A population of Esherichia coli is given time to grow for eight generations. The final population is 1,996,800 cells. What was the initial population in cells?

7800 cells

Explanation:

The number of Escherichia coli cells N after n generations is given by

N = 2ⁿ × N₀ where N₀ = initial population

Now making N₀ subject of the formula, we have

N₀ = N/2ⁿ

Since n = 8 and N = 1,996,800 cells, substituting the values of the variables into the equation, we have

N₀ = N/2ⁿ

N₀ = 1,996,800 cells/2⁸

N₀ = 1,996,800 cells/256

N₀ = 7800 cells

So, the initial population of Escherichia coli cells is 7800 cells.

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