everything in you should find that m=35kg
I hope this helps.
Answer:
Total pressure 5.875 atm
Explanation:
The equation for above decomposition is

rate constant 
Half life 
Initial pressure 
Pressure after 3572 min = P
According to first order kinematics


solving for P we get
P = 2.35 atm

initial 4.70 0 0
change -2x +2x +x
final 4.70 -2x 2x x
pressure of
after first half life = 2.35 = 4.70 - 2x
x = 1.175
pressure of
after first half life = 2x = 2(1.175) = 2.35 ATM
Total pressure = 2.35 + 2.35 + 1.175
= 5.875 atm