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11Alexandr11 [23.1K]
3 years ago
9

A solution contains some or all of the ions Cu2+,Al3+, K+,Ca2+, Ba2+,Pb2+, and NH4+. The following tests were performed, in orde

r, on the solution.1. Addition of 6 M HCl produced no reaction.2. Addition of H2S with 0.2 M HCl produced a black solid.3. Addition of (NH4)2S produced a white solid.4. Addition of (NH4)2HPO4 in NH3 produced no reaction.5. The final supernatant when heated produced a purple flame.Identify which of these ions were present in the solution (P), which were absent (A), and which were inconclusive (I)a. Cu2+b. Al3+c. K+d. Ca2+e. Ba2+f. Pb2+g. NH4+
Chemistry
1 answer:
xeze [42]3 years ago
6 0

Answer:

See below explanation

Explanation:

When having a mixture of metals in solution, you may perform an analytical study (using selective chemical conditions), that may help you to determine whether a metal (cation) is present or not

Using selective analytes (or conditions), leads to consecutive precipitations, until most of the cations are separated in precipitates

With this technique, you may identify metals in different groups, each group will have its analyte (or condition), which will help to have a different precipitate:

- Group I: Ag⁺, Pb⁺², Hg⁺²;  Analyte: HCL ; Precipitate: AgCl (white) , PbCl₂, HgCl₂

- Group II: As⁺³ , Bi⁺³, Cd⁺², Cu⁺² , Sb⁺³, Sn⁺² ; Analyte: H₂S (g) with HCL ; Precipitate: As₂S₃ , Bi₂S₃ , CdS (yellow) , CuS (black), Sb₂S₃, SnS

- Group III: Co⁺², Fe⁺², Fe⁺³, Mn⁺², Ni⁺², Zn⁺², Al⁺³, Cr⁺³; Analyte: NaOH or NH₃ with (NH₄)₂S (ac) ; Precipitate: CoS (black) , FeS, MnS , NiS (black), ZnS (white) , Al(OH)₃ (white), Cr(OH)₃  

- Group IV: Mg⁺², Ca⁺², Sr⁺², Ba⁺²; Analyte: Na₂CO₃ (ac) or (NH₄)₂HPO₄ (ac); Precipitate: respective carbonate or phosphate MgCO₃/MgHPO₄, CaCO₃/CaHPO₄ , SrCO₃/SrHPO₄, BaCO₃/BaHPO₄

- Group V: Li⁺, K⁺, Na⁺, Rb⁺, Cs⁺, NH₄⁺ ; will remain all in final solution

According to the original statement:

A solution contains one or more of the following: Cu⁺², Al⁺³, K⁺, Ca⁺², Ba⁺², Pb⁺², NH₄⁺

1) Addition on HCl 6M produces no change: we can say the sample does not contain Pb⁺² (group I)

2) Addition of H₂S with 0.2 M HCL produced a black solid: we could say sample contains Cu⁺²(group II)

3) Addition of (NH₄)₂HPO₄ in NH₃ produces no reaction: we could say we don´t have Ca⁺² and /or Ba⁺²  (group IV)

4) The final supernatant, when heated produced a purple flame: in the final solution, we have K⁺ (group V), which produces a purple flame (based on its characteristic emission spectrum when subjected to flame)

This analysis will be inconclusive for NH₄⁺ (according to above describe technique)

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Magnesium hydroxide is added to a solution of hydrochloric acid. A reaction occurs and magnesium chloride and water are formed.
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hydrochloric acid (HCI) is a - reactant

Explanation: It’s in my notes

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3 years ago
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3 years ago
Catalyst
Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

\tt \dfrac{4}{5}\times 0.5=0.4

volume NO at 1273 K and 1 atm

\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L

b. 15 L NH3 at STP ( 1mol = 22.4 L)

\tt \dfrac{15}{22.4}=0.67~mol

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

\tt \dfrac{6}{4}\times 0.67=1

mass H2O(MW = 18 g/mol) :

\tt mass=mol\times MW=1\times 18=18~g

c. mol NO at 1273 K and 1 atm :

\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34

mol ratio of NO : O2 = 4 : 5, so mol O2 :

\tt \dfrac{5}{4}\times 0.34=0.425

Volume O2 at STP :

\tt 0.425\times 22.4=9.52~L

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The correct answer is (A) from 2nd to 3rd shell.


The explanation :


when a gain of energy is the shift of the electrons from a shell of low energy to the shell of high energy


and we have here 2nd shell is the shell of low energy, and 3rd shell is the shell of high energy.


∴ (A) from 2nd to 3rd shell is the correct answer.

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