Suppose that 40 mL of water is poured simultaneously into three containers, each having a different sediment. After 10 seconds, 5 mL remains atop Sediment A, 20 mL of water remains atop Sediment B, and 28 mL of water remains atop Sediment C. Water continues to seep into each of the substances. Which substance has the lowest permeability?

The answer to your question is Sediment C

Explanation:

Permeability is the property of passing through a material without physically or chemically affecting it.

If a substance is very permeable, it will pass through a material rapidly others substances can be permeable but can not pass through the material quickly.

In the data given, we can conclude that the third substance is the least permeable because less quantity of water has passed through it.

2b2t

explanation:

2b2t

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