Answer: The organic compound used to create fragrances in perfume industry are ester.
Explanation:
Organic compounds are defined as the compounds in which a covalent bond is present between a hydrogen and a carbon atom. These are also known as hydrocarbons.
Esters are an organic compound which have a pleasant and fruity smell and is used in the perfume industry.
Therefore, the organic compound used to create fragrances in perfume industry are ester.
Orbitals are the regions where the possibility of finding the electrons are the most. -2, -1, 0, 1, 2, are the possible values of ml for an electron in a d orbital.
The ml will be 2 for the d -orbitals and is defined by the principal quantum number or the magneticquantum number. It states the orientation of the orbital and ranges from – l ≤ ml ≤ l.
As, ml for d orbitals = 2, then, -2, -1, 0, 1, 2 are the possible values that ranges from -l to +l.
Therefore, -2, -1, 0, 1, 2 are the values of ml.
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Specialized cells are generally found only in multicellular organisms.
A multicellular organism can be described as an organism that contains more than one cell, in contrast to a unicellular organism. Multicellularity has been independently at least 25 times in eukaryotes and in some prokaryotes, like myxobacteria, cyanobacteria, and actinomycetes.
All species of animals, plants, and most fungi are multicellular organisms, whereas a few organisms are partially uni- and partially multicellular, such as slime molds and social amoebae.
Multicellular organisms develop in many ways such as by cell division or by aggregation of many single cells. Colonial organisms can be defined as identical individuals joining together to form a colony.
Unicellular organisms are divided, and the daughter cells failed to separate which results in a conglomeration of identical cells in one organism, which could develop specialized tissues.
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Answer:
Specialized cells are found only in multicellular organisms.
Explanation:
Or organisms are made up of more than one cell.
(2) compound because the atoms of the elements are combined in a proportion that varies
(3) mixture because the atoms of the elements are combined in a fixed proportion
(4) mixture because the atoms of the elements are combined in a proportion that varies
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.
Answer:the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.
Explanation:The given chemical reaction is:
Pb^4+(aq) + 2 Ce^3+(aq) -> Pb^2+(aq) + 2 Ce^4+(aq)
The standard cell potential (E°cell) for this electrochemical reaction is 0.06 V.
The standard cell potential for a galvanic cell can be calculated using the Nernst equation:
E°cell = E°cathode - E°anode
In this reaction, Pb^4+(aq) is being reduced to Pb^2+(aq), so it is the reduction half-reaction, and Ce^3+(aq) is being oxidized to Ce^4+(aq), so it is the oxidation half-reaction.
The standard reduction potentials (E°) for the half-reactions are as follows:
For the reduction half-reaction:
Pb^4+(aq) + 2 e^- -> Pb^2+(aq) E°red = x (we'll solve for x)
For the oxidation half-reaction:
2 Ce^3+(aq) -> 2 Ce^4+(aq) + 2 e^- E°red = 1.44 V (This value is usually given)
Now, plug these values into the Nernst equation:
E°cell = E°cathode - E°anode
0.06 V = x - 1.44 V
Now, solve for x:
x = 0.06 V + 1.44 V
x = 1.50 V
So, the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.
The answer is bimass power plant.