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aniked [119]
3 years ago
7

1. The element copper has naturally occurring isotopes with mass numbers of 63 and 65. The relative abundance and atomic masses

are 69.2% for a mass of 62.93amu and 30.8% for a mass of 64.93amu. Calculate the average atomic mass of copper.
Chemistry
1 answer:
Basile [38]3 years ago
8 0

<u>Answer:</u> The average atomic mass of copper is 63.546 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i  .....(1)

  • <u>For isotope 1 (Cu-63):</u>

Mass of isotope 1 = 62.93 amu

Percentage abundance of isotope 1 = 69.2 %

Fractional abundance of isotope 1 = 0.692

  • <u>For isotope 2 (Cu-65):</u>

Mass of isotope 2 = 64.93 amu

Percentage abundance of isotope 2 = 30.8 %

Fractional abundance of isotope 2 = 0.308

Putting values in equation 1, we get:

\text{Average atomic mass of Copper}=[(62.93\times 0.692)+(64.93\times 0.308)]\\\\\text{Average atomic mass of Copper}=63.546amu

Hence, the average atomic mass of copper is 63.546 amu

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What type of bond would occur between carbon (C) and nitrogen (N)
FrozenT [24]

Answer:

covalent

Explanation:

The carbon and the nitrogen very often form bonds in nature, carbon-nitrogen bonds, which are covalent types of bonds. In fact, the bonds between the carbon and nitrogen are one of the most abundant in the biochemistry and the organic chemistry. The bonds between these two can be double bonds, as well as triple bonds. The carbon-nitrogen bonds have the tendency to be strongly polarized toward the nitrogen.

8 0
3 years ago
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands
Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

E=\frac{R_y}{n^2}

R_y=2.18\times 10^{-18} J =  Rydberg energy

n =  principal quantum number of the orbital

Energy of 11th orbit = E_{11}

E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J

Energy of 10th orbit = E_{10}

E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J

=E'=0.38\times 10^{-20} J

\lambda =\frac{hc}{E'} (Planck's' equation)

\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}

\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

3 0
2 years ago
What is the answer please tell me
Nastasia [14]

Answer:

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Explanation:

5 0
3 years ago
A scientist wants to determine which fertilizer is more effective—Fertilizer X or Fertilizer Y. The best way for her to proceed
Alecsey [184]

Answer:

The correct answer is "three groups of plants—a group fertilized by X, a group fertilized by Y, and a control group with no fertilizer".

Explanation:

I had to look for the problem to know the options.

The best way to determine which fertilizer is most effective is to have three evaluation groups. One group will be tested with fertilizer X and another with fertilizer Y. By leaving the third group without applying fertilizer, this will be the general pattern for comparing the effectiveness of the other two.

Have a nice day!

4 0
3 years ago
Which one of these compounds is an ionic compound? question 18 options: n2 na2o co2?
spayn [35]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Covalent

                

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                    E.N of Nitrogen      =   3.04

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For Na₂O,

                    E.N of Oxygen       =   3.44

                    E.N of Sodium       =   0.93

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For CO₂,

                    E.N of Oxygen       =   3.44

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4 0
3 years ago
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