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TEA [102]
2 years ago
11

How many moles of KF would need to be added to 2500 ml of water to make 1.2 M solution?

Chemistry
1 answer:
Oduvanchick [21]2 years ago
4 0

The number of moles of KF needed to prepare the solution is 3 moles

<h3>What is molarity?</h3>

Molarity is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

<h3>How to determine the mole of KF </h3>
  • Volume = 2500 mL = 2500 / 1000 = 2.5 L
  • Molarity = 1.2 M
  • Mole of KF =?

Molarity = mole / Volume

1.2 = mole of KF / 2.5

Cross multiply

Mole of KF = 1.2 × 2.5

Mole of KF = 3 moles

Learn more about molarity:

brainly.com/question/9468209

#SPJ1

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Considering the following precipitation reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Which ion would NOT be present in
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Answer:

The question is incomplete and confusing.

  • In the complete ionic equation you write all the ions that are formed. Those are: Pb²⁺, NO₃⁻, K⁺, and I⁻. They all are present in the complete ionic equation.

  • In the net ionic equation, the spectator ions do not appear. They are: NO₃⁻ and K⁺. They would not be present in the net ionic equation, but they do in the complete ionic equation.

See below the details.

Explanation:

Which compound will not form ions?

<u />

<u>1. Write the balanced molecular equation:</u>

  • Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

<u />

<u>2. Write the ionizations for the ionic aqueous compounds:</u>

<u />

  • Pb(NO₃)₂(aq) →  Pb⁺²(aq) + 2NO₃⁻(aq)

  • 2KI(aq) → 2K⁺(aq) + 2I⁻(aq)

  • 2KNO₃(aq) → 2K⁺(aq) + 2NO₃⁻(aq)

<u />

<u>3. Write the complete ionic equation:</u>

Pb⁺²(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) +  2K⁺(aq) + 2NO₃⁻(aq)

Hence, since PbI₂(s) does not ionize, but stays in solid form, it will not form ions.

All, Pb⁺², NO₃⁻, K⁺, and I⁻ will be present in the total ionic equation.

It is in the net ionic equation that the spectator ions are removed. Those, are NO₃⁻ and K⁺, because they are on both sides of the complete ionic equation.

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