The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 20. kg of sucrose (C₁₂H₂₂O₁₁) per hour per hectare (1 ha = 10,000 m²). Assuming that sucrose is produced by the reaction\[ 12\text{CO}_2(g) + 11\text{H}_2\text{O}(l) \rightarrow \text{C}_{12}\text{H}_{22}\text{O}_{11}(s) + 12\text{O}_2(g) \quad \Delta H = 5640 \text{ kJ} \]Calculate the percentage of sunlight used to produce the sucrose – that is, determine the efficiency of photosynthesis in this field.Percent efficiency = [ ] %The text includes two interactive buttons labeled "Submit Answer" and "Try Another Version," with a note at the bottom indicating "3 item attempts remaining."

The fraction of light energy that is converted into chemical energy during the process of…

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 20. kg of sucrose (C12H22O11) per hour per hectare (1 ha = the reaction 10,000 m²). Assuming that sucrose is produced by 12CO2 (g) + 11H20(1) · → C12 H22 O11 (s) + 1202(g) AH = 5640 kJ calculate the percentage of sunlight used to produce the sucrose – that is, determine the efficiency of photosynthesis in this field. Percent efficiency =

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 20. kg of sucrose (C12H22O11) per hour per hectare (1 ha = the reaction 10,000 m²). Assuming that sucrose is produced by 12CO2 (g) + 11H20(1) · → C12 H22 O11 (s) + 1202(g) AH = 5640 kJ calculate the percentage of sunlight used to produce the sucrose – that is, determine the efficiency of photosynthesis in this field. Percent efficiency =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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The Sun supplies about 1.0 kilowatt of energy for each squaremeter of surface area (1.0 kW/m2, where a watt = 1 J/s).Plants produce the equivalent of about 0.20 g of sucrose(C12H22O11) per hour per square meter. Assuming that thesucrose is produced as follows, calculate the percentage ofsunlight used to produce sucrose.12 CO2(g)+ 11 H2O(l)-----> C12H22O11 + 12 O2(g)ΔH = 5645 kJ
The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 23 kg of sucrose (C12H22 O11) per hour per hectare (1 ha = 10,000 m²). Assuming that sucrose is produced by the reaction 12CO₂(g) + 11H₂O(l) → C12H22 O11 (8) + 1202 (9) AH = 5640 kJ calculate the percentage of sunlight used to produce the sucrose - that is, determine the efficiency of photosynthesis in this field. Percent efficiency = %
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