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zhenek [66]
3 years ago
13

The average atomic mass of nitrogen is 14.0067. The atomic masses of the two stable isotopes of nitrogen.^14 N and^15 N, are 14.

003074002 and 15.00010897 amu, respectively. Use this information to determine the percent abundance of^14 N. % Determine the percent abundance of^15 N. %
Chemistry
1 answer:
Ymorist [56]3 years ago
6 0

Answer:

The answer to your question is Abundance N-14 = 99.63%

                                                    Abundance N-15 = 0.37%

Explanation:

Data

Average atomic mass = 14.0067 amu

N-14 = 14.003074002

N-15 = 15.00010897

Abundance N-14 = ?   x

Abundance N-15 = ?   1 - x

Formula

Average atomic mass = (mass of N-14)x + (mass of N-15)(1 - x)

Substitution

14.0067 = 14.003074002x + 15.00010897(1 - x)

Solve for x

14.0067 = 14.003074002x + 15.00010897 - 15.00010897x

14.0067 - 15.00010897 = 14.003074002x - 15.00010897x

-0.9934 = -0.99703x

x = -0.9934 / -0.99703

x = 0.9963

Conclusion

Abundance N-14 = 0.9963 x 100 = 99.63 %

Abundance N-15 = 1 - 0.9963 = 0.0037 x 100 = 0.37 %

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Represent the decomposition of aluminum oxide using the same number of atoms of molecules as are
pantera1 [17]

Answer:

Decomposition of aluminium oxide forms  aluminium atoms and  oxygen atoms.

Explanation:

<u>Decomposition reaction:</u>

When a single compound break down into two or more simpler products.

For example "AB" reactant undergoes decomposition to form "A" and "B" products.

The chemical reaction is as follows.

AB\rightarrow A+B

The given compound is aluminium oxide.

The decomposition reaction of aluminium oxide is a follows.

Al_{2}O_{3}\rightarrow Al+O_{2}

The balanced equation is as follows.

2Al_{2}O_{3}\rightarrow 4Al+3O_{2}

Therefore, Decomposition of aluminium oxide forms aluminium atoms and  oxygen atoms.

4 0
3 years ago
Write 2,469,100 in Engineering Notation with 3 significant figures.
Gala2k [10]

Answer: 2.47\times 10^{6}

Explanation:

Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power.  It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.

The engineering notation written in the form:

a\times 10^b

where,

a = the number which is greater than 0 and less than 999

b = an integer multiple of 3

Now converting the given value of 2,469,100 into engineering notation, we get 2.47\times 10^{6}

Hence, the correct answer is, 2.47\times 10^{6}

6 0
2 years ago
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
Lady_Fox [76]

Answer:

The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)

Explanation:

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

An oxidizing element or oxidizing agent is one that reaches a stable energy state as a result of which the oxidant is reduced and gains electrons. The oxidizing agent causes oxidation of the reducing agent generating the loss of electrons of the substance and, therefore, oxidizes in the process.

In other words, the oxidizing agent is that chemical species that in a redox process accepts electrons released by the reducing agent and, therefore, is reduced in said process. The oxidizing agent is reduced because, upon receiving electrons from the reducing agent, a decrease in the value of the charge or oxidation number of one of the atoms of the oxidizing agent is induced .

Electrochemical cells, galvanic cells or batteries are called devices that are capable of transforming chemical energy originated in a spontaneous redox process into electrical energy.

The cellular potential is generally in standard conditions, that is, 1 M with respect to solute concentrations in solution and 1 atm for gases.

In this case you have the reaction:

3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq)

In this case the following half-reactions occur:

Semi-reaction of oxidation ( an atom or group of atoms loses electrons, or increases its positive charges): Fe³⁺(aq) + 3 e- -->Fe(s); E⁰ = -0.04 V

Semi-reaction of reduction (an atom or group of atoms gains electrons, increasing its negative charges): Cl₂(g) + 2 e- --> 2 Cl-(aq); E⁰=1.36 V

In an electrochemical cell at 25°C  the potentials of the  semi-reactions are usually measured  in the sense of reduction  and generally the standard potential between both electrochemical cells will be:

E^{0} =E^{0} _{reduction} -E^{0} _{oxidation}

E⁰=1.36 V - (-0.04 V)

E⁰=1.36 V + 0.04 V

<em>E⁰=1.40 V</em>

<em><u>The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)</u></em>

7 0
3 years ago
A student wants to dissolve the maximum amount of CaF2 (Ksp = 3.2 x 10^−11) to make 1 L of aqueous solution.
Mekhanik [1.2K]

Answer:

The answer to A would be iv) 0.01 M HCL, but I'm not sure what the answer to part B is. Hope this helps.

Explanation:

7 0
3 years ago
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Mice21 [21]
C plants only have a cell wall, however animal cells have a cell wall and a cell membrane
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