Answer:
number of Al atom =54÷27=2atom of Al. number of o atom = 160÷16=10 atom of o . =214U MASS of ALO2.
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
POH = - log [ OH-]
pOH = - log [ 1 x 10⁻¹²]
pOH = 12
6.022 x 1023 atoms are in 14 grams of NO2
