Answer: 8556 mm, or 855.6 cm (8560 mm to 3 sig figs)
Explanation: Convert mm to cm by dividing by 10 (1cm/10mm)
Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2
Calculate the volume occupied by 1.40 kg of foil in cm^3. 1.40kg = 1400g
1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au
Volume = Area (of the face) * Length
We want Length:
Length = Volume/Area
L = (518.5 cm^3/0.606 cm^2)
L = 855.6 cm (8556 mm) Round to 3 sig figs (856 cm and 8560 mm)
Opposites attract, like for example magnets, one is positive and the other is negatively charged, they will attract
Electrolysis can be used to separate a substance into its original components/elements and it was through this process that a number of elements have been discovered and are still produced in today's industry.
Answer: -
Concentration of PbI₂ = 1.5 x 10⁻³ M
PbI₂ dissociates in water as
PbI₂ ⇄ Pb²⁺ + 2 I⁻
So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.
Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =
= 1.5 x 10⁻³ x 2 M
= 3 x 10⁻³ M
PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.
[Pb²⁺] = 1.5 x 10⁻³ M
So solubility product for PbI₂
Ksp = [Pb²⁺] x [ I⁻]²
=1.5 x 10⁻³ x (3 x 10⁻³)²
= 4.5 x 10⁻⁹
Molar mass :
Li₂S = <span>45.947 g/mol
AlCl</span>₃ = <span>133.34 g/mol
</span><span>3 Li</span>₂<span>S + 2 AlCl</span>₃<span> = 6 LiCl + Al</span>₂S₃
3 * 45.947 g Li₂S ----------> 2 * <span>133.34 g AlCl</span>₃
1.084 g Li₂S ----------------> ?
Mass Li₂S = 1.084 * 2 * 133.34 / 3 * 45.947
Mass Li₂S = 289.08112 / 137.841
Mass Li₂S = 2.0972 g
hope this helps!
