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vadimmilevsky559
@vadimmilevsky559
August 2021
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какой объем (н.у.) углекислого газа расходуется при образовании в процессе фотосинтеза глюкозы массой 360г
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Alexei78
Verified answer
Дано
m(C6H12O6) = 360 g
---------------------------------
V(CO2)-?
6CO2+6H2O-- Ф/С-->C6H12O6+6O2
M(C6H12O6) = 180 g
n(C6H12O6) = m/M = 360/180 = 2 mol
6n(CO2)= n(C6H12O6)
n(CO2) = 2*6 / 1= 12 mol
v(CO2) = n*Vm = 12*22.4 = 268.8 L
ответ 268.8 л
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Answers & Comments
Verified answer
Даноm(C6H12O6) = 360 g
---------------------------------
V(CO2)-?
6CO2+6H2O-- Ф/С-->C6H12O6+6O2
M(C6H12O6) = 180 g
n(C6H12O6) = m/M = 360/180 = 2 mol
6n(CO2)= n(C6H12O6)
n(CO2) = 2*6 / 1= 12 mol
v(CO2) = n*Vm = 12*22.4 = 268.8 L
ответ 268.8 л