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
sasho [114]
3 years ago
10

Which shows the correct order of experiments from oldest to most recent

Chemistry
1 answer:
IceJOKER [234]3 years ago
8 0
A bc it’s right to your welcome
You might be interested in
Carbonic acid reacts with water to yield bicarbonate ions and hydronium ions: h2co3+h2o?hco3?+h3o+ identify the conjugate acid-b
Levart [38]
Conjugate base pairs are acid and bases having common features. These features are the equal gain or loss of protons of the pairs. Conjugate pairs should always be one base and one acid. One would not exist without the other. Conjugate acids are the substances that gains protons while conjugates bases are those that loses protons. <span>The substances in the equilibrium reaction that is given is identified as follows: 
HCO3^-      +     H2O <----->   CO3^2-          +             H3O^+ 
 acid                   base         conjugate base           conjugate acid 

HCO3^- ion is an intermediate molecule of CO2 and CO3^2-. When we add OH- to HCO3^-, we produce CO3^2-. And when we add H+ to HCO3, we produce CO2. </span>
7 0
3 years ago
Which event most likely occurs at point V?<br><br> cooling<br> erosion<br> heating<br> melting
marta [7]

Answer: cooling

Explanation:

5 0
3 years ago
Read 2 more answers
2. Hydrogen gas at a temperature of 22.0°C that is confined in a 5.00L cylinder exerts a pressure of 4.20atm. If the gas is rele
Umnica [9.8K]

Answer: n∗R=22+273.15/4.2∗5n

P2=n∗R∗T2/V2=n∗R∗33.6+273.15/10

Explanation:

4 0
3 years ago
Why does brining a turkey make it juicier?
olya-2409 [2.1K]

Answer:

hope this answer will help you.

6 0
3 years ago
Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and
blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g;     Hydrogen H = 2.24 g     Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon

For Hydrogen:

=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen

For Oxygen:

=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon

For Hydrogen:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen

For Oxygen:

\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen

Thus, the empirical formula is HCO₂

4 0
3 years ago
Other questions:
  • As an environmental engineer how would you apply the process of separating mixtures to clean up a polluted beach? What would you
    13·1 answer
  • The tilt of the Earth on its axis does not change? <br> Agree or Disagree?
    9·1 answer
  • A small amount of a solid is added to water. The observation made after fifteen minutes is shown in the figure. Which of these s
    12·1 answer
  • 13. Which of the following has the highest amount of kinetic energy?
    13·1 answer
  • What is the molarity of a potassium triiodide solution, KI3(aq), if 30.00 mL of the solution is required to completely react wit
    12·1 answer
  • Who was the first person to propose that atoms existed?
    13·2 answers
  • PLEASE HELP
    5·1 answer
  • Pls help me on this I have 5 minutes left to submit
    5·2 answers
  • True or False:
    13·1 answer
  • take two carbon atoms and join by two springs making a double bond. complete the ethene moleeule by adding two hydrogen atoms to
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!