That is because Alchemy is not based on real research and facts, but rather on magic and occultism. Sure, many things in it have been used as basics for some chemistry, but most is false and incorrect.
The vapor pressure is obtained as 23.47 torr.
<h3>What is the vapor pressure?</h3>
Given that; p = x1p°
p = vapor pressure of the solution
x1 = mole fraction of the solvent
p° = vapor pressure of the pure solvent
Δp = p°(1 - x1)
Δp =x2p°
Δp = vapor pressure lowering
x2 = mole fraction of the of the solute
Number of moles of glycerol = 32.5 g/92 g/mol = 0.35 moles
Number of moles of water = 500.0 g/18 g/mol = 27.8 moles
Total number of moles = 0.35 moles + 27.8 moles = 28.15 moles
Mole fraction of glycerol = 0.35 moles/28.15 moles = 0.012
Mole fraction of water = 27.8 moles/28.15 moles =0.99
Δp = 0.012 * 23.76 torr
Δp = 0.285 torr
p1 = p° - Δp
p1 = 23.76 torr - 0.285 torr
p1 = 23.47 torr
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The original or accepted value for the percent by mass of water in a hydrate = 36%
Percen by mass of water in the hydrate determined by the student
in the laboratory = 37.8%
So the difference between the actual and the percent by mass in water determined by student = (37.8 - 36.0)%
= 1.8%
So the percentage of error made by the student = (1.8/36) * 100 percent
= (18/360) * 100 percent
= ( 1/20) * 100 percent
= 5 percent
So the student makes an error of 5%. Option "1" is the correct option.
Answer:
All elements of period 7 are radioactive. This period contains the actinides, which includes plutonium, the naturally occurring element with the heaviest nucleus; subsequent elements must be created artificially.
Explanation:
hope u understand :)
Answer:
Answers are in the explanation
Explanation:
Ksp of CdF₂ is:
CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)
Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²
When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:
6.44x10⁻³ = [X] [2X]²
6.44x10⁻³ = 4X³
X = 0.1172M
<h3>[F⁻] = 0.2344M</h3><h3 />
Ksp of LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]
When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:
1.84x10⁻³ = [X] [X]
1.84x10⁻³ = X²
X = 0.0429
<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
