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3 years ago

How many atoms are in 5.64 moles of iron

1 answer:

7 0

Hello devin ...The atomic mass of iron (Fe) is 55.85 amu. The atomic mass of radium (Ra) is 226 amu. One mole of atoms of any element contains 6.022 X 1023 atoms, regardless of the type of element. The mass of one mole of an element depends on what that element is, and is equal to the atom mass of that element in grams....i hope this helps you out

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2 years ago

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2 years ago

What is the freezing point (in degrees Celcius) of 4.14 kg of water if it contains 235.1 g of butanol, C 4 H 9 O H

Karolina [17]

Answer:

Explanation:

Molal freezing point depression constant of butanol Kf = 8.37⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

mol weight of butanol = 70 g

235.1 g of butanol = 235.1 / 70 = 3.3585 moles

3.3585 moles of butanol dissolved in 4.14 kg of water .

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= 6.79⁰C

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freezing point of solution = - 6.79⁰C .

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2 years ago

Part of Dr. Ritchey’s work in her previous job involved boiling and condensing hydrocarbon mixtures. In one of her experiments,

Hitman42 [59]

Answer:

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Explanation:

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3 years ago

Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

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:

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

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

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2 years ago