Asolution was prepared by mixing 50 μl of 8 mg/ml octadecane internal standard with 100 μl of 12 mg/ml of a dodecane standard with 350 μlhexane. the octadecane peak area was 357 while the dodecane area was 516. calculate the response factor. the same internal standard was used to prepare an unknown sample of dodecane. the sample was prepared by mixing 50 μl of 8 mg/ml octadecane and 100 μl dodecaneof unknown concentration. the areas were 382 for octadecane and 463 for dodecane. calculate the unknownconcentration. remember to use significant figures.

For cyclohexane compounds drawn in a chair conformation:   -axial bonds are vertical  -equatorial bonds are parallel to two bonds within the ring.  thus, as i assume the chair conformation is drawn:   -the x group is in an axial position, as are the five other vertical bonds.  -three axial bonds are on the top face of the ring  -the other three axial bonds are on the bottom face  the position of the second substituent on each ring carbon is equatorial.  gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other.  drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1.  click this link and watch some academic videos that may you more understand the topic 

Moles of co₂ = 2.35/44 = 0.0534 mole.from ideal gas equation pv=nrt,  t = 18+273k = 291 kelvin.∴px5 = 0.0534x0.0821x291∴p = 0.0534x0.0821x291/5 = 0.2551 atm.now, 1 atm = 760 mm of hg  ∴ 0.02551 atm = 0.02551 x 760 = 19.38 mm of hg