The sun supplies energy at a rate of about 10 kilowatt per square meter of surface area (1 watt = 1 y/s). The plants in an agricultural field produce the equivalent of 18.2 kg
sucrose (C12H22011) per hour per hectare (1 hectare = 10.000 m?). Assume that sucrose is produced by the reaction
12 CO2(8) + 11 H20 - C12H22011(s) 12 (8) AH = 5640 kJ
calculate the percentage of sunlight used to produce the sucrose-that is determine the efficiency of photosynthesis.
The reaction
SO3(8) + H2001) --- H250 (aq)

You have to calculate the oxidation estates of the atoms in each compound.

I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.

In K2Cr2O7:

- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.

- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.

That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.

In Cr2O3:

- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6

- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.

So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.

Cr has a change in oxidation number of - 3.