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Troyanec [42]
2 years ago
9

What is the formula for the precipitate when solutions of manganese (II) nitrate and

Chemistry
1 answer:
natita [175]2 years ago
6 0

The precipitate formed when a solution of manganese(II) nitrate and

potassium sulfide react is Manganese sulfide (MnS)

To obtain the precipitate formed when a solution of manganese(II) nitrate and potassium sulfide react, we shall write the balanced ionic equation. This is illustrated below:

Manganese (II) nitrate => Mn(NO₃)₂

Potassium sulfide => K₂S

In solution, they will react as follow:

Mn(NO₃)₂ (aq) —> Mn²⁺(aq) + 2NO₃¯(aq)

K₂S(aq) —> 2K⁺(aq) + S²¯(aq)

Mn(NO₃)₂ (aq)  + K₂S(aq) —>  

Mn²⁺(aq) + 2NO₃¯(aq) + 2K⁺(aq) + S²¯(aq) —> MnS(s) + 2K⁺(aq) + 2NO₃¯(aq)

Cancel out the spectator ions (i.e K⁺ and NO₃¯)

Mn²⁺(aq) + S²¯(aq) —> MnS(s)

From the net ionic equation above, we can see that MnS is insoluble.

Therefore, the precipitate formed when a solution of manganese(II) nitrate and potassium sulfide react is Manganese sulfide (MnS)

Learn more: brainly.com/question/21280827

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Answer:

The amount of solute added.

Explanation:

The amount of solute added is directly proportional to the number of ions.

The higher the amount added the higher the number of moles.

The number of moles is multiplied by the Avogadro's constant to get the number ions.

No of ions= No of moles × L

L is the Avogadro's number.

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<h3>Answer:</h3>

2 L Ne

<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
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.07 mol Ne (g)

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

STP - 22.4 L per mole

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 0.07 \ mol \ Ne(\frac{22.4 \ L \ Ne}{1 \ mol \ Ne})
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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

1.568 L Ne ≈ 2 L Ne

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