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

How does Dalton's atomic theory account for the fact that when \(1.000 \mathrm{~g}\) of water is decomposed into its elements, \(0.111 \mathrm{~g}\) of hydrogen and \(0.889 \mathrm{~g}\) of oxygen are obtained regardless of the source of the water?

Short Answer

Expert verified
Dalton's atomic theory explains the decomposition of 1.000 g of water into 0.111 g of hydrogen and 0.889 g of oxygen, regardless of its source, through mass conservation and consistent mass ratios of elements. The decomposition reaction conserves mass, as the sum of hydrogen and oxygen produced equals the initial mass of water: \(0.111g_{H_{2}} + 0.889g_{O_{2}} = 1.000g_{H_{2}O}\). Additionally, the mass ratio of hydrogen to oxygen (\(\approx 0.125\)) remains the same for any source of water, further supporting Dalton's atomic theory.

Step by step solution

01

Understand the problem

We are given that when 1.000 g of water is decomposed, 0.111 g of hydrogen and 0.889 g of oxygen are obtained. We need to show that this outcome is consistent with Dalton's atomic theory, which states that the mass of the reactants should equal the mass of the products formed.
02

Mass conservation during the decomposition

If we decompose 1.000 g of water into its elements, the sum of the masses of hydrogen and oxygen produced should equal the mass of water that we started with, according to Dalton's theory. So, we have: Mass of hydrogen + Mass of oxygen = Mass of water \(0.111g_{H_{2}} + 0.889g_{O_{2}} = 1.000g_{H_{2}O}\) Thus, the mass is conserved during the decomposition of water, which is supported by Dalton's atomic theory.
03

Relative proportions of hydrogen and oxygen in water

Another aspect to consider is how the ratio of masses of hydrogen and oxygen remains the same, regardless of the source of the water. Let's calculate the mass ratio of hydrogen to oxygen: Ratio = (Mass of hydrogen) / (Mass of oxygen) \(= \frac{0.111g_{H_{2}}}{0.889g_{O_{2}}}\) \(≈ 0.125\) The mass ratio of hydrogen to oxygen is approximately 0.125. This holds true for any source of water, meaning that these proportions are consistent, which also supports Dalton's atomic theory. In conclusion, Dalton's atomic theory accounts for the fact that when 1.000 g of water is decomposed into its elements, 0.111 g of hydrogen and 0.889 g of oxygen are obtained regardless of the source of the water through mass conservation during reactions and consistent mass ratios of elements.

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!

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

Predict the chemical formula for the ionic compound formed by (a) \(\mathrm{Ca}^{2+}\) and \(\mathrm{Br}^{-}\) (b) \(\mathrm{K}^{+}\) and \(\mathrm{CO}_{3}^{2-}\), (c) \(\mathrm{Al}^{3+}\) and \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), (d) \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{SO}_{4}^{2-}\), (e) \(\mathrm{Mg}^{2+}\) and \(\mathrm{PO}_{4}{ }^{3-}\)

(a) What isotope is used as the standard in establishing the atomic mass scale? (b) The atomic weight of boron is reported as \(10.81\), yet no atom of boron has the mass of \(10.81\) amu. Explain

Write the empirical formula corresponding to each of the following molecular formulas: (a) \(\mathrm{Al}_{2} \mathrm{Br}_{6}\) (b) \(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{10}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\) (d) \(\mathrm{P}_{4} \mathrm{O}_{10}\) (e) \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\) (f) \(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}\).

Give the name or chemical formula, as appropriate, for each of the following binary molecular substances: (a) \(\mathrm{SF}_{6},(\mathrm{~b}) \mathrm{IF}_{5},(\mathrm{c}) \mathrm{XeO}_{3}\), (d) dinitrogen tetroxide, \((\mathrm{e}) \mathrm{hy}\) - drogen cyanide, (f) tetraphosphorus hexasulfide.

Use Coulomb's law, \(F=k Q_{1} Q_{2} / d^{2}\), to calculate the electric force on an electron \(\left(Q=-1.6 \times 10^{-19} \mathrm{C}\right)\) exerted by a single proton if the particles are \(0.53 \times 10^{-10} \mathrm{~m}\) apart. The constant \(k\) in Coulomb's law is \(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\). (The unit abbreviated \(\mathrm{N}\) is the Newton, the SI unit of force.)

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