How many moles of co2 must enter the calvin cycle for the synthesis of one mole of glucose?
1 answer:
Answer:3 moles
Explanation:
For every three molecules of CO2 that enters the Calvin cycle, one molecule of the three carbon glyceraldehyde 3-phosphate (G3P) is produced. Two molecules of G-3-P are required to produce one molecule of glucose. Therefore, the Calvin cycle needs to make a total of 6 turns to produce two molecules of G-3-P.
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Answer:
43.89 min
Explanation:
Given that:-
The speed of light =
The distance =
The conversion of distance in km to distance into m is shown below as:-
1 km = 1000 m
So,
Distance =
The relation between speed distance and time is shown below as:-
Thus,
Time = 2633.33 seconds
Also, 1 s = 1/60 min
So,
Time=
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,
, the reaction will be
and we know, pH = -log[H+] and pOH = -log[OH-]
And![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
</span>
For triehtylamine,
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4And
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
Answer:
sample B contains the larger density
Explanation:
Given;
volume of sample A, V = 300 mL = 0.3 L
Molarity of sample A, C = 1 M
volume of sample B, V = 145 mL = 0.145 L
Molarity of sample B, C = 1.5 M
molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol
Molarity is given as;
The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g
The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g
The density of sample A
The density of sample B
Therefore, sample B contains the larger density