Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}, 23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.

Short Answer

Expert verified
The samples ranked in order of increasing number of atoms are 0.50 mol H_2O (\(3.011 \times 10^{23} \: \text{atoms}\)), 23 g Na (\(6.022 \times 10^{23} \: \text{atoms}\)), and \(6.0 \times 10^{23}\) N_2 molecules (\(1.2 \times 10^{24} \: \text{atoms}\)).

Step by step solution

01

Determine the molar masses

Consult the periodic table to find the atomic weights of each element involved. Keep in mind that the molecular weight is the sum of the atomic weights of all the atoms in a molecule. - Molar mass of H_2O: \(2 \times 1.01 (\text{H}) + 16.00 (\text{O}) \approx 18.02 \:\text{g/mol}\) - Molar mass of Na: \(22.99 \:\text{g/mol}\) - Molar mass of N_2: \(2 \times 14.01 (\text{N}) \approx 28.02 \:\text{g/mol}\)
02

Convert quantities to atoms

Use the molar mass of each sample and Avogadro's number (\(6.022 \times 10^{23} \: \text{particles/mol}\)) to convert the given quantities into the number of atoms present: - Number of atoms in 0.50 mol H_2O: \(0.50 \: \text{mol} \times 6.022 \times 10^{23} \: \text{atoms/mol} = 3.011 \times 10^{23} \: \text{atoms}\) - Number of atoms in 23 g Na: \(\frac{23 \: g}{22.99 \: g/mol} \times 6.022 \times 10^{23} \: \text{atoms/mol} = 6.022 \times 10^{23} \: \text{atoms}\) - Number of atoms in \(6 \times 10^{23} \: \text{N}_2} \: \text{molecules} : (2 atoms per molecule) \(\times 6.0 \times 10^{23} \: \text{molecules} = 1.2 \times 10^{24} \: \text{atoms}\)
03

Rank samples in increasing number of atoms

Now that we have the number of atoms for each sample, rank them in order of increasing number of atoms: 1. 0.50 mol H_2O: \(3.011 \times 10^{23} \: \text{atoms}\) 2. 23 g Na: \(6.022 \times 10^{23} \: \text{atoms}\) 3. \(6.0 \times 10^{23} \: \text{N}_2 \: \text{molecules} : 1.2 \times 10^{24} \: \text{atoms}\) In conclusion, the samples ranked in order of increasing number of atoms are 0.50 mol H_2O, 23 g Na, and \(6.0 \times 10^{23}\) N_2 molecules.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
When dealing with chemical substances, it's vital to understand the concept of molar mass. The molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of the atoms that compose the molecule. This makes it easier to convert between the mass of a substance and the number of moles.
For example, water (H_2O) has a molar mass determined by adding up the atomic masses of its composite atoms: two hydrogen atoms and one oxygen atom. Using a periodic table:
  • Hydrogen (H) has an atomic mass of approximately 1.01 g/mol.
  • Oxygen (O) has an atomic mass of about 16.00 g/mol.
Thus, the molar mass of water is \(2 \times 1.01 + 16.00 \approx 18.02\ g/mol\).
Understanding molar mass helps chemical calculations and conversions from grams to moles, essential for understanding chemical reactions and stoichiometry.
Avogadro's Number
Avogadro's number, denoted as \(6.022 \times 10^{23}\), is a fundamental constant in chemistry. It represents the number of atoms, ions, or molecules in one mole of a substance.
This concept connects the macroscopic world (measurable quantities like grams or liters) with the microscopic world (atoms and molecules). For instance, if you have one mole of water, you literally have \(6.022 \times 10^{23}\) molecules of H_2O.
Here are some practical applications of Avogadro's number:
  • Determining the number of particles in a given sample, facilitating the conversion from moles to individual particles.
  • Understanding the scale of chemical reactions by relating mole quantities to actual particle counts.
Leveraging Avogadro's number is crucial for chemists to understand composition and reaction partakers at a molecular scale, simplifying calculations involving enormous particle quantities.
Atomic Weights
Atomic weights act as the stepping stone for various chemical calculations, serving as a crucial property of elements. The atomic weight of an element is an average mass of an atom, taking into account the natural abundance of isotopes. It is generally expressed in atomic mass units (amu) but often used in grams per mole in molar mass calculations.
For example, sodium ( Na) has an atomic weight of approximately 22.99 amu or g/mol. This reflects the isotopic composition of sodium as found in nature.
  • Atomic weights allow us to calculate molar masses of compounds, aiding in the quantification of substances in stoichiometric equations.
  • By consulting peripheral tables, atomic weights provide invaluable data required to calculate other chemical properties and facilitate conversions between moles and grams.
Atomic weights are indispensable in chemical analysis, allowing chemists to comprehend and predict the behavior of elements in reactions and compounds accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which reagent is the limiting reactant when \(1.85 \mathrm{~mol}\) \(\mathrm{NaOH}\) and \(1.00 \mathrm{~mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

Very small crystals composed of 1000 to 100,000 atoms, called quantum dots, are being investigated for use in electronic devices. (a) A quantum dot was made of solid silicon in the shape of a sphere, with a diameter of \(4 \mathrm{~nm}\). Calculate the mass of the quantum dot, using the density of silicon \(\left(2.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) (b) How many silicon atoms are in the quantum dot? (c) The density of germanium is \(5.325 \mathrm{~g} / \mathrm{cm}^{3} .\) If you made a \(4 \mathrm{~nm}\) quantum dot of germanium, how many Ge atoms would it contain? Assume the dot is spherical.

Balance the following equations: (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)\) (c) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)\) (d) \(\mathrm{Al}_{4} \mathrm{C}_{3}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{4}(g)\) (e) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (f) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (g) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) \(\mathrm{MgSO}_{4}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)\)

(a) What scientific principle or law is used in the process of balancing chemical equations? (b) In balancing equations, why should you not change subscripts in chemical formulas? (c) How would one write out liquid water, water vapor, aqueous sodium chloride, and solid sodium chloride in chemical equations?

(a) Define the terms limiting reactant and excess reactant. (b) Why are the amounts of products formed in a reaction determined only by the amount of the limiting reactant? (c) Why should you base your choice of what compound is the limiting reactant on its number of initial moles, not on its initial mass in grams?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free