SEBA Class 10 Science Important Chapter 4 Carbon and its compounds Solutions English Medium As Per SEBA New Syllabus to each chapter is provided in the list so that you can easily browse through different chapters SEBA Class 10 Science Additional Solutions in English and select need one. SEBA Class 10 General Science Additional Notes English Medium Download PDF. SEBA Important Solutions for Class 10 Science.
SEBA Class 10 Science Important Chapter 4 Carbon and its compounds
Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. Assam SEBA Board Class 10 General Science Additional Question Answer are part of All Subject Solutions. Here we have given SEBA Class 10 Science Important Question Answer English Medium for All Chapters, You can practice these here.
Carbon and its compounds
Chapter: 4
| IMPORTANT QUESTION ANSWER |
Very short Answers type questions:
Q.1. Name the element whose one of the allotropic forms is buck minsterfullerene.
Ans: Carbon.
Q.2. What are the two properties of carbon which lead to the formation of a large number of carbon compounds?
Ans: Catenation and Tetravalency.
Q.3. State whether the following statement is true of false: Diamond and graphite are the covalent compounds of carbon element.
Ans: False.
Q.4. Name the scientist who disproved the ‘vital force theory’ for the formation of organic compounds.
Ans: Friedrich wohler.
5. Name the element whose allotropic form is graphite.
Ans: Carbon.
Q.6. In addition to some propane and ethane, LPG cylinders contain mainly two isomers of another alkane. Name the two isomers and write their condensed structural formula.
Ans: n-butane and iso-butane.
Structural formula:
Q.7. Buckminsterfullerene is a spherical molecule in which 60 carbon atoms are arranged in interlocking hexagonal and pentagonal rings of carbon atoms.
(a) How many hexagons of carbon atoms are present in one molecule of buck minsterfullerence?
(b) How many pentagons of carbon atoms are present in one molecule of buck minsterfullerence?
Ans: (a) 20 hexagons.
(b) 12 pentagone.
Q.8. Name the hardest natural substance known.
Ans: Diamond.
Q.9. Which of the following molecule is called buck minsterfullerene ? C₉₀ ,C₆₀ , C ₇₀, C₁₂ O
Ans: C₆₀
Q.10. What is the next higher homologue of methanol?
Ans: Ethanol.
Q.11. Identify the functional group present in the following compound and name it according to IUPAC system: CH₃ OH
Ans: Alcohol group; Methanol.
Q.12. What is the common name of methanol?
Ans: Formaldehyde.
Q.13. Write the name of the following functional groups:
Ans: (a) Alkyne.
(b) Alkene.
Q.14. Which of the following will give brisk efferves cence with sodium hydrogen carbonate and why? CH₃ COOH, CH₃ CH₂ OH
Ans: CH3COOH(ethanoic acid) gives brisk effervescence with sodium hydrogen carbonate.
Q.15. Name the functional group present in an organic compound which gives brisk effervescence with NaHCO₃
Ans: Carboxylic acid group (−COOH) is the functional group which gives brisk effervescence with NaHCO3.
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