Answer: Multiple covalent bonds may occur in atoms that contain carbon,nitrogen, or <u><em>oxygen</em></u>.
Kinda confused what worksheet your on
2-ethyl-4,4 -dimethyl hex-1-ene.
Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c

0.511
Now using the equilibrium constant 
![K_c = \dfrac{[F6P]}{[G6P]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5BF6P%5D%7D%7B%5BG6P%5D%7D)
![0.511 = \dfrac{[F6P]}{[2.65]}](https://tex.z-dn.net/?f=0.511%20%3D%20%20%5Cdfrac%7B%5BF6P%5D%7D%7B%5B2.65%5D%7D)
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM
Used the Ideal gas law to get the grams of carbon dioxide
PV=nRT Where p is pressure, v is volume, n is a number of moles, t is temperature and r is the ideal gas constant. The value of R is 0.0821.
Solution
First, find the mole of carbon dioxide
n= PV/RT
= 1 x 1.0 / 0.081 x (2.73 x 102)
= 1 / 0.081 x 278.46
= 1 / 22.56
= 0.045 mol
Convert the moles into grams
Grams of CO2 = 0.045 mol x 44 g/mol = 1.98 g
So the grams of CO2 is 1.98g