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 18 kg of sucrose (C12H22O11) per hour per hectare (1 ha=10,000meters squared). Assuming that sucrose is produced by the reaction 12CO2(g)+11H2O(l)>>>C12H22O11(s)+12O2(g) ?H= 5640kJ 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 18 kg of sucrose (C12H22O11) per hour per hectare (1 ha=10,000meters squared). Assuming that sucrose is produced by the reaction 12CO2(g)+11H2O(l)>>>C12H22O11(s)+12O2(g) ?H= 5640kJ 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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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 18 kg of sucrose (C12H22O11) per hour per hectare (1 ha=10,000meters squared). Assuming that sucrose is produced by the reaction
12CO2(g)+11H2O(l)>>>C12H22O11(s)+12O2(g) ?H= 5640kJ
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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