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Satoetoe24
22.08.2019 •
Chemistry
Emission factors biodiesel combustion in the production and transporta plant - 0.1 kg co, e-/ litre biodiesel production and transportation of fertilizer - 40 kg co, e-/ kg fertilizer electricity from the nation of herbicide - 23 kg co, e-/ kg herbicide hal grid - 1.032 kg co, e-/kwh roduction stage = 0.014 kg co, el piesel combustion stage = 73.25 g co, e / mj the calculations are for the production to biodiesel entire life cycle of the production process - from canola electricity combustion. given the above data, calculate the following: iv. i. total electricity requirement ii. total bio-diesel required for ements for all shire residents (kwh) for generator electricity production (kg) iii. total quantity of canola se total area of land required to produce the volume of ca a seeds required to produce desired electricity output(kg) quired to produce the volume of canola seed required as biodiesel input (ha) y total quantity of fertilize er required to generate required volume of canola (kg) vi. total quantity of herbicide reau ide required to generate required volume of canola (kg) ectricity from grid required for conversion of canola into biodiesel (kwh) total quantity of diesel required a el required during on-farm and post-farm stages (ltrs) total ghg emissions created given the electricity, diesel, bio-dies herbicide consumed in biodiesel production? total ghg emissions in tonne co, e? vii. total - viij. ix. x. 2. using a bar diagram, show which stage how which stage is creating the most ghg emissions? 3. draw a pie diagram showing the breakdown of ghg emissions in terms of inputs. 4. identify the ghg production hotspot and suggest possible mitigation strategies.
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Ответ:
See explanation
Explanation:
Energy of the photon;
E = hc/λ
h= Planks constant
c= speed of light
λ = wavelength
E= 6.6 × 10^-34 × 3 × 10^8/4 x10 -7
E = 4.95 × 10^-19 J
If 1ev=1.602x10-19 J
x = 4.95 × 10^-19 J
x= 3.1 ev
From Einstein's photoelectric equation;
KE = E - Wo
Where;
KE = kinetic energy of ejected photoelectron
E= energy of the photon
Wo= work function of the metal
KE = 3.1 eV - 2.13 eV
KE= 0.97 eV
KE = 0.97 eV × 1.602x10-19 J
KE = 1.55 × 10^-19 J
KE = 1/2 mv^2
1.55 × 10^-19= 1/2 × 9.1 × 10^-31 × v^2
v= √2 × 1.55 × 10^-19/9.1 × 10^-31
v= 5.8 × 10^5 m/s