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hammer [34]
3 years ago
5

Use the solubility rules from the Lab 4 introduction and your knowledge of qualitative separation schemes from the lab to answer

the following questions.
The qualitative analysis experiment you did is actually an abbreviated version of a much larger analysis scheme in which many different cations are separated and identified. Suppose a mixture contains Ag+, K+, NH4+, Hg22+, Pb2+, Mg2+, Sr2+, Ba2+, Cu2+, Al3+ and Fe3+.
(a) Which of the following ions could you separate, by causing them to precipitate, with the addition of HCl? (Hint: HCl is a source of chloride ions. Select all that apply.)
Ag+ K+ NH4+
Hg22+ Pb2+ Mg2+
Sr2+ Ba2+ Cu2+
Al3+ Fe3+
(b) After the addition of HCl, the above sample is centrifuged and decanted. Which of the following cations remaining in the supernatant could you separate, by causing them to precipitate, with the addition of H2SO4? (Hint: H2SO4 is a source of sulfate ions. Select all that apply.)
Ag+ K+ NH4+
Hg22+ Pb2+ Mg2+
Sr2+ Ba2+ Cu2+
Al3+ Fe3+
(c) After the addition of H2SO4, the above sample is centrifuged and decanted. Which of the following cations remaining in the supernatant could you separate, by causing them to precipitate, with the addition of H2CO3? (Hint: H2CO3 is a source of HCO3− and carbonate ions. Select all that apply.)
Ag+ K+ NH4+
Hg22+ Pb2+ Mg2+
Sr2+ Ba2+ Cu2+
Al3+ Fe3+
(d) Choose one of the cations above and write the net precipitation reaction that occurs. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
Cu2+(aq) + (CO3)2-(aq) → CuCO3(s)
Your answer contains an improperly or incompletely formatted chemical formula. Your answer contains improper superscript or subscript
Chemistry
1 answer:
frez [133]3 years ago
8 0

Answer:

Explanation:

a ) silver chloride AgCl and lead chloride PbCl₂  Hg₂Cl₂ ,  are not dissolved in ware

so Ag⁺ and Pb⁺² Hg₂ ⁺² will precipitate out .

b ) BaSO₄ , SrSO₄  are insoluble

Ba⁺² and Sr⁺² will precipitate out .

c )

Al₂( CO₃)₃ ,  Fe₂(CO₃)₃  , CuCO₃ insoluble .

Al⁺³ and Fe⁺³ and Cu⁺² will be insoluble .

d )  

2Fe⁺³(aq) + 3CO₃⁻²(aq) = Fe₂(CO₃)₃ (s)

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mote1985 [20]

Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+ H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).

pKb(C₆H₅NH₂) = 9.40.

Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.

c₀(C₆H₅NH₂) = 0.45 M.

c(C₆H₅NH₃⁺) = c(OH⁻) = x.

c(C₆H₅NH₂) = 0.45 M - x.

Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).

4·10⁻¹⁰ = x² /  (0.45 M - x). 

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7 0
3 years ago
What is the density of CHCL3 vapor at 1.00atm and 298K?
Advocard [28]

Answer:

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

Explanation:

By ideal gas equation:

PV=nRT

Number of moles (n)

can be written as: n=\frac{m}{M}

where, m = given mass

M = molar mass

PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT

where,

\frac{m}{V}=d which is known as density of the gas

The relation becomes:

PM=dRT    .....(1)

We are given:

M = molar mass of chloroform= 119.5 g/mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the gas = 298K

P = pressure of the gas = 1.00 atm

Putting values in equation 1, we get:

1.00atm\times 119.5g/mol=d\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\d=4.88g/L

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

7 0
3 years ago
What can you infer from the fact that nitrogen gas is useless to animals?
makvit [3.9K]

Answer:

Denitrification changes nitrogen into a gaseous form.

Explanation:

3 0
2 years ago
The charge on a magnesium ion is +2. When it formed an ionic compound with another element, it lost 2... what?
Naya [18.7K]

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6 0
2 years ago
Determine whether a precipitate will form when 0.96g of Na2CO3 is combined with 0.20g BaBr2 in a 10 L solution.
Mashcka [7]

Answer:

Qsp > Ksp, BaCO3 will precipitate

Explanation:

The equation of the reaction is;

Na2CO3 + BaBr2 -------> 2NaBr + BaCO3

Since BaCO3 may form a precipitate we can determine the Qsp of the system.

Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles

concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M

Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles

concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M

Hence;

[Ba^2+] = 6.7 * 10^-5 M

[CO3^2-] = 9.1 * 10^-4 M

Qsp = [6.7 * 10^-5] [9.1 * 10^-4]

Qsp = 6.1 * 10^-8

But, Ksp for BaCO3 is 5.1*10^-9.

Since Qsp > Ksp, BaCO3 will precipitate

3 0
3 years ago
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