# Chemistry

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What is the minimum number of photons with wavelength 645nm needed to prepare 135 grams of glucose?
When the thermochemical equation is
C6H12O6(s)+ 6O2(g)--> 6CO2(g)+ 6H2O(g) Reaction is -2539.01 Kj/mol

• Chemistry -

How much energy is needed for a photon of 645 nm? That is delta E = hc/wavelength - 6.626 x 10^-34*3 x 10^8/645 x 10^-9 = about 3 x 10^-19 J/photon (but you need to do this exactly).

How much energy is needed to prepare 135 g glucose? That will be 135/180 (you need to convert the molar mass exactly since my 180 is an estimate) = about 0.75 mol glucose so 0.75 x 2539.01 x 10^3 J/mol = about 1904 J.
So how many packets of energy of 3 x 10^-19 do you need to have a total of 1904 J? Check my work.

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