If there is a higher concentration of sodium on the left side of a beaker and the membrane is permeable to the sodium, which of the following will occur? \

sodium is going to migrate the left to the right side through the membrane

Explanation:

Differences in concentration of sodium produce a movement of sodium through the membrane until reach the equilibrium

A permeable membrane is a barrier that only allows certain particles to pass through them. Diffusion is a physical process whereby the particles move from a region of higher concentration to a region of lower concentration, and this movement occurs until the equilibrium is reached (same concentration in every region).

If the membrane is permeable to the sodium, it means that only sodium atoms will be able to migrate from one side to another. And according to difussion phenomenon, sodium atoms will migrate from the side with higher concentration to the side of lower concentration. So, a migration of sodium will occur from the left side of the beaker to the right side of the beaker until the concentration of sodium is the same in both sides.  

im not exactly sure but i think its b!

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2