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Yanka
9 days ago
7

Sodium only has one naturally occuring isotope, 23 Na , with a relative atomic mass of 22.9898 u . A synthetic, radioactive isot

ope of sodium, 22 Na , is used in positron emission tomography. 22 Na has a relative atomic mass of 21.9944 u. A 1.8385 g sample of sodium containing a mixture of 23 Na and 22 Na has an apparent "atomic mass" of 22.9573 u .Find the mass of 22 Na contained in this sample?
Chemistry
1 answer:
KiRa [971]9 days ago
6 0

Answer:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

The total mass of the isotope mixture is 1.8385g.

It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.

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8 days ago
A student checks the air in her bicycle tires early in the morning when it is cool outside. If she measures it again later in th
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Answer:

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Explanation:

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10 days ago
In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a
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The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

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To determine the enthalpy change, the following equation applies

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To calculate the heat (Q):

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Next, we need to calculate the number of moles involved in the reaction (n):

n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles

With these two values, we can substitute them into the first equation:

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5 days ago
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Question 1.

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The correct count is 5. The outer layer of electrons is termed the valence shell, containing what we refer to as valence electrons. Here, the outermost shell corresponds to n=3, meaning all electrons in this layer are classified as valence electrons. While we do have subshells (3s and 3p), they indicate the spatial region where an electron may be located but do not clarify valence.

Question 3.

Chlorine possesses 7 valence electrons, Neon has 8 valence electrons, Phosphorus contains 5 valence electrons, Sulfur has 6 valence electrons. The position of an element in the periodic table aids in determining its electron configuration, and the group number provides insights into the number of valence electrons. By merely checking the periodic table's group number, we can ascertain the valence electrons.

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1 day ago
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4) Balance the following redox reaction in an acidic solution. What are the coefficients in front of H⁺ and Fe3+ in the balanced
eduard [944]

Answer:

- The coefficients in front of H⁺ and Fe³⁺ are 8 and 5 respectively.

- A total of 5 moles of electrons are exchanged.

Explanation:

This reaction is represented as:

Fe²⁺(aq) + MnO₄⁻(aq) → Fe³⁺(aq) + Mn²⁺(aq)

Analyzing the oxidation states:

Fe²⁺ transitions to Fe³⁺

This indicates an increase in oxidation state → OXIDATION

Meanwhile, Mn in MnO₄⁻ starts with +7 and transforms into Mn²⁺

This suggests a decrease in oxidation state → REDUCTION

Let's formulate the half reactions:

Fe²⁺ → Fe³⁺  +  1e⁻    (it loses 1 mole of electrons)

MnO₄⁻ + 5e⁻ →  Mn²⁺  (it gains 5 moles of electrons)

Next, we will balance the oxygen atoms. In an acidic environment, water is added to balance the oxygens on the opposite side. Since there are 4 oxygens on the reactant side, we add 4 H₂O to the product side.

MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Now, to balance the hydrogen atoms, we have 8 hydrogens in the products, necessitating the inclusion of 8H⁺ in the reactants, yielding the complete half-reaction:

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Notably, there's 1e⁻ in the oxidation and 5e⁻ in the reduction. To cancel electrons, we must multiply the oxidation half-reaction by 5.

(Fe²⁺ → Fe³⁺  +  1e⁻) x 5

5Fe²⁺ → 5Fe³⁺  +  5e⁻  

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

By adding both half reactions, we have:

5Fe²⁺  + 8H⁺  + MnO₄⁻ + 5e⁻ →  5Fe³⁺  +  5e⁻   + Mn²⁺  + 4H₂O

The electrons cancel out, resulting in the balanced equation:

5Fe²⁺  + 8H⁺  + MnO₄⁻  →  5Fe³⁺  + Mn²⁺  + 4H₂O

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10 days ago
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