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4 years ago

What is the mass of 2.47 moles of fluorine ( f. atoms? 0.130 g 7.69 g 46.9 g my answer 282 g

Chemistry

1 answer:

mixer [17]4 years ago

3 0

The mass of 2.47 moles of fluorine atom is 46.9 g

calculation

mass of Fluorine= moles x molar mass

moles = 2.47 moles

molar mass of fluorine = 18.998 g/mol

mass is therefore= 2.47 moles x 18.008 g/mol = 46.9 g

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On another planet, the isotopes of titanium have the given natural abundances. Isotope Abundance Mass (u) 46Ti 70.600% 45.95263

Stells [14]

Answer: The average atomic mass of element titanium is 46.889 amu.

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)

For isotope 1 (Ti-46) :

Mass of isotope 1 = 45.95263 amu

Percentage abundance of isotope 1 = 70.600 %

Fractional abundance of isotope 1 = 0.70600

For isotope 2 (Ti-48) :

Mass of isotope 2 = 47.94795 amu

Percentage abundance of isotope 2 = 11.900 %

Fractional abundance of isotope 2 = 0.11900

For isotope 3 (Ti-50) :

Mass of isotope 3 = 49.94479 amu

Percentage abundance of isotope 3 = 17.500 %

Fractional abundance of isotope 3 = 0.17500

Putting values in equation 1, we get:

$\text{Average atomic mass of titanium}=[(45.95263\times 0.70600)+(47.94795\times 0.11900)+(49.94479\times 0.17500)]$

$\text{Average atomic mass of titanium}=46.889amu$

Hence, the average atomic mass of element titanium is 46.889 amu.

6 0

4 years ago

A chemistry student needs 50.00g of acetic acid for an experiment. He has available 450. g of a 31.9%w solution of acetic acid i

Yuliya22 [10]

Answer:

The student should use 156.7 g of solution

Explanation:

31.9 %w solution of acetic acid in ethanol means 100 g of solution contains 31.9 g of acetic acid.

So, 450. g of solution contains $(\frac{31.9\times 450.}{100})g$ of acetic acid or 143.55 g of acetic acid

Alternatively, 143.55 g of acetic acid is present in 450. g of solution

So, 50.00 g of acetic acid is present in $(\frac{450.\times 50.00}{143.55})g$ solution or 156.7 g of solution

Hence the student should use 156.7 g of solution

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Rina8888 [55]

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Explanation:

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3 years ago

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