A 125 μm film of rubber having an area of 0.6 m2 separates air from pure O2, with both gases at 1 atm and 298K. If oxygen has a diffusion coefficient in rubber of 0.21 × 10-9 m2 /s, calculate the mass flowrate (mg/min) of oxygen through the surface. Plot the concentration profile as a function of z. (cA vs. z, where A = O2).

the child will have heterozygous dominant chromosomes as characters of his father are dominant over the characters of his mother.

the child is heterozygous

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be