Answer:
In order to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25 x 10⁻² M, you need to weight 0.4055 g of FeCl₃ and add to 200.0 mL of water.
Explanation:
Concentration: 1.25 x 10⁻² M
1,25 x 10⁻² mol FeCl₃ ___ 1000 mL
x ___ 200.0 mL
x = 2.5 x 10⁻³ mol FeCl₃
Mass of FeCl₃:
1 mol FeCl₃ _____________ 162.2 g
2.5 x 10⁻³ mol FeCl₃ _______ y
y = 0.4055 g FeCl₃
1 mol ------------- 6.02x10²³ atoms
1.14 moles ------ ( atoms )
atoms = 1.14 x ( 6.02x10²³) / 1
atoms = 6.86 x 10²³ / 1
= <span> 6.86 x 10²³ atoms
hope this helps!</span>
Per hour-68x60=4080
Per day-4080x24=97920
Per week-97920x7=685440
