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mamaluj [8]
3 years ago
13

You know that an unlabeled bottle contains an aqueous solution of one of the following: AgNO3, KCl, or (NH4)2SO4. A friend sugge

sts that you test a portion of the solution with Ba(NO3)2 and then with NaCl solutions. Explain how these two tests together would be sufficient to determine which salt is present in the solution.
Chemistry
1 answer:
Afina-wow [57]3 years ago
6 0

Answer:

These tests determine the solubility of the compounds formed upon adding the test solution

Explanation:

Addition of Ba(NO₃)₂ will cause a precipitate ((Ba)₂SO₄) to form in the solution of (NH₄)₂SO₄. No precipitates will form in the other unknown solutions. Thus, whether or not the solution is ammonium sulfate can be determined.

Addition of NaCl solution will cause a precipitate (AgCl) to form in the solution of AgNO₃. No precipitates will form in the other unknown solutions. Thus, whether or not the solution is silver nitrate can be determined.

If no precipitates form, then the unknown solution must be KCl.

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How many moles are in 8.63 x 103 atoms of Li?
Ira Lisetskai [31]
<h3>Answer:</h3>

1.43 × 10⁻²⁰ mol Li

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

8.63 × 10³ atoms Li

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 8.63 \cdot 10^3 \ atoms \ Li(\frac{1 \ mol \ Li}{6.022 \cdot 10^{23} \ atoms \ Li})
  2. Multiply/Divide:                \displaystyle 1.43355 \cdot 10^{-20} \ moles \ Li

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li

4 0
3 years ago
If 2.0 mol of Zn is mixed with 3.0 mol<br> of HCl, which reactant will be limiting?
Soloha48 [4]

The following Balanced Reaction will take place:

Zn + 2HCl → ZnCl₂ + H₂

In the question, we have 2 moles of Zinc and 3 moles of HCl for this reaction

<u>Amount of HCl required to completely react with 2 moles of Zn:</u>

Since we need 2 moles of HCl for every mole of Zn, we will need 2(2) = 4 moles of HCl for every 2 moles of Zn

<u>Identifying the Limiting Reagent:</u>

But we are only given 3 moles of HCl where we need 4 moles to completely react.

So, since HCl is in less amount, it is the Limiting Reagent

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user100 [1]
Iodine has an electronegativity of 2.5, and potassium has an electronegativity of 0.8, so the difference is:
2.5 - 0.8 = 1.7
7 0
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