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Dmitry_Shevchenko [17]
3 years ago
11

Describe a method to calculate the average atomic mass of the sample in the previous question using only the atomic masses of li

thium -6 and lithium -7 without using the simulation.
Chemistry
1 answer:
alexira [117]3 years ago
3 0

Answer:

Explanation:

To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.

The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.

This can be expressed below:

        RAM = Σmₙαₙ

where mₙ is the mass of isotope n

           αₙ is the abundance of isotope n

for this problem:

RAM of Li = m₆α₆ + m₇α₇

       m₆ is mass of isotope Li-6

        α₆ is the abundance of isotope Li-6

       m₇ is mass of isotope Li-7

        α₇ is the abundance of isotope Li-7

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Answer:

Its a physical

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Explanation:

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Why is there so little evidence of the Period of Heavy Bombardment on Earth?
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C. Weathering, plate movements and volcanoes destroyed many of

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Explanation:

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6 0
3 years ago
Read 2 more answers
How many mL of a 0.193 M aqueous solution of ammonium bromide, , must be taken to obtain 8.42 grams of the salt?
PilotLPTM [1.2K]

Answer:

445 mL NH₄Br

Explanation:

(Step 1)

Convert grams ammonium bromide (NH₄Br) to moles using its molar mass.

Molar Mass (NH₄Br): 14.007 g/mol + 4(1.008 g/mol) + 79.904 g/mol

Molar Mass (NH₄Br): 97.943 g/mol

8.42 grams NH₄Br                1 mole
-----------------------------  x  ------------------------  =  0.0860 moles NH₄Br
                                         97.943 grams

(Step 2)

Calculate the volume of NH₄Br using the molarity equation. Then, convert liters (L) to milliliters (mL).

Molarity (M) = moles / volume (L)

0.193 M = 0.0860 moles / volume

(0.193 M) x volume = 0.0860 moles

volume = 0.445 L

1,000 mL = 1 L

0.445 L NH₄Br           1,000 mL
------------------------  x  ------------------  =  445 mL NH₄Br
                                        1 L

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