Describe the importance of plaster of paris
1 answer:
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pH of the acetyl choline solution before incubation = 7.65
pH of the solution after incubation = 6.87
The difference in concentration of hydronium ion before and after incubation
=-
=
This difference in hydronium ion concentration can be attributed to the increase in the concentration of acetic acid, which is formed when acetylcholine is hydrolyzed by acetycholinesterase. The mole ratio of acetylcholine to acetic acid is 1:1.
Therefore the moles of acetylcholine =
Answer:
The answer to the question is;
The first ion to precipitate out is the Al³⁺ ion and the concentration of the Al³⁺ ion when the Ca²⁺ ion begins to precipitate is 1.12 × 10⁻⁵ M.
Explanation:
To solve the question, we note that
aluminum nitrate, Al(NO₃)₃ will dissociate as follows
Al(NO₃)₃ → Al³⁺ (aq) + 3NO₃⁻ (aq)
Therefore when sodium phosphate is added to a solution that contains aluminum nitrate we have the following system of aluminium phosphate which is
AlPO₄(s) ⇄ Al³⁺(aq) + PO₄³⁻(aq)
The solubility product for the above reaction is
Ksp = [Al³⁺][PO₄³⁻] = 9.84×10⁻²¹
The solubility product for calcium phosphate is expressed as
Ca₃(PO₄)₂(s) ⇄ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)
With Ksp = [Ca²⁺]³[PO₄³⁻]² = 2.07×10⁻³³
From the solubility product, we can find the concentration of [PO₄³⁻] at which precipitation starts as follows
The phosphate concretion for Al³⁺ when precipitation starts is
[PO₄³⁻] = = 9.84×10⁻²¹ / 0.007 = 1.406×10⁻¹⁸ M
The phosphate concretion for Ca²⁺ when precipitation starts is
[PO₄³⁻] = =
= 8.75×10⁻¹⁶ M
(Aluminium phosphate precipitates out first)
The reaction favors the precipitation of the aluminum phosphate first due to the lower concentration of the [PO₄³⁻] ions in the [Al³⁺][PO₄³⁻] system which is lower than the relative [PO₄³⁻] in the [Ca²⁺]³[PO₄³⁻]².
Therefore, the more sodium phosphate added serves to precipitate the remaining aluminium phosphate.
The process continues and the concentration of Al³⁺ decreases as more precipitates form. The process continues until the equilibrium conditions satisfies the precipitation threshold level for the calcium phosphate system concentration whereby the concentration of the Al³⁺ in the solution is given by.
[Al³⁺] = = 1.12 × 10⁻⁵ M
Therefore the concentration of this aluminium ion when the calcium ion begins to precipitate = 1.12 × 10⁻⁵ M.
Answer:
43.13Kg of melamine
Explanation:
The problem gives you the mass of urea and two balanced equations:
First we need to calculate the number of moles of urea that are used in the reaction, so:
molar mass of urea =
The problem says that you have 161.2Kg of urea, so you take that mass of urea and find the moles of urea:
161.2Kg of urea2684 moles of urea
Now from the stoichiometry you have:
2684 moles of urea = 447 moles of melamine
The molar mass of the melamine is so we have:
= 5637.64 g of melamine
Converting that mass of melamine to Kg:
5637.64 g of melamine * = 56.38 Kg of melamine, that is the theoretical yield of melamine.
Finally we need to calculate the mass of melamine with a yield of 76.5%, so we have:
%yield = 100*(Actual yield of melamine / Theoretical yield of melamine)
Actual yield of melamine = = 43.13Kg of melamine