Answer:
Passive Transport
Explanation:
The three examples of passive transport are
Diffuison
Osmosis
facilated diffuison
So the answer can be A or B
The given concentration of boric acid = 0.0500 M
Required volume of the solution = 2 L
Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.
Calculating the moles of 0.0500 M boric acid present in 2 L solution:

Converting moles of boric acid to mass:

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.
First convert 0.163 grams of N2O to mol by dividing it with
the molecular weight. The molecular weight of N2O is 44 grams/mol. The answer
would be 3,79x10^-3. Then multiply it with 2 since there are 2 Nitrogen in one
mole of N2O. Therefore, there are 7.41x10^-3 moles of Nitrogen.
F-1 because there is a one after it.