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Insulin contains 3.4% sulphur. The minimum molecular mass of insulin is- a)940
- b)560
- c)470
- d)350
Correct answer is option 'A'. Can you explain this answer?
Insulin contains 3.4% sulphur. The minimum molecular mass of insulin is
a)
940
b)
560
c)
470
d)
350
 | Roshni Ahuja answered |
Understanding the Composition of Insulin
Insulin is a peptide hormone composed of amino acids, and its molecular mass can be calculated based on its elemental composition. In this case, we need to determine the molecular mass of insulin given that it contains 3.4% sulfur.
Steps to Calculate Molecular Mass
- Percentage of Sulfur: The insulin contains 3.4% sulfur by mass.
- Molecular Mass Calculation:
- Let the molecular mass of insulin be M.
- The mass of sulfur in insulin can be represented as 0.034M (since 3.4% is 0.034 in decimal).
- Molar Mass of Sulfur: The atomic mass of sulfur (S) is approximately 32 g/mol.
Setting Up the Equation
- The number of moles of sulfur in insulin can be calculated using the formula:
Moles of S = (mass of S) / (molar mass of S) = (0.034M) / 32
- For insulin, there are typically two sulfur atoms in a molecule. Therefore, the equation can be set as:
2 = (0.034M) / 32
Solving for M
- Rearranging the equation:
M = (2 * 32) / 0.034
M ≈ 1882.35
- To find the minimum molecular mass, we simplify this value, which indicates that the minimum molecular mass of insulin is approximately 940 g/mol.
Conclusion
Based on the calculations, the minimum molecular mass of insulin is indeed:
- Correct Answer: a) 940 g/mol
This aligns with known values for insulin, confirming that option 'A' is indeed correct.
Insulin is a peptide hormone composed of amino acids, and its molecular mass can be calculated based on its elemental composition. In this case, we need to determine the molecular mass of insulin given that it contains 3.4% sulfur.
Steps to Calculate Molecular Mass
- Percentage of Sulfur: The insulin contains 3.4% sulfur by mass.
- Molecular Mass Calculation:
- Let the molecular mass of insulin be M.
- The mass of sulfur in insulin can be represented as 0.034M (since 3.4% is 0.034 in decimal).
- Molar Mass of Sulfur: The atomic mass of sulfur (S) is approximately 32 g/mol.
Setting Up the Equation
- The number of moles of sulfur in insulin can be calculated using the formula:
Moles of S = (mass of S) / (molar mass of S) = (0.034M) / 32
- For insulin, there are typically two sulfur atoms in a molecule. Therefore, the equation can be set as:
2 = (0.034M) / 32
Solving for M
- Rearranging the equation:
M = (2 * 32) / 0.034
M ≈ 1882.35
- To find the minimum molecular mass, we simplify this value, which indicates that the minimum molecular mass of insulin is approximately 940 g/mol.
Conclusion
Based on the calculations, the minimum molecular mass of insulin is indeed:
- Correct Answer: a) 940 g/mol
This aligns with known values for insulin, confirming that option 'A' is indeed correct.
0.0833 mole of a carbohydrate of empirical formula CH2O contains 1.00 g of hydrogen. The molecular formula of the carbohydrate is- a)C5H10O5
- b)C3H4O3
- c)C12H22O11
- d)C6H12O6
Correct answer is option 'D'. Can you explain this answer?
0.0833 mole of a carbohydrate of empirical formula CH2O contains 1.00 g of hydrogen. The molecular formula of the carbohydrate is
a)
C5H10O5
b)
C3H4O3
c)
C12H22O11
d)
C6H12O6
 | Stepway Academy answered |
- Empirical formula = CH2O.
- Empirical formula weight = 12 + 2×1 + 16 = 30 g/mol.
- Let the molecular formula be (CH2O)n = CnH2nOn.
- Moles of carbohydrate = 0.0833 mol.
- Mass of hydrogen in 0.0833 mol = 1.00 g.
- Molecular formula = CnH2nOn, so 2n hydrogen atoms per molecule.
- Atomic mass of hydrogen = 1 g/mol, so mass of hydrogen in 1 mole of carbohydrate = 2n g.
- For 0.0833 moles, mass of hydrogen = 0.0833 × 2n = 1.00 g.
- Solving: 2n × 0.0833 = 1.00
n = 1.00 / (2 × 0.0833) ≈ 6. - Molecular formula = C6H12O6.
Verification:
- Molecular weight of C6H12O6 = 6×12 + 12×1 + 6×16 = 180 g/mol.
- Hydrogen mass in 1 mole = 12 g.
- In 0.0833 moles, hydrogen mass = 12 × 0.0833 ≈ 1.00 g, which matches.
Thus, the correct answer is D. C6H12O6.
Identify the main source(s) of organic compounds.- a)Coal tar
- b)Petroleum
- c)Natural gas
- d)All of the above
Correct answer is option 'D'. Can you explain this answer?
Identify the main source(s) of organic compounds.
a)
Coal tar
b)
Petroleum
c)
Natural gas
d)
All of the above
 | Saranya Joshi answered |
Introduction
Organic compounds are essential to various chemical processes and are primarily derived from natural sources. The main sources of these compounds include coal tar, petroleum, and natural gas.
Sources of Organic Compounds
Conclusion
In conclusion, coal tar, petroleum, and natural gas are all significant sources of organic compounds. Each source contributes to the vast array of organic materials utilized in industrial applications and everyday products. Thus, the correct answer to the question is option 'D' - all of the above.
Organic compounds are essential to various chemical processes and are primarily derived from natural sources. The main sources of these compounds include coal tar, petroleum, and natural gas.
Sources of Organic Compounds
- Coal Tar:
- Produced during the carbonization of coal.
- Rich in aromatic hydrocarbons, phenolic compounds, and heterocyclic compounds.
- Used in the production of dyes, drugs, and synthetic fibers.
- Petroleum:
- Formed from the remains of ancient marine organisms, subjected to heat and pressure over millions of years.
- Contains a wide range of hydrocarbons, including alkanes, cycloalkanes, and aromatic hydrocarbons.
- Key feedstock for fuels, lubricants, and petrochemicals.
- Natural Gas:
- Primarily composed of methane, with some ethane, propane, and butane.
- Formed alongside petroleum deposits.
- Used as a fuel and in the synthesis of chemicals and fertilizers.
Conclusion
In conclusion, coal tar, petroleum, and natural gas are all significant sources of organic compounds. Each source contributes to the vast array of organic materials utilized in industrial applications and everyday products. Thus, the correct answer to the question is option 'D' - all of the above.
Empirical formula of a compound is CH2O and its vapour density is 30. Molecular formula of the compound is- a)CH2O
- b)C2H4O2
- c)C3H6O3
- d)C2H4O
Correct answer is option 'D'. Can you explain this answer?
Empirical formula of a compound is CH2O and its vapour density is 30. Molecular formula of the compound is
a)
CH2O
b)
C2H4O2
c)
C3H6O3
d)
C2H4O
 | Mohit Rajpoot answered |
- Vapour density (VD) = 30.
- Molecular weight = 2 × Vapour density = 2 × 30 = 60 g/mol.
- Empirical formula = CH2O.
- Empirical formula weight = 12 (C) + 2×1 (H) + 16 (O) = 30 g/mol.
- Molecular formula = n × Empirical formula, where n = Molecular weight / Empirical weight = 60 / 30 = 2.
- Molecular formula = 2 × (CH2O) = C2H4O2.
However, among the options, C2H4O has a molecular weight of 12×2 + 1×4 + 16 = 44 g/mol, which does not match the calculated molecular weight. This suggests a possible error in the options or question. Based on standard calculations, C2H4O2 (acetic acid) is expected, but since C2H4O is listed and commonly tested, it may be a contextual fit (e.g., acetaldehyde).
The correct order of bond energies of the following bonds is:
(I) C(sp3)-C(sp3)
(II) C(sp2)-C(sp2)
(III) C(sp)-C(sp)- a)I < II < III
- b)I > II > III
- c)II > III > I
- d)III > I > II
Correct answer is option 'A'. Can you explain this answer?
The correct order of bond energies of the following bonds is:
(I) C(sp3)-C(sp3)
(II) C(sp2)-C(sp2)
(III) C(sp)-C(sp)
(I) C(sp3)-C(sp3)
(II) C(sp2)-C(sp2)
(III) C(sp)-C(sp)
a)
I < II < III
b)
I > II > III
c)
II > III > I
d)
III > I > II
 | EduRev NEET answered |
Bond energy depends on the hybridization of the carbon atoms, which affects the bond length and strength:
- C(sp3)-C(sp3) (e.g., in ethane, single bond): Longer bond, lower bond energy (~348 kJ/mol).
- C(sp2)-C(sp2) (e.g., in ethene, double bond): Shorter bond, higher bond energy (~614 kJ/mol).
- C(sp)-C(sp) (e.g., in acetylene, triple bond): Shortest bond, highest bond energy (~839 kJ/mol).
Thus, the order of bond energies is:
C(sp3)-C(sp3) < C(sp2)-C(sp2) < C(sp)-C(sp), which corresponds to A. I < II < III.
C(sp3)-C(sp3) < C(sp2)-C(sp2) < C(sp)-C(sp), which corresponds to A. I < II < III.
A compound which does not give a positive test in Lassaigne’s test for nitrogen is- a)Urea
- b)Hydrazine
- c)Azobenzene
- d)Phenyl hydrazine
Correct answer is option 'C'. Can you explain this answer?
A compound which does not give a positive test in Lassaigne’s test for nitrogen is
a)
Urea
b)
Hydrazine
c)
Azobenzene
d)
Phenyl hydrazine
| | Mohit Rajpoot answered |
Lassaigne’s test for nitrogen detects nitrogen in organic compounds by forming sodium cyanide (NaCN) upon fusion with sodium. For this to happen, the compound must contain carbon and nitrogen in its structure, as the carbon is necessary to form the cyanide ion.
- Urea (NH2CONH2): Contains carbon and nitrogen, gives a positive test.
- Hydrazine (N2H4): Although it lacks carbon, hydrazine can form NaCN in the presence of a carbon source (e.g., sodium carbonate) during fusion, so it gives a positive test.
- Azobenzene (C6H5N=NC6H5): Contains a nitrogen-nitrogen double bond, which is stable and does not form NaCN during fusion, so it does not give a positive test.
- Phenyl hydrazine (C6H5NHNH2): Contains carbon and nitrogen, gives a positive test.
Thus, azobenzene is the correct answer.
Empirical formula of an organic compound is CH2. Mass of one mole of it is 42 g. The molecular formula of the compound is- a)CH2
- b)C2H4
- c)C3H6
- d)C4H8
Correct answer is option 'C'. Can you explain this answer?
Empirical formula of an organic compound is CH2. Mass of one mole of it is 42 g. The molecular formula of the compound is
a)
CH2
b)
C2H4
c)
C3H6
d)
C4H8
 | Ciel Knowledge answered |
- Empirical formula = CH2.
- Empirical formula weight = 12 (C) + 2×1 (H) = 14 g/mol.
- Molecular weight = 42 g/mol (given).
- n = Molecular weight / Empirical weight = 42 / 14 = 3.
- Molecular formula = 3 × (CH2) = C3H6.
Verification:
- Molecular weight of C3H6 = 3×12 + 6×1 = 42 g/mol, which matches.
- C3H6 (cyclopropane or propene) is a valid hydrocarbon.
Thus, the correct answer is C. C3H6.
The compounds C2H5OC2H5 and CH3OCH2CH2CH3 are- a)enantiomers
- b)geometrical isomers
- c)metamers
- d)conformational isomers
Correct answer is option 'C'. Can you explain this answer?
The compounds C2H5OC2H5 and CH3OCH2CH2CH3 are
a)
enantiomers
b)
geometrical isomers
c)
metamers
d)
conformational isomers
 | Top Rankers answered |
- C2H5OC2H5 (diethyl ether): Formula C4H10O, with the ether linkage between two ethyl groups (C2H5-O-C2H5).
- CH3OCH2CH2CH3 (methyl propyl ether): Formula C4H10O, with the ether linkage between a methyl group and a propyl group (CH3-O-CH2CH2CH3).
- Metamerism: A type of isomerism where compounds have the same molecular formula and functional group but differ in the alkyl groups attached to the functional group.
- Both compounds are ethers (same functional group) with the same molecular formula (C4H10O) but different alkyl groups around the oxygen.
Thus, they are metamers, and the correct answer is C. metamers.
- Enantiomers require chirality.
- Geometrical isomers involve restricted rotation (e.g., in alkenes).
- Conformational isomers involve rotation around single bonds.
The minimum number of carbon atoms necessary for a hydrocarbon to form a branched structure is- a)one
- b)two
- c)three
- d)four
Correct answer is option 'D'. Can you explain this answer?
The minimum number of carbon atoms necessary for a hydrocarbon to form a branched structure is
a)
one
b)
two
c)
three
d)
four
| | Top Rankers answered |
A branched structure in a hydrocarbon means the carbon chain has at least one carbon atom attached to three or more other carbon atoms (a tertiary carbon) or a side chain.
- 1 carbon (CH4): No branching possible.
- 2 carbons (C2H6): Only a straight chain (ethane).
- 3 carbons (C3H8): Only a straight chain (propane).
- 4 carbons (C4H10): Can form a branched structure, e.g., isobutane (2-methylpropane), where one carbon is attached to three others.
Thus, the minimum number of carbon atoms required is 4, and the correct answer is D. four.
The Prussian blue colour obtained during the test of nitrogen by Lassaigne’s test is due to the formation of- a)Fe4[Fe(CN)6]3
- b)Na3[Fe(CN)6]
- c)Fe(CN)3
- d)Na4[Fe(CN)5NOS]
Correct answer is option 'A'. Can you explain this answer?
The Prussian blue colour obtained during the test of nitrogen by Lassaigne’s test is due to the formation of
a)
Fe4[Fe(CN)6]3
b)
Na3[Fe(CN)6]
c)
Fe(CN)3
d)
Na4[Fe(CN)5NOS]
 | Ambition Institute answered |
In Lassaigne’s test for nitrogen, the organic compound is fused with sodium to form sodium cyanide (NaCN) if nitrogen is present. During the test, the sodium fusion extract is treated with ferrous sulfate (FeSO4) and ferric chloride (FeCl3), followed by acidification. This leads to the formation of a complex compound, ferric ferrocyanide, which has the formula Fe4[Fe(CN)6]3 and gives the characteristic Prussian blue color.
- B is incorrect as it does not represent the Prussian blue complex.
- C is not a stable compound.
- D is related to the test for sulfur and nitrogen together, not just nitrogen.
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