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liq [111]
2 years ago
12

A certain element consists of two stable isotopes. The first has atomic mass of 7.02 amu and a percent natural abundance of 92.6

%. The second has an atomic mass of 6.02 amu and a percent natural abundance of 7.42%. What is the atomic weight of the element? in amu
Chemistry
1 answer:
statuscvo [17]2 years ago
8 0

<u>Answer:</u> The average atomic mass of the element is 6.95 amu

<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)

We are given:

Mass of isotope 1 = 7.02 amu

Percentage abundance of isotope 1 = 92.6 %

Fractional abundance of isotope 1 = 0.926

Mass of isotope 2 = 6.02 amu

Percentage abundance of isotope 2 = 7.42 %

Fractional abundance of isotope 2 = 0.0742

Putting values in equation 1, we get:

\text{Average atomic mass of element}=[(7.02\times 0.926)+(6.02\times 0.0742)]

\text{Average atomic mass of element}=6.95amu

Hence, the average atomic mass of the element is 6.95 amu

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2 years ago
A gas takes up a volume of 35 L, and has a pressure of 4.8 atm. What is the new pressure of
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3 years ago
What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua
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Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Na_2S

Given mass = 15.5 g

Molar mass of Na_2S = 78.0452 g/mol

<u>Moles of Na_2S = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>

Given: For CuSO_4

Given mass = 12.1 g

Molar mass of CuSO_4 = 159.609 g/mol

<u>Moles of CuSO_4 = 12.1 g / 159.609 g/mol = 0.0758 moles</u>

According to the given reaction:

Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS

1 mole of Na_2S react with 1 mole of CuSO_4

0.1986 mole of Na_2S react with 0.1986 mole of CuSO_4

Available moles of CuSO_4 = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, CuSO_4 is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of CuSO_4 gives 1 mole of CuS

0.0758 mole of CuSO_4 gives 0.0758 mole of CuS

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

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% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

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For Row 1 according to the respective Column is

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For Row 2 according to the respective Column is

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For Row 3 according to the respective Column is

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For Row 4 according to the respective Column is

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For Row 5 according to the respective Column is

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From the question we are told

Ions, isotope, average atomic mass fill in the blank

Generally

Where

  • Z signifies the Proton Number of the atom
  • A signifies Mass Number of the atom

Therefore

For Row 1 according to the respective Column is

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For Row 2 according to the respective Column is

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For Row 4 according to the respective Column is

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For Row 5 according to the respective Column is

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