Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.
Answer:
the correct answer to your question is 20
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.
After the first 2 min the 20 g are only 10g, (1 half-life)
after other 2 more min 5g (2 half-lives)
after other 2 more min 2.5 g (3 half-lives)
after 2 more 1.25 g (4 haf lifes)
