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Sveta_85 [38]
3 years ago
5

Calculate the average atomic mas of an unknown element that has two naturally-occurring isotopes.

Chemistry
2 answers:
professor190 [17]3 years ago
8 0
The average atomic mass if the element above is calculated by the sum of the product of the isotope abundance and its atomic mass unit. It is expressed as:

Average atomic mass = Σ xi(Mi)
<span>Average atomic mass = (.7547 x 248.7) + (.2453 x 249.4) = 248.87
</span>
Hope this helps.
jonny [76]3 years ago
7 0

Answer:

248.9 uma

Explanation:

We calculate the average atomic mass using the following formula

m= \frac{(m_1.\%_1)+(m_2.\%_2)}{100} \\

This formula depends on how many isotopes the atom has.

If the atom had three isotopes another term would be added to the formula

m1= 248.7 uma\\\%_1= 75.47\%\\m_2= 249.4 uma\\\%_2= 24.53\%

m= \frac{(248.7)(75.47)+(249.4)(24.53)}{100}\\ m=248.9 uma

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Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
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Answer:

\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}

Explanation:

1. Density from mass and volume

\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}

2. Volume from density and mass

V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}

3. Mass from density and volume

\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                =<u> 12.8 mL</u>

Volume of object               = 11.8 mL

\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

 

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