Answer:
ΔE = 5.02 x 10⁻¹⁹ j
Explanation:
ΔE (photon) = h·f = (6.63 x 10⁻³⁴ j·s)(7.57 x 10¹⁴ s⁻¹) = 5.02 x 10⁻¹⁹ j
h = Planck's Constant = 6.63 x 10⁻³⁴ j·s
f = frequency (given) = 7.57 x 10¹⁴ s⁻¹
There are 3.98 × 10^23 atoms of oxygen in the sample.
Given that;
1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen
x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen
x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms
x = 0.0368 moles
The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18
Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.
Learn more: brainly.com/question/9743981
The group is called the noble gas i think