The net reaction of the calvin cycle is the conversion of co2 into the three-carbon sugar g3p. along the way, reactions rearrange carbon atoms among intermediate compounds and use the atp and nadph produced by the light reactions. in this exercise, you will track carbon atoms through the calvin cycle as required for the net production of one molecule of g3p. for each intermediate compound in the calvin cycle, identify the number of molecules of that intermediate and the total number of carbon atoms contained in those molecules. as an example, the output g3p is labeled for you: 1 molecule with a total of 3 carbon atoms. labels may be used once, more than once, or not at all. hints

a) 3 molecule of carbon dioxide (CO2) - 3 carbon atoms

b) 6 molecule of 3 - PGA - 18 carbon atoms ( 3 cabon atoms in 1   3 PGA molecule)

c)  6 molecule og G3P - 18 carbon atom ( 3 carbon atoms in G3P molecule )

d) 5 molecules of G3P - 15 carbon atom

e)  3 molecule of R5P   -   15 carbon atoms (  5 carbon atom of R5P  olecule )

f) 3 molecule of RuBP  -    15 carbon atoms ( r carbon atom of RuBp molecule )

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol