1) What kind of macromolecule is shown here?
(Carbohydrates, Proteins or Lipids)
2) Identify the bond between 1 and 2.
3) Identify the bond between 2 and 3.
Answers
1) The macromolecule shown is a carbohydrate.
2) The bond between 1 and 2 would be a glycosidic bond.
3) The bond between 2 and 3 would also be a glycosidic bond.
Carbohydrates are macromolecules composed of carbon, hydrogen, and oxygen atoms. They are commonly found in foods and serve as a source of energy in living organisms. Carbohydrates are made up of monosaccharide units, which can be linked together through glycosidic bonds to form larger carbohydrate molecules.
The glycosidic bond is a type of covalent bond that forms between the hydroxyl (-OH) groups of two monosaccharide units. It involves the condensation reaction, where a molecule of water is eliminated as the bond forms.
The glycosidic bond plays a crucial role in joining monosaccharide units and creating polysaccharides, such as starch, cellulose, and glycogen.
In the given structure, the bond between 1 and 2 represents a glycosidic bond because it joins two monosaccharide units together. Similarly, the bond between 2 and 3 also represents a glycosidic bond, indicating the linkage between additional monosaccharide units.
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Related Questions
QUESTION 7 What is the pH of water? O pH12 O pH9 O pH7 O pH5 QUESTION 8 What is the pH when fish die from pollution? O pH12 O pH9 O pH7 O pH4 QUESTION 9 A solution with a pH less than 7 is basic. O True O False
Answers
7. The pH of water is pH7.
The pH scale measures the acidity or alkalinity of a substance. It ranges from 0 to 14, with pH7 considered neutral. Water has a pH of 7, indicating that it is neither acidic nor basic. It is important to note that the pH of pure water can vary slightly due to the presence of dissolved gases and minerals, but it generally remains close to pH7.
8. When fish die from pollution, the pH is typically around pH4.
Pollution can introduce harmful substances into water bodies, leading to a decrease in pH. Acidic pollutants, such as sulfur dioxide and nitrogen oxides, can cause the pH of water to drop significantly. When fish are exposed to highly acidic water, their physiological processes are disrupted, and they may die as a result. A pH of around pH4 is considered highly acidic and can be detrimental to aquatic life.
9. A solution with a pH less than 7 is acidic.
This statement is false. A solution with a pH less than 7 is actually considered acidic, not basic. The pH scale ranges from 0 to 14, with pH7 being neutral. Solutions with a pH below 7 are acidic, indicating a higher concentration of hydrogen ions (H+) in the solution. On the other hand, solutions with a pH above 7 are basic or alkaline, indicating a higher concentration of hydroxide ions (OH-) in the solution.
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Select the following terms to describe the relative concentrations of the molecules listed below if TAC cycle is completely inactive: assuming there is no electron shuttle and no other metabolic ways involved. 00 [mitochondrial FADH2] [cytosolic NADH] [pyruvate] [mitochondrial ATP] Acetyl-CoA [mitochondrial ADP] 1. Normal 2. Higher than normal 3. Lower than normal 4. None
Answers
For the given relative concentrations of the molecule we have: option 1, Normal, option 2, Higher than normal, option 3, Lower than normal and option 4, None, is the correct answer.
Given terms are: [mitochondrial FADH2] [cytosolic NADH] [pyruvate] [mitochondrial ATP] Acetyl-CoA [mitochondrial ADP].
The relative concentrations of the molecules listed below if TAC cycle is completely inactive are:
None [mitochondrial FADH2][cytosolic NADH][pyruvate]Higher than normal [mitochondrial ATP]
Lower than normal Acetyl-CoA[mitochondrial ADP]
The TAC cycle is responsible for the production of high energy ATP molecules.
If the TAC cycle is inactive, then there will be no energy generated. Therefore, the concentration of mitochondrial ATP will be None, and the concentration of mitochondrial FADH2 and cytosolic NADH will be higher than normal.
However, without the TAC cycle, the concentration of Acetyl-CoA will be lower than normal and the concentration of mitochondrial ADP will also be lower than normal.
Thus, the relative concentrations of the molecules listed below if the TAC cycle is completely inactive will be: None [mitochondrial FADH2] [cytosolic NADH] [pyruvate]Higher than normal [mitochondrial ATP]
Lower than normal Acetyl-CoA[mitochondrial ADP].
Therefore, option 1, Normal, option 2, Higher than normal, option 3, Lower than normal and option 4, None, is the correct answer.
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write the balance chemical equation and identify the reaction type
Write the balance chemical equation and identify the reaction type 1: sodium bicarbonate \( + \) acetic acid \( \rightarrow \) sodium acetate \( + \) carbonic acid carbonic acid \( \rightarrow \) carb
Answers
NaHCO3 + CH3COOH ⇒ CH3COONa + H2CO3,
it is a double displacement reaction (acid-base reaction)
In the given reaction, sodium bicarbonate (NaHCO3) reacts with acetic acid (CH3COOH) to produce sodium acetate (CH3COONa) and carbonic acid (H2CO3). To balance the equation, we need to ensure that the number of atoms of each element is equal on both sides. The balanced equation shows that one molecule of sodium bicarbonate reacts with one molecule of acetic acid to produce one molecule of sodium acetate and one molecule of carbonic acid. This balancing ensures that the number of atoms of each element (Na, H, C, O) is the same on both sides of the equation. The reaction type is identified as a double displacement reaction because the positive ions (Na+ and H+) and the negative ions (HCO3- and CH3COO-) exchange places to form the products. In this case, sodium from sodium bicarbonate replaces the hydrogen ion from acetic acid, forming sodium acetate. Simultaneously, the bicarbonate ion combines with the hydrogen ion from acetic acid to form carbonic acid. Overall, the reaction between sodium bicarbonate and acetic acid is a double displacement reaction, precisely an acid-base reaction.
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Question 12 of 24 Submit What is the correct common name for the compound shown here? methyl iso propyl ether ether
Answers
The correct common name for the compound shown below is Methyl isopropyl ether. So, the option "methyl iso propyl ether" is correct.
Common names are not standardized names, and they may differ from one place to another. The IUPAC (International Union of Pure and Applied Chemistry) system is the standard way of naming chemical compounds. UPAC is best known for its works standardizing nomenclature in chemistry, but IUPAC has publications in many science fields including chemistry, biology and physics. Some important work IUPAC has done in these fields includes standardizing nucleotide base sequence code names; publishing books for environmental scientists, chemists, and physicists; and improving education in science The names can be long, but they are precise and identify the chemical compound exactly. The IUPAC name for the compound shown below is 1-methoxy-2-methylpropane or alternatively methyl 2-methoxypropane.
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Provide the key fragment structures of the mass spectrometry
data. The possible molecular formula is:
C5H9O2Br
Relative Intensity 100 80 40 20- o fim 20 40 60 80 Titr 100 120 m/z 140 160 180 200 15.0 28.0 37.0 38.0 39.0 42.0 43.0 49.0 50.0 51.0 52.0 61.0 62.0 63.0 73.0 74.0 75.0 76.0 77.0 89.0 90.0 91.0 91.5 1
Answers
Mass spectrometry is a scientific technique used for the identification of unknown compounds, determination of isotopic composition, and determination of the structure of compounds, among others. The fragments generated in mass spectrometry can help in determining the molecular formula of the compound. In this case, the key fragment structures of the mass spectrometry data with a possible molecular formula of C5H9O2Br are as follows:
15.0, 28.0, 37.0, 38.0, 39.0, 42.0, 43.0, 49.0, 50.0, 51.0, 52.0, 61.0, 62.0, 63.0, 73.0, 74.0, 75.0, 76.0, 77.0, 89.0, 90.0, 91.0, 91.5
The relative intensity of each of the fragments is also given as 100, 80, 40, 20, and so on. The relative intensity of each fragment provides information about the abundance of that fragment in the sample.
The molecular formula C5H9O2Br indicates that the compound has 5 carbon atoms, 9 hydrogen atoms, 2 oxygen atoms, and 1 bromine atom. By analyzing the fragment structures and their relative intensity, we can propose the following possible fragment structures:
- 15.0: CH3O2Br
- 28.0: C2H5Br
- 37.0: C2H5O2
- 38.0: C2H6Br
- 39.0: C2H6O
- 42.0: C3H5OBr
- 43.0: C3H5O
- 49.0: C4H9Br
- 50.0: C4H10O2
- 51.0: C4H9O2Br
- 52.0: C4H10O
- 61.0: C5H9O
- 62.0: C5H10Br
- 63.0: C5H10O
- 73.0: C5H9BrO2
- 74.0: C5H10O2Br
- 75.0: C5H9O2
- 76.0: C5H10BrO
- 77.0: C5H9BrO
- 89.0: C5H9BrO2
- 90.0: C5H10O2Br
- 91.0: C5H9O2Br
- 91.5: C5H10BrO
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a Following are three organic acids and the pk, of each: pyruvic acid, 3.08; benzoic acid, 4.19; and citric acid, 2.10. What is the K, of each acid? Ka( citric acid) - Which of the three is the strong
Answers
The equilibrium constant (Kₐ) can be calculated using the pKₐ values of the acids. The Kₐ values for pyruvic acid, benzoic acid, and citric acid are approximately 10⁻¹¹, 10⁻⁴, and 10⁻¹ respectively. Among the three acids, citric acid has the highest Kₐ and therefore is the strongest acid.
The equilibrium constant (Kₐ) is related to the pKₐ by the equation Kₐ = 10^(-pKₐ). Using this relationship, we can calculate the Kₐ values for each acid based on their given pKₐ values.
For pyruvic acid with a pKₐ of 3.08, the Kₐ is approximately 10^(-3.08), which is around 10⁻¹¹. This indicates that pyruvic acid is a relatively weak acid.
For benzoic acid with a pKₐ of 4.19, the Kₐ is approximately 10^(-4.19), which is around 10⁻⁴. Benzoic acid is stronger than pyruvic acid but weaker than citric acid.
For citric acid with a pKₐ of 2.10, the Kₐ is approximately 10^(-2.10), which is around 10⁻¹. Citric acid has the highest Kₐ value among the three acids, indicating that it is the strongest acid.
Therefore, based on the Kₐ values, citric acid is the strongest acid among pyruvic acid, benzoic acid, and citric acid.
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Water has the following composition: pH = 7.8 HCO32 = 85 mg/L as CaCO3 Ca²+ = 32 mg/L as CaCO3 Mg2+ = 40 mg/L as CaCO3 The following three questions pertain to this water. What is the highest theoretical concentration of Ca2+ (M) that can be dissolved at this pH in equilibrium with Ca(OH)₂(s) assuming no other calcium solids will form? Note: Don't be alarmed - it will be a large number! Ca(OH)(s) <--> Ca²+ + 2OH Kp-10:53
Answers
The first step in solving this problem is to calculate the activity product of calcium ions in the water to determine the saturation state of calcium with respect to Ca(OH)₂ (s).Then, using the solubility product (Ksp) of calcium hydroxide, we can calculate the theoretical maximum concentration of calcium ions in the water.
For Ca(OH)₂(s), the equilibrium expression is Ca(OH)(s) <--> Ca²+ + 2OH Kp-10:53The equilibrium constant, Kp-10:53, for this reaction is equal to the solubility product of Ca(OH)₂ (s) because it is an ionic solid. The Ksp of Ca(OH)₂ (s) is given as Ksp= [Ca²+][OH]². Using this, we can calculate the activity product, Q, for calcium ions in the water at equilibrium with Ca(OH)₂ (s):Q = [Ca²+][OH]²
the activity product of calcium ions in the water is:Q = [Ca²+][OH-]²= [Ca²+](1.58 x 10-8)²= 3.97 x 10-17The equilibrium constant, Kp-10:53, is equal to Ksp= [Ca²+][OH-]², so we can write:Ksp = [Ca²+](1.58 x 10-8)²Ksp/(1.58 x 10-8)² = [Ca²+]= (10-10.53)/(1.58 x 10-8)² = 3.24 x 10-6 mol/LThis is the theoretical maximum concentration of calcium ions that can exist in the water without precipitation of calcium solids. Note that this is an extremely high concentration of calcium ions.
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