5.6 x 10^3 cm = __

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

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:

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

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

Lithium arsenate (Li3AsO4) and iron nitrate (Fe(NO3)3) are dissolved in water. What precipitate would you expect to form? what i

Alla [95]

Answer:

CO₃²⁻ + H₂O <----------------> HCO₃⁻ + OH⁻

The chemical formula of the precipitate is Fe(OH)₃

Explanation:

Fe(NO₃)₃ and K₂CO₃ are strong electrolytes and completely dissociate in water. Carbonate ions is a weak base and combine with water to form hydroxide ions (OH⁻), CO₃²⁻ + H₂O <----------------> HCO₃⁻ + OH⁻
Ferric, Fe (III), combines with these hydroxide ions to form insoluble precipitates. Fe(OH)₃ is only partially soluble i.e., it does not completely dissociate in water. When the solutions of Fe(NO₃)₃ and K₂CO₃ are mixed, Fe(OH)₃ precipitates out due to the strong electrostatic attraction between Fe (III) and hydroxide ions.

8 0

3 years ago

One hundred million copper atoms are arranged in a line. Which is a reasonable estimate for the length of this line?

Anettt [7]

The size of an atom is roughly 10^-10 metres

10^-10 • 100,000,000 = 0.01 metres

Since 1 metre = 100 centimetre

0.01 x 100 = 1 centimetre

There’s your answer

8 0

3 years ago

determine the frequency and wavelength (in nm) of the light emitted when the e- fell from n=4 and n=2

Lostsunrise [7]

Answer:

Frequency = 6.16 ×10¹⁴ Hz

λ = 4.87×10² nm

Explanation:

In case of hydrogen atom energy associated with nth state is,

En = -13.6/n²

For n = 2

E₂ = -13.6 / 2²

E₂ = -13.6/4

E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

E₄ = -13.6 / 4²

E₄ = -13.6/16

E₄ = -0.85 ev

Kinetic energy of electron = -E₄ = 0.85 ev

Wavelength of radiation emitted:

E = hc/λ = E₄ - E₂

hc/λ = E₄ - E₂

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev - (-3.4ev )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J

λ = 4.87×10⁻⁷ m

m to nm:

4.87×10⁻⁷ m ×10⁹nm/1 m

4.87×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /4.87×10⁻⁷ m

Frequency = 6.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 6.16 ×10¹⁴ Hz

3 0

3 years ago

Cooking an egg is an example of what type of change

Katen [24]

Answer: chemical change

7 0

2 years ago

5.6 x 10^3 cm = ___cg