Answer:
Solubility indicates the maximum amount of a substance that can be dissolved in a solvent at a given temperature. Such a solution is called saturated,it is calculated by dividing the mass of the compound by mass of the solvent then multiply by the molar concentration.
molar mass of solvent (Agl)= 107.8682 + 126.90447 = 234.77267g/mol
For 0.038M of Nal
molar mass of Nal compound= 22.99 + 126.90447 = 149.89447g/mol
solubility = [ 149.89447 / 234.77267] * 0.038 = 0.024
For 0.05M of AgNO3
molar mass of AgNO3 compound = 107.8682 + 14.01 + 3(16) = 169.8782g/mol
solubility = [169.8782 / 234.77267] * 0.05 = 0.036
For 0.025M of KI
molar mass of KI compound = 39.0983 + 253.8089 = 292.9072g/mol
solublity = [292.9072 / 234.77267] * 0.025 = 0.031
For 0.0125M of Lil
molar mass of Lil compound = 6.941 + 253.8089 = 260.7499g/mol
solubility = [260.7499 / 234.77267] * 0.0125 = 0.014
Therefore the solution with the highest solubility is 0.05M of AgNO3
<span>Protons and neutrons are both centrally located in an atom's nucleus, while negative electrons surround the nucleus in designated shells (which are commonly thought of as "orbits" as labeled in the PhET simulation).
Nucleus of an atom has positive charge, electrons are negative and atom has neutral net charge.</span>
That depends. there are 2 possible answers.
H
C - C = C - H gives a different answer on the right than on the left.
One the left side, the second Carbon is attached to a double bond and has but one hydrogen attached to it.
The Carbon on the right of the double bond has 2
H
C- C = C - H
H
I'm not sure what you should put. It's one of those things that I would repeat my argument and submit it.
When P1/P2 = C1/C2
and C is the molarity which = moles/volume
so, P1/P2 = [(mass1/mw)/volume] / [(mass2/mw)/volume]
P1/P2 = (mass1/mw)/1.5L / (mass2/mw)/1.5L
so, Mw and 1.5 L will cancel out:
∴P1/P2 = mass1 / mass2
∴ mass 2 = mass1*(P2 / P1)
= 0.278g * (78 bar / 62 bar)
= 0.35 g
∴ the quantity of argon that will dissolve at 78 bar = 0.35 g
Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
