Consider the following acid and bases HCO2H ka = 1.8 x 10^-4 HOBr Ka = 2.0 x 10^-9
(C2H5)2NH kb = 1.3 x 10-3
HONH2 kb = 1.1 x 10^-8
choose sobstances to create ph = 4 buffer solutions:
select all tha apply
HONH3NO3
HOBr
NaOBr
(C2H5)2NH2Cl
(C2H5)2NH
HCO2H
KHCO2
HONH2

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 

A) the enthralling of formation for pure element in its standard state is always positivehope this : )