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sveta [45]
3 years ago
7

The density of sodium is 968 g/cm3. Suppose you need 250 gm of sodium for an experiment. Which volume of sodium do you need? a 2

,420 cm3 b 0.26 cm3 c 3.87 cm3
Chemistry
1 answer:
Flauer [41]3 years ago
5 0
Density = 968 g/cm³

mass = 250 g

Volume  = ?

Therefore:

D = m / V

968 = 250 / V

V = 250 / 968 => 0.26 cm³  

Answer C

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Molar mass = 305.42 g/mol

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Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine
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<u>Answer:</u> The percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 75.77% and 24.23% respectively.

<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)

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'

  • <u>For _{17}^{35}\textrm{Cl} isotope:</u>

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

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

  • <u>For _{17}^{37}\textrm{Cl} isotope:</u>

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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3 years ago
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