Most lakes have rivers flowing out of them, carrying water to the ocean. however, some lakes, including great salt lake and the dead sea, have no outlet. water flowing into these lakes leaves only through evaporation. considering that the water flowing into the lakes contains many dissolved substances, how does the lack of an outlet affect the composition of these lakes? what would you expect to happen after many more years of inflow and evaporation?

The lakes when flow towards ocean carrying water, they carry salt with them. So they enter the oceans with salt content, hence the concentration of salt in them remains almost constant.

In case of lakes with no outlet (like Great salt lake), they do not transfer any salt or water to oceans. Infact due to evaporation the amount of water gets decreased in them. This decrease the solvent amount and hence the concentration of dissolved salt increases due to decrease in volume of water.

The lakes thus become more salty after many more years of inflow and evaporation.

In terms of chemistry

Concentration = amount of solute (the salts here) /  volume of solvent (water here)

Evaporation : causes decrease in volume of water

inflow of water: caused increase in volume of water and salt

overall due to decrease in volume of water the ratio of amount of solute to volume of water increases making the lake more salty.

Since the only way of water flow to these lakes or bodies of water is through evaporation, I would expect an increase in unknown substances in the composition of the lakes due to the amount of contamination that globalization produces and affects terribly the surroundings when these unknown substances travel through evaporation as the outlet of these bodies of water. Therefore I think continuous contamination is what to expect after many more years of inflow and evaporation.

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