Answer:
9-10 ppm.
0.2-0.4 ppm.
Explanation:
The proton on the aldehyde group will appear at approximately 9-10 ppm whereas the methylene peak on the alcohol is the only peak 0.2-0.4 ppm for either compound. Aldehydes and aromatics are quite distinctive in the Nuclear magnetic resonance (NMR). Aldehydes show up from 9-10 ppm, usually as a small singlet; aromatic protons show up from 6.5-8.5 ppm. NMR spectroscopy is the use of NMR to study the physical, chemical, and biological properties of matter.
Answer:
The plant and animal cells are eukaryotic and contain well developed cellular organelles.
The cell membrane, cytoplasm, chromosomes, and mitochondria are the structures that are present in both the plant and the animal cells.
The cell wall and chloroplast are present only in the plant cell.
Answer:
- 2K(s) + (1/8) S₈ (s) + (3/2) O₂(g) → K₂SO₃ (s)
Explanation:
The<em> standard enthalpy of formation </em>of a substance is the change in enthalpy that happens when one mole of the substance is formed from the elements in their standard states.
Thus, to calculate the standard state of formation of a compound you must:
- 1. Identify the elements that form the compound
- 2. Identify the standard form of each element
- 3. Set the equation to form one mole of the compound, which may require to use fractional coefficients for some of the elements.
Applying that to our compound K₂SO₃
<u>1. Elements:</u>
- potassium, K;
- sulfur, S; and
- oxygen, O.
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<u>2. Standard forms of the elements:</u>
- potassium: solid, K(s)
- sulfur: solid, octatomic molecules, S₈ (s)
- oxygen: diatomic gas, O₂(g)
<u>3. Reaction:</u>
- K(s) + S₈ (s) + O₂(g) → K₂SO₃ (s)
Balance, keeping one mole of K₂SO₃. You will need to use fractional coefficients for some elements:
- 2K(s) + (1/8) S₈ (s) + (3/2) O₂(g) → K₂SO₃ (s) ← answer
Lichens are an examples of Symbiosis
