Answer:
Mass% Cr = 85.5%
Explanation:
<u>Given:</u>
Mass of CrBr3 sample = 0.8409 g
Mass of the AgBr precipitate = 1.0638 g
<u>To determine:</u>
The mass percent of Cr in the sample
<u>Calculation:</u>
The reaction of CrBr3 with silver nitrate results in the precipitation of the bromide ion as silver chloride (AgBr) and Cr as soluble Cr(NO3)2
CrBr3(aq) + 3AgNO3(aq)→ 3AgBr(s) + Cr(NO3)3(aq)
Molecular weight of AgBr =187.77 g/mol
Moles of AgBr precipitated is:
Since 1 mole of AgBr contains 1 mole of Cl, therefore:
# moles of Cl = 0.004478 moles
At wt of Cl = 35.45 g/mol
Mass%(Cr) = 100 - 14.50=85.5%
Answer:
Here's what I get
Explanation:
You have an equilibrium reaction between Fe³⁺/ SCN⁻ and FeSCN²⁺.
When you add AgNO₃, the Ag⁺ reacts with the SCN⁻. It forms a colourless precipitate of Ag(SCN).
Ag⁺(aq) + SCN⁻(aq) ⟶ AcSCN(s)
According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
If you add Ag⁺ to the equilibrium solution, it removes the SCN⁻ [as an Ag(SCN) precipitate].
The system responds by trying to replace the missing SCN⁻:
The Fe(SCN)²⁺ dissociates to form SCN⁻, so the position of equilibrium shifts to the left,
You now have more Fe³⁺ and SCN⁻ and less of the highly coloured Fe(SCN)²⁺ at the new equilibrium.
The deep red colour becomes less intense.
This question is a dilution problem. In order to prepare a certain concentration and volume from a stock solution, you have to use the equation,
M1V1 = M2V2
From the equation, you need to calculate the volume you need from the stock solution to be able to have the diluted concentration then dilute it to a certain volume.
(1.0 M) (V1) = (0.1 M) (100 mL)
V1 = 0.1 mL is needed from the 1.0 M HCl solution
1) All matter is made of atoms. 2) Atoms are indivisible and indestructible.
