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yuradex [85]
2 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]2 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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5 plates is the highest amount that can be served

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2 years ago
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Small beads of iridium-192 are sealed in a plastic tube and inserted through a needle into breast tumors. If an Ir-192 sample ha
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Answer:

296.1 day.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).

t is the time of the reaction (t = ??? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).

<em>∴ kt = lna/(a-x)</em>

(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).

(9.365 x 10⁻³ day⁻¹)(t) = 2.773.

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5 0
2 years ago
Calculate the number of chloride ions in 6.8 g of zinc chloride
Bumek [7]

Answer:

The correct answer is 0.300 * 10^23 ions.

Explanation:

Based on the given question, there is a need to find the number of chloride ions in the mentioned 6.8 grams of zinc chloride compound.  

The moles of zinc chloride (ZnCl2) is,  

= mass of zinc + 2 mass of chlorine

= 65.38 + 2 (35.45)

=65.38 + 70.90

= 136.28 grams (The molecular mass of zinc is 65.38 and the molecular mass of chlorine is 35.45)

Thus, 136.28 g of ZnCl2 contains 70.90 grams of chlorine

Therefore, 6.8 grams of ZnCl2 will comprise = (70.90/136.28) * 6.8

= 3.537 g of chlorine

70.90 g of Cl comprise 6.022*10^23 chlorine, thus, 3.537 g of Cl will comprise (6.022*10^23/70.90) * 3.537

= 0.300 * 10^23 ions of chlorine.  

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3 0
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Write a balanced equation for the transmutation that occurs when a scandium-48 nucleus undergoes beta decay.
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Answer:

A scandium-48 nucleus undergoes beta-minus decay to produce a titanium-48 nucleus.

\rm ^{48}_{21}Sc \to ^{48}_{22}Ti + ^{\phantom{1}\,0}_{-1}e^{-} + \bar{\mathnormal{v}}_e.

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In both decay modes, the mass number of the nucleus stays the same.

However, in a beta-minus decay, the atomic number of the nucleus increases by one. In a beta-plus decay, the atomic number decreases by one.

Each beta-minus decay releases one electron and one electron antineutrino. Each beta-plus decay releases one positron and one electron neutrino.

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This question did not specify whether the decay here is beta-plus or a beta-minus. However, the relative atomic mass of this element can give a rough estimate of the mode of decay.

Each element (e.g, \rm Sc) can have multiple isotopes. These isotopes differ in mass. The relative atomic mass of an element is an average  across all isotopes of this element. This mass is weighted based on the relative abundance of the isotopes. Its value should be closest to the most stable (and hence the most abundant) isotope.

The mass number of scandium-48 is significantly larger than the relative atomic mass of this element. In other words, this isotope contains more neutrons than isotopes that are more stable. There's a tendency for that neutron to convert to a proton- by beta-minus decay, for example.

The atomic number of the nucleus will increase by 1. 21 + 1 = 22. That corresponds to titanium. The mass number stays the same at 48. Hence the daughter nucleus would be titanium-48. Note that two other particles: one electron and one electron \rm e^{-} and one antineutrino \bar{v}_{\text{e}} (note the bar.) The neutrino helps balance the lepton number of this reaction.

6 0
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