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yuradex [85]
3 years ago
13

A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

ionic equation for this reaction by filling in the blanks. The phase symbols and charges on species are already provided.
Chemistry
1 answer:
Margarita [4]3 years ago
4 0

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

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Answer:

Ionic size increases from top to bottom within the group.

Explanation:

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Explanation:

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The pathway by which carbon is transferred from living organisms to the atmosphere is called:
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According to the Arrhenius concept, which of the following substances is NOT a base in aqueous solutionsA. NH3B. LiOHC. NaOHD. H
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Explanation:

Arrhenius theory explained that the acids are the ones that have H⁺, either H in its formula. Following this, the bases are the ones that have OH⁻ , either OH and its formula.

It can be used only with compounds with H, or OH.

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5 0
3 years ago
The mole fraction of iodine, i2, dissolved in dichloromethane, ch2cl2, is 0.115. what is the molal concentration, m, of iodine i
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The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:

0.115 mol I₂

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We need moles of solute, which we have, and must convert our moles of solvent to kg:

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We can now calculate the molality:

m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂

The molality of the iodine solution is 1.53.
5 0
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