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

What ­if you discovered Avogadro’s number was not 6×1023at a rate of one number each second? Determine your hourly wage. Would you do it? Could you do it?

Short Answer

Expert verified

As the mass of carbon- 12 thenthe hourly wage is 6 g.

Step by step solution

01

Definition of mole

A mole is a sample of a natural element with a mass equal to the element's atomic mass expressed in grams and contains 1 mole of atoms. This number of atoms is called Avogadro's number.

02

Calculation of Avogadro’s number

If the Avogadro's numberis 3.01×1023, mass of 3.01×1023atoms of carbon-12 (atomic mass of 12 u) willbe 12g.

3.01×1023atoms12uatoms=12g3.01×1023u=1g

As carbon 12 is taken as standard value for reference, the relative masses of all the elements also remain the same.So, relative masses given in periodic table remain same.

03

Calculation of Avogadro’s number

If the Avogadro’s number is 3.01×1023, it is not considered on the basic on the mole, mass of 3.01×1023atoms of carbon-12 (atomic mass of 12 u) can be calculated as shown below:

6.02×1023atoms12uatom=12gmassof3.01×1023atoms=12g6.02×1023atoms3.01×1023

massof3.01×1023atoms=6g

04

Conclusion

As the mass of carbon- 12 is 6g,the hourly wage is 6 g.

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

Aspartame is an artificial sweetener that is 160 times sweeter than sucrose (table sugar) when dissolved in water. It is marketed as NutraSweet. The molecular formula of aspartame is C14H18N2O5

a. Calculate the molar mass of aspartame.

b. How many moles of molecules are in 10.0 g of aspartame?

c. What is the mass in grams of 1.56 moles of aspartame?

d. How many molecules are in 5.0 mg of aspartame?

e. How many atoms of nitrogen are in 1.2 g of aspartame?

f. What is the mass in grams of 1.0 x 109 molecules of aspartame?

g. What is the mass in grams of one molecule of aspartame?

You may have noticed that water sometimes drips from an exhaust pipe of a car as it is running.Is this evidence that there is at least a small amount of water originally present in the gasoline?Explain.

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction:

2C6H5Cl+C2HOCl3C14H9Cl5+H2O

In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral.
a. What mass of DDT is formed, assuming 100% yield?
b. Which reactant is limiting? Which is in excess?
c. What mass of the excess reactant is left over?
d. If the actual yield of DDT is 200.0 g, what is the percent yield?

The molecular formula of acetylsalicylic acid (aspirin), one of the most commonly used pain relievers, is C9H8O4

a. Calculate the molar mass of aspirin.

b. A typical aspirin tablet contains 500. mgC9H8O4. What amount (moles) ofC9H8O4molecules and what number of molecules of acetylsalicylic acid are in a 500. mg tablet?

The reaction between potassium chlorate and red phosphorus takes place when you strike amatch on a matchbox. If you were to react52.9gof potassium chlorate(KClO3)with excess redphosphorus, what mass of tetra phosphorus decoxiderole="math" localid="1648562536382" (P4O10)could be produced?

role="math" localid="1648562558748" KClO3(s)+P4(s)→P4O10(s)+KCl(s) (Unbalanced).

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