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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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A compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what
Charra [1.4K]
1) mass composition

N: 30.45%
O: 69.55%
   -----------
   100.00%

2) molar composition

Divide each element by its atomic mass

N: 30.45 / 14.00 = 2.175 mol

O: 69.55 / 16.00 = 4.346875

4) Find the smallest molar proportion

Divide both by the smaller number

N: 2.175 / 2.175 = 1

O: 4.346875 / 2.175 = 1.999 = 2

5) Empirical formula: NO2

6) mass of the empirical formula

14.00 + 2 * 16.00 = 46.00 g

7) Find the number of moles of the gas using the equation pV = nRT

=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)

=> n = 0.01769 moles

8) Find molar mass

molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol

9) Find how many times the mass of the empirical formula is contained in the molar mass

92.14 / 46.00 = 2.00

10) Multiply the subscripts of the empirical formula by the number found in the previous step

=> N2O4

Answer: N2O4
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3 years ago
Can anybody help me with these 2 ohms law
Fynjy0 [20]

Answer:

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Almost 99% of earths atmosphere is made up of two gases. What are the two gases and the percents of each?
Natali5045456 [20]
The Answer Is A 21% oxygen and 78%nitrogen
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A solution contains an unknown amount of dissolved magnesium. Addition of
Scrat [10]

Taking into account the reaction stoichiometry, 2.13 grams of magnesium was dissolved in the solution.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Mg²⁺: 1 mole
  • Na₂CO₃: 1  mole
  • MgCO₃: 1 mole
  • Na⁺: 2 moles

The molar mass of the compounds is:

  • Mg²⁺: 24.3 g/mole
  • Na₂CO₃: 106 g/mole
  • MgCO₃: 84.3 g/mole
  • Na⁺: 23 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Mg²⁺: 1 mole ×24.3 g/mole= 24.3 grams
  • Na₂CO₃: 1 mole ×106 g/mole= 106 grams
  • MgCO₃: 1 mole ×84.3 g/mole=84.3 grams
  • Na⁺: 2 moles ×23 g/mole= 46 grams

<h3>Mass of magnesium dissolved</h3>

The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?

mass of magnesium=\frac{0.0877 moles of Na_{2}C O_{3}x24.3 grams of magnesium }{1 mole of Na_{2}C O_{3}}

<u><em>mass of magnesium= 2.13 grams</em></u>

Finally, 2.13 grams of magnesium was dissolved in the solution.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

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Answer:

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