Which gas exerts the greatest pressure? (A) He (B) Ne (C) NO (D) All gases exert the same amount of pressure.

Short Answer

Expert verified
(D) All gases exert the same amount of pressure, according to the ideal gas law.

Step by step solution

01

Understand the nature of gases

First, it's essential to remember that, according to the ideal gas law (\(PV=nRT\)), the pressure exerted by a gas depends on the temperature, volume, and number of gas molecules present and not the nature or type of the gas.
02

Apply the gas law to all the given options

Applying the ideal gas law to all the options, we find that regardless of whether the gas is He, Ne, or NO, under the same conditions (the same volume, temperature, and number of particles), they all exert the same pressure.

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

Silver sulfate, \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) , has a solubility product constant of \(1.0 \times 10^{-5} .\) The below diagram shows the products of a precipitation reaction in which some silver sulfate was formed. (Diagram Can't Copy) Which ion concentrations below would have led the precipitate to form? (A) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (B) \(\left[\mathrm{Ag}^{+}\right]=0.10 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (C) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.10 M\) (D) This is impossible to determine without knowing the total volume of the solution.

Use the following information to answer questions 14-16 The radius of atoms and ions is typically measured in Angstroms \((A),\) which is equivalent to \(1 * 10^{-10} \mathrm{m} .\) Below is a table of information for three different elements. TABLE NOT AVAILABLE Neon has a smaller atomic radius than phosphorus because: (A) Unlike neon, phosphorus has electrons present in its third energy level. (B) Phosphorus has more protons than neon, which increases the repulsive forces in the atom. (C) The electrons in a neon atom are all found in a single energy level. (D) Phosphorus can form anions, while neon is unable to form any ions.

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ If the \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}(s)\) half-reaction has a standard reduction potential of \(+0.52 \mathrm{V},\) what is the standard reduction potential for the \(\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}(s)\) half-reaction? (A) \(+0.76 \mathrm{V}\) (B) \(-0.76 \mathrm{V}\) (C) \(+0.24 \mathrm{V}\) (D) \(-0.24 \mathrm{V}\)

\(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightarrow 2 \mathrm{HF}(g)\) Gaseous hydrogen and fluorine combine in the reaction above to form hydrogen fluride with an enthalpy change of \(-540 \mathrm{kJ}\) . What is the value of the heat of formation of \(\mathrm{HF}(g) ?\) (A) \(-1,080 \mathrm{kJ} / \mathrm{mol}\) (B) \(-270 \mathrm{kJ} / \mathrm{mol}\) (C) 270 \(\mathrm{kJ} / \mathrm{mol}\) (D) 540 \(\mathrm{kJ} / \mathrm{mol}\)

A solution of \(\mathrm{Co}^{2+}\) ions appears red when viewed under white light. Which of the following statements is true about the solution? (A) A spectrophotometer set to the wavelength of red light would read a high absorbance. (B) If the solution is diluted, the amount of light reflected by the solution will decrease. (C) All light with a frequency that is lower than that of red light will be absorbed by it. (D) Electronic transmissions within the solution match the wavelength of red light.

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