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3 years ago

All digits shown on measuring device, plus one estimated digit, are considered_______.

Physics

1 answer:

vichka [17]3 years ago

3 0

Significant

Explanation:

All the digits shown on a measuring device and one estimated digit are all considered to be significant.

A significant digit is a set of values that shows how precise a measurement is reported.

Measuring devices such as calculators gives their values in significant digits.

Non- zero digits in a measurement are always significant.
Zero's before a decimal are not significant. Those after a number in a decimal are significant.
Zero's between digits are significant.

learn more:

Significant digits brainly.com/question/2743055

#learnwithBrainly

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3 years ago

The boiling point of water is 91.30 °C on the

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Answer:

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Explanation:

To convert from degrees celsius to degrees fahrenheit, use this equation:

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3 years ago

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lidiya [134]

Answer:

1.4m/s

Explanation:

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3 years ago

An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99

sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes 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:

For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882

Putting values in equation 1, we get:

$\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]$

$\text{Average atomic mass}=20.169amu$

Hence, the average atomic mass of the given element is 20.169 amu.

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