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Dafna1 [17]
3 years ago
5

How many atoms are in 3.4 moles of iron

Chemistry
1 answer:
nikitadnepr [17]3 years ago
4 0
<h3>Answer:</h3>

2.047 × 10^24 atoms

<h3>Explanation:</h3>
  • We need to know that 1 mole of an element contains atoms equivalent to the Avogadro's constant.
  • The Avogadro's constant is 6.022 × 10^23.
  • Therefore, 1 mole of Fe contains 6.022 × 10^23 atoms.

In this case we are given 3.4 moles of Iron.

we are required to find the number of atoms in 3.4 moles.

Therefore, since 1 mole of Fe contains 6.022 × 10^23 atoms.

Then, 3.4 mole will contain;

= 3.4 moles × 6.022 × 10^23 atoms/mole

= 2.047 × 10^24 atoms

Thus, 3.4 moles of Iron contains 2.047 × 10^24 atoms

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gregori [183]

The atomic mass of an element is the average relative mass of its atoms as compared with an atom of carbon-12 whose mass is taken as 12. In terms of percentages of different isotopes, the average atomic mass can be determined as follows: A_{average} = ∑(p_{i}X A_{i})/100. Where, p_{i} is the percentage abundance of isotope with atomic number A_{i}.

So, average atomic mass of neon = \frac{(90.51 X 19.99) + (0.27 X 20.99) + (9.22 X 21.99)}{100} = 20.177 amu.

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3 years ago
Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

Therefore,  \lambda = 1 / 6580.62 = 0.000152 m

(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

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2 years ago
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Nana76 [90]

Wavelength of the light is 2.9 × 10⁻⁷ m.

<u>Explanation:</u>

Planck - Einstein equation shows the relationship between the energy of a photon and its frequency, and they are directly proportional to each other and  it is given by the equation as E = hν,

where E is the energy of the photon

h is the Planck's constant = 6.626 × 10⁻³⁴ J s

ν is the frequency

From the above equation, we can find the frequency by rearranging the equation as,

ν = $ \frac{E}{h} = $ \frac{6.8 \times 10^{-19}}{6.626\times10^{-34}} = 1.03\times10^{15} s^{-1}

Now the frequency and the wavelength are in inverse relationship with each other.

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It can be rearranged to get λ as,

λ = c / ν

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So wavelength is 2.9 × 10⁻⁷ m.

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4 years ago
The surface area of a star can be estimated based on the stars
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