One of the first labs to analyze oxygen consumption was

Answers

Answer 1
Answer:

Answer:

One of the first labs to analyze oxygen consumption was established by Lavoisier and Laplace in the late 18th century.

Explanation:

Answer 2
Answer:

Final answer:

Initial laboratories studying oxygen consumption used innovative techniques, such as isotope tracers, and advanced equipment, such as respirometers, to examine various organisms and processes. Renowned scientist Mildred Cohn played a crucial role in this early research, providing insights into metabolic pathways and enzyme mechanisms.

Explanation:

The first labs to investigate oxygen consumption studied a variety of concepts and organisms. Pioneering this study, Mildred Cohn made significant discoveries utilizing isotopes as tracers in reactions. This strategy provided invaluable insights into metabolic pathways and enzyme mechanisms. Specifically, this strategy was used to understand the phosphorylation of glucose in mitochondria.

The respirometer is an essential tool in measuring oxygen consumption. In an experiment conducted in a laboratory, mice were used to observe the respiration levels at different temperatures. This was done by tracking the volume of oxygen each mouse consumed over multiple five-minute trials.

Also, oxygen consumption in humans can be combined with our understanding of our digestive process. The energy humans and other organisms need to function normally is generated through the gradual oxidation of chemical compounds in the body. Oxygen acts as the ultimate oxidizing agent in these reactions, linking to our digestion and energy conversion processes.

Learn more about Oxygen Consumption here:

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single replacement
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Answers

Single replacement.

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The melting and freezing points of a substance are the same temperature.a. True
b. False

Answers

That's false. A melting point is hot/warm, while the freezing point is cold.
False! The melting point and freezing points would differ based on their properties. 

What is the predicted order of first ionization energies from highest to lowest for lithium (Li), sodium (Na), potassium (K), and rubidium (Rb)?

Answers

Answer;

From highest to lowest is Li>Na>K>Rb.

Explanation;

- Ionization energy is the energy that is required to dislodge or remove an electron from the outer most energy level or energy shell. The first ionization energy is the energy required to remove the first electrons from the outermost energy shell.

- Lithium will have the highest first ionization energy, as the outer electron is closer to the pull of the nucleus and not shielded by full shells. This means that rubidium will be the lowest.


Answer;

From highest to lowest is Li>Na>K>Rb.

Explanation;

- Ionization energy is the energy that is required to dislodge or remove an electron from the outer most energy level or energy shell. The first ionization energy is the energy required to remove the first electrons from the outermost energy shell.

- Lithium will have the highest first ionization energy, as the outer electron is closer to the pull of the nucleus and not shielded by full shells. This means that rubidium will be the lowest.


The first ionization energy is the energy that the atom lost its first electrons. The energy decrease and the atom is more reactive. So from highest to lowest is Li>Na>K>Rb.

Metals experience plastic deformation when _____.

Answers

Metals experience plastic deformation when a specific amount of pressure and temperature is applied to them. Most metals have low specific heat. Because of their low specific heat, they easily get hot when heat is applied to them.

26. All atoms area. positively charged, with the number of protons exceeding the number of electronsb, negatively charged, with the number of electrons exceeding the number of protonswith the number of protons equaling the number of electronsd. neutral, with the number of protons equaling the number of electrons. which is equal to thenumber of neutrons27. The particles that are found in the nucleus of an atom arec. protons and neutronsa, neutrons and electronsd. protons and electronsb. electrons only28. As a consequence of the discovery of the nucleus by Rutherford, which model of the atom is thought to betrue?a. Protons. electrons, and neutrons are evenly distributed throughout the volume of the atom.b. The nucleus is made of protons, electrons, and neutrons.c. Electrons are distributed around the nucleus and occupy almost all the volume of the atom.d. The nucleus is made of electrons and protons.29. The nucleus of an atom isa. the central core and is composed of protons and neutronsb. positively charged and has more protons than neutronsc. negatively charged and has a high densityd. negatively charged and has a low density

Answers

(26) All atoms area...with the number of protons equaling the number of electrons

(27) 
The particles that are found in the nucleus of an atom are... protons and neutrons.

(28) 
As a consequence of the discovery of the nucleus by Rutherford, which model of the atom is thought to be true?...Protons. electrons, and neutrons are evenly distributed throughout the volume of the atom.

(29) 
The nucleus of an atom is...the central core and is composed of protons and neutrons

A transformer has a primary voltage of 115 V and a secondary voltage of 24 V. If the number of turns in the primary is 345, how many turns are in the secondary? A. 1,653
B. 690
C. 72
D. 8

Answers

There are numerous information's of immense importance already given in the question.
Primary voltage of the transformer (V1) = 115 V
Secondary voltage of the transformer (V2) = 24 V
Number of turns in the primary (T1) = 345
Let us assume the number of turns in the secondary = T2
Then
(V1/V2) = (T1/T2)
(115/24) = 345/T2
115T2 = 345 * 24
115 T2 = 8280
T2 = 8280/115
     = 72
So the correct option among all the options given in the question is option "C".

Answer:

The is 1653

Explanation:

Primary Voltage = # of turns in primary

Secondary Voltage  # of turns in secondary

115 V = 345

24 V     # of turns in secondary

115\25=4.79167

4.79167 = 345

               # of turns in secondary

4.79167 * 345 = 1653.12

When rounded 1653