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ycow [4]
1 year ago
14

The mass of a single strontium atom is 1.46×10-22 grams. How many strontium atoms would there be in a sample of strontium with a

volume of 178 cm3 if the density of strontium is 2.60 g/cm3?
Chemistry
1 answer:
Keith_Richards [23]1 year ago
7 0

The number of atoms that would be in a sample of strontium with a volume of 178 cm³ is 3.17×10²⁴ atoms

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to determine the mass of strontium </h3>
  • Volume of strontium = 178 cm³
  • Density of strontium = 2.60 g/cm³
  • Mass of strontium =?

The mass of strontium can be obtained as illustrated below:

Density = mass / volume

Cross multiply

Mass = Density × Volume

Mass of strontium = 2.60 × 178

Mass of strontium = 462.8 g

<h3>How to determine the number of atoms</h3>

The number of atoms in 462.8 g sample of strontium can be obtained as follow:

1.46×10⁻²² g of strontium = 1 atom

Therefore,

462.8 g of strontium = (462.8 g × 1 atom) / 1.46×10⁻²² g

462.8 g of strontium = 3.17×10²⁴ atoms

Thus, 3.17×10²⁴ atoms is present in the sample.

Learn more about density:

brainly.com/question/952755

Learn more about Avogadro's number:

brainly.com/question/26141731

#SPJ1

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Living things are made mostly of what four main elements
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Which statement describes a property of covalent compounds?
Marat540 [252]

They have a low boiling point because of weak intermolecular forces. They

Explanation:

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3 0
3 years ago
Antimony has 2 stable isotopes 121-sb mass 120.9038 and 123-sb mass 122.9042 amu. calculate the percent abundances of these isot
iragen [17]
The atomic mass of an element is a result of the weighted average of the masses of its corresponding isotopes. 

<span>We are given: </span>
<span>Sb-121: mass of 120.9038 amu, x abundance </span>
<span>Sb-123: mass of 122.9042 amu, 1-x abundance </span>

<span>To be able to calculate the atomic mass of antimony, we multiply the percent abundances of the isotopes by their respective atomic masses. Then, we add,</span>

<span>Atomic mass = (Atomic Mass of Sb-121) (% Abundance) + (Atomic Mass of Sb-123) (% Abundance )</span>

121.760 amu  = (120.9038 amu)(x) + (122.9042 amu)(1 -x)

<span>Solving for x,</span>
<span>120.9038x + 122.9042 -122.9042x = 121.760</span>
<span>x = 0.57096 or 57.096% </span>
1-x = 1-0.57096 = 0.42904 or 42.90% 
6 0
4 years ago
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