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Kryger [21]
4 years ago
9

The volume of solution in a test tube is measured to be 0.0067 liter. Which digits in the measurement are significant figures?

Chemistry
2 answers:
puteri [66]4 years ago
8 0

<u>Given data:</u>

Volume of solution measured = 0.0067 L

<u>To determine:</u>

The number of significant figures

<u>Explanation:</u>

Significant figures are the number of digits that accurately describe a measured value.

As per the rules:

All non-zero digits are significant

Leading zero's i.e. the zero's after a decimal which comes before a non-zero digit are not significant

Hence in the given value 0.0067, there are 2 significant figures (6 &7)


frozen [14]4 years ago
6 0

Answer:

Digits in the measurement are significant figures are '6' and '7'.

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for 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: 604, 6.49 and 65.4 all have three significant figures.

All zero’s between integers are always significant. For example: 5005, 4.805 and 50.06 all have four significant figures.

All zero’s preceding the first integers are never significant. For example: 0.0058 has two significant figures.

All zero’s after the decimal point are always significant. For example: 4.500, 47.00 and 550.0 all have four significant figures.

Significant figures in measurement 0.0067 liter = 2

Digits in the measurement are significant figures are '6' and '7'.

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Answer:

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Explanation:

<u>SOLUTION :-</u>

Balance it by using 'hit & trial' method , and you'll get the answer :-

<u>2</u>H₂S + <u>1</u>O₂ → <u>2</u>H₂O + <u>2</u>S

<u></u>

<u>VERIFICATION :-</u>

<em>In reactant side of equation :-</em>

  • Number of atoms in H = 2×2 = 4
  • Number of atoms in S = 2×1 = 2
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<em>In product side of equation :-</em>

  • Number of atoms in H = 2×2 = 4
  • Number of atoms in O = 2×1 = 2
  • Number of atoms in S = 2×1 = 2

Number of atoms of each element is equal in both reactant & product side of equation. Hence , the equation is balanced.

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Answer:

The mass of oxygen in 45.0 g sample of calcium carbonate is approximately 21.58 g

Explanation:

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The molar mass of CaCO₃ = 100.0869 g/mol

The mass of oxygen in one mole of CaCO₃ = 3 × 15.999 g = 47.997 g

The mass of oxygen in 45.0 g sample of CaCO₃, <em>m</em>, is goven as follows;

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