E = hc/(lamda)
The lamda symbol is wavelength, which this site does not have. I can represent it with an "x" instead.
Plancks constant, h = 6.626×10^-32 J·s
Speed of light, c = 3.00×10^8 m/s
The energy must be greater than or equal to 1×10^-18 J
1×10^-18 J ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / x
x ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / (1×10^-18 J)
x ≤ 1.99×10^-7 m or 199 nm
The wavelength of light must be greater than or equal to 199 nm
Answer:
177.277amu
Explanation:
the total occuring isotopes for Hafnium is =6.
First isotope had an atomic weight of 173.940amu
Second isotope =175.941amu
Third isotope =176.943amu
Fourth isotope=177.944amu
Fifth isotope. =178.946amu
sixth isotope .179.947amu
<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>
Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu
Average atomic weight= 1063.661amu /6 = 177.2768333amu
= 177.277amu to 3 decimal places.
Naproxen is known to be a weak acid. In order to calculate
its ka and pka, use the equation of getting the ph of weak acid which is ph=
-1/2 log [(Ka)(Mwa)]. The Ka value is 3.18x10^-5. The pKa can be obtained
through pKa = - log Ka. The pKa is 4.5.
It would 47.7 because you would have to both minus the number together.
