To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
2 M x V1 = 0.1 M x .5 L
<span>V1 = 0.025 L or 25 mL of the
2 M KCl solution is needed</span>
The moles of fluorine present are 71/19 = 3.74
Now, we know that one mole of gas at 273 K and 101.3 kPa (S.T.P.) occupies 22.4 liters
Volume of 3.74 moles at S.T.P = 3.74 x 22.4
Volume = 83.776 L = 83,776 mL
Now, we use Boyle's law, that for a given amount of gas,
PV = constant
P x 6843 = 101.3 x 83776
P = 1,240 kPa
Hydrofluoric acid is not a strong acid despite being a hydrohaulic acid (hydrogen and a halogen). HF is highly corrosive and can be used to dissolve most oxides
All of them are correct! good!!
