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Ilia_Sergeevich [38]
3 years ago
6

The mass spectrum of a hypothetical element shows that 78.21% of the atoms have a mass of 27.977 u, 1.346% have a mass of 28.976

u, and the remaining have a mass of 29.974 u. Calculate the average atomic mass of this element (in units of u). (Use 4 Sig. Fig. Don't round until end.)
Chemistry
1 answer:
Crank3 years ago
4 0

Answer: 28.40

Explanation:

For A(27.977 u)

Mass number = 27.977

Abundance = 78.21%

For B (28.976 u)

Mass number = 28.976

Abundance = 1.346%

For C(29.974 u)

Mass number = 29.974

Abundance = 20.444%

Averages atomic mass

= (Mass of A x abundance of A)/100 + (Mass of B x abundance of B)/100 + (Mass of C x abundance of C)/100

= (27.977x78.21)/100 + (28.976x1.346)/100 + (29.974x20.444)/100

= 21.8808117 + 0.39001696 + 6.12788456

= 28.40

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Perform the calculation and record the answer with the correct number of significant figures.
kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

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<u>For A:</u> The answer becomes 0.1

<u>For B:</u> The answer becomes 0.4884

<u>Explanation:</u>

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

  • Digits from 1 to 9 are always significant and have infinite number of significant figures.
  • All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
  • All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
  • All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
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<u>Rule applied for addition and subtraction:</u>

The least precise number present after the decimal point determines the number of significant figures in the answer.

<u>Rule applied for multiplication and division:</u>

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

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This a a problem of subtraction and division.

First, the subtraction is carried out.

\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}

Here, the least precise number after decimal was 1.

\Rightarrow \frac{0.4}{3.19}=0.125

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

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This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

\Rightarrow \frac{43.72}{89.516}=0.48840

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

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