1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
soldi70 [24.7K]
1 year ago
12

In each pair, choose the larger of the indicated quantities or state that the samples are equal:

Chemistry
1 answer:
masha68 [24]1 year ago
5 0

<u>0.4 mol of </u>O_3 molecules has larger mass than 0.4 mol of O atoms

<h3>What are molecules?</h3>

The smallest component of a substance that possesses both its chemical and physical characteristics. One or more atoms make up molecules.

They may have the same atoms (for example, an oxygen molecule has two oxygen atoms) or different atoms if they have more than one (a water molecule has two hydrogen atoms and one oxygen atom). Biological molecules like DNA and proteins can include thousands of atoms.

the lowest recognised unit into which a pure material can be divided while retaining its chemical makeup and attributes is made up of two or more atoms.

Learn more about Molecules

brainly.com/question/1078183

#SPJ4

You might be interested in
Only oppositely charged objects can attract each other. true false
Delicious77 [7]
No as neutral object will attract and be attracted by a positive and negative charge 

hope that helps 
4 0
3 years ago
1. Calculate how many moles of glycine are in a 130.0-g sample of glycine.2. Calculate the percent nitrogen by mass in glycine.
Alexxx [7]

Answer:

n=1.732mol

\% N=18.7\%

Explanation:

Hello!

In this case, since the molecular formula of glycine is C₂H₅NO₂, we realize that the molar mass is 75.07 g/mol; thus, the moles in 130.0 g of glycine are:

n=130.0g*\frac{1mol}{75.07 g}\\\\ n=1.732mol

Furthermore, we can notice 75.07 grams of glycine contains 14.01 grams of nitrogen; thus, the percent nitrogen turns out:

\% N=\frac{14.01}{75.07}*100\% \\\\\% N=18.7\%

Best regards!

4 0
3 years ago
A certain first-order reaction 45% complete in 65seconds, determine the rate constant and the half life for the process ​
masha68 [24]

The rate constant : k = 9.2 x 10⁻³ s⁻¹

The half life : t1/2 = 75.3 s

<h3>Further explanation</h3>

Given

Reaction 45% complete in 65 s

Required

The rate constant and the half life

Solution

For first order ln[A]=−kt+ln[A]o

45% complete, 55% remains

A = 0.55

Ao = 1

Input the value :

ln A = -kt + ln Ao

ln 0.55 = -k.65 + ln 1

-0.598=-k.65

k = 9.2 x 10⁻³ s⁻¹

The half life :

t1/2 = (ln 2) / k

t1/2 = 0.693 : 9.2 x 10⁻³

t1/2 = 75.3 s

3 0
3 years ago
How many moles are in 6.80 x 10^23 atoms <br> of gold, Au?
frosja888 [35]

Answer:

1.13 moles Au

Explanation:

Moles Au = 6.80x10²³atoms / 6.023x10²³atoms/mole = 1.13 moles Au

8 0
3 years ago
How would u describe the cells that make up the parts of your eye?​
aleksandrvk [35]

Answer:

optical (optic is the word used for referring to the eye ie optic nerve)

Explanation:

4 0
3 years ago
Other questions:
  • Based on what you have learned about intermolecular forces, would you say that matter is fundamentally attracted or repulsed by
    9·1 answer
  • Which element has the highest melting point? ​
    6·2 answers
  • An organism had 1,000 grams of carbon-14 (a radioactive form of carbon) in it when it died. How much remains after five half-liv
    13·1 answer
  • What is the molar mass of helium? (show your work)
    7·1 answer
  • Which side of the mountain faces
    9·1 answer
  • What is the mass in grams of one mole of any substance know as
    14·1 answer
  • Please help quick im on a test and i only got an few mins left
    15·2 answers
  • 14.A 4.25 gram sample of an unknown gas is found to occupy a volume of 1.70 L at a pressure of 883 mm Hg and a temperature of 58
    15·1 answer
  • Why do you think they stilk proruxe bright flowers to attract bees?
    13·2 answers
  • Tính số nguyên tử oxi
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!