Answer:
Oxidized and reducing agent: manganese.
Reduced and oxidizing agent: mercury.
Explanation:
Hello!
In this case, for the reaction:

We keep in mind that the species that increase the oxidation state is the oxidized one whereas the one that decrease the oxidation state is the reduced one; therefore manganese is the oxidized one as well as the reducing agent because it goes from 0 to +2 and mercury the reduced one as well as the oxidizing agent because it goes from 2+ to 0.
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<u>Answer:</u>

∆x is the Uncertainty in the position of the particle
∆p is the Uncertainty in the momentum of the particle



is the answer
Number of photons can be calculated by dividing the needed energy by the energy per photon.
The minimum energy needed is given as 2 x 10^-17 joules
Energy per photon = hc / lambda where h is planck's constant, c is the speed of light and lambda is the wavelength
Energy per photon = (<span>6.626 x 10^-34 x 3 x 10^8) / (475 x 10^-9)
= 4.18 x 10^-19 J
number of photons = (2 x 10^-17) / (4.18 x 10^-19)
= 47.79 photons which is approximately 48 photons</span>