Answer:
We know that
ħf = ф + Ekmax
where
ħ = planks constant = 6.626x10^-34 J s
f = frequency of incident light = 1.3x10^15 /s (1 Hz =
1/s)
ф = work function of the cesium = 2.14 eV
Ekmax = max kinetic energy of the emmitted electron.
We distinguish that:
1 eV = 1.602x10^-19 J
So:
2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J
So,
Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J
= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J
Answer:
5.19x10^-19 J
Kinetic energy:
In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.
Answer details:
Subject: Chemistry
Level: College
Keywords:
• Energy
• Kinetic energy
• Kinetic energy of emitted electrons
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Answer:
The percentage abundance of Eu isotopes are 52 % and 48 %
.
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope,:
% = x %
Mass = 151.0 amu
For second isotope :
% = 100 - x
Mass = 153.0 amu
Given, Average Mass = 151.96 amu
Thus,
Solving for x, we get that:
x = 52 %
<u>Thus percentage abundance of Eu isotopes are 52 % and 48 %
.</u>
A. 6 moles
B. 9 moles
C. 3 moles
D. 20 moles
I think please check me, in case I am wrong
Answer:
114 kPa
Explanation:
Using Gay-Lussac's law you get the equation 
and converting celcius you get the final equation of 
. After dividing 85.5 by 27+273(300) you get 0.285 and then you multiply 0.285 by 127+273 (400). You finally get 114 kPa
Answer: A. True
Explanation:
I'm not all the way sure, so please don't hate on me. I looked it up to double check and it should be true.
!PLEASE NOT HATE IF IT'S WRONG!
