Question:
$6$ litres of an alkene require $27$ litres of oxygen at constant temperature and pressure for complete combustion. The alkene is:
$6$ litres of an alkene require $27$ litres of oxygen at constant temperature and pressure for complete combustion. The alkene is:
Updated On: Jun 22, 2025
- Ethene
- Propene
- 1-Butene
- 2-Butene
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
CnH2n + 1.5nO2 → nCO2 + nH2O
6X1.5n= 27
n= 3
So , the correct answer is (B) : Propene
Was this answer helpful?
2
3
Top Questions on Some basic concepts of chemistry
- Explain the mechanism of cleaning action of soap (micelle formation).
- JKBOSE Class X Board - 2026
- Science
- Some basic concepts of chemistry
- Explain the neutralisation reaction with an example and its utility in daily life.
- JKBOSE Class X Board - 2026
- Science
- Some basic concepts of chemistry
- Explain the formation of Ionic Compounds (e.g., \(NaCl\) or \(MgCl_2\)) and state their properties.
- JKBOSE Class X Board - 2026
- Science
- Some basic concepts of chemistry
- Differentiate between metals and non-metals based on their chemical properties.
- JKBOSE Class X Board - 2026
- Science
- Some basic concepts of chemistry
- What are amphoteric oxides? Give two examples.
- JKBOSE Class X Board - 2026
- Science
- Some basic concepts of chemistry
View More Questions
Questions Asked in JEE Main exam
- Let \( f \) be a polynomial function such that \[ f(x^2+1)=x^4+5x^2+2,\quad \text{for all } x\in\mathbb{R}. \] Then \[ \int_0^3 f(x)\,dx \] is equal to:
- JEE Main - 2026
- Functions
- A rectangle is formed by lines \( x = 0, y = 0, x = 3 \) and \( y = 4 \). A line perpendicular to \( 3x + 4y + 6 = 0 \) divides the rectangle into two equal parts. Then the distance of the line from the point \( \left( -1, \frac{3}{2} \right) \) is:
- JEE Main - 2026
- Coordinate Geometry
If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where \( \alpha, \beta \in \mathbb{N} \), then the value of \( \alpha + \beta \) is ___________.
- JEE Main - 2026
- Trigonometric Identities
- Let \( f: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f''(x)>0 \) for all \( x \in \mathbb{R} \) and \( f'(a-1) = 0 \), where \( a \) is a real number. Let \( g(x) = f(\tan^2 x - 2\tan x + a) \), \( 0<x<\frac{\pi}{2} \).
Consider the following two statements :
(I) \( g \) is increasing in \( (0, \frac{\pi}{4}) \)
(II) \( g \) is decreasing in \( (\frac{\pi}{4}, \frac{\pi}{2}) \)
Then,- JEE Main - 2026
- Applications of Derivatives
- Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to
- JEE Main - 2026
- Arithmetic Progression
View More Questions
Concepts Used:
Some Basic Concepts of Chemistry
Chemistry is a vast subject and for understanding its significance we can take help of following points:
- Chemistry plays an important role in understanding various subjects like physics, geology and biology.
- Chemistry is a core branch of science that explains us about the various compositional properties and interaction of matter. It also helps to understand various chemical reactions.
- Chemistry is important in order to understand the behaviour of fertilizers, alkenes, acids, salts, dyes, polymers, drugs, soaps and alloys in organic and inorganic chemistry.
- Chemistry plays an important role in various fields like healthcare, industrial, research, food, and farm activities.
Read More: Some Basic Concepts of Chemistry
Classification of Matter
There are two ways of classifying the matter:
(A) Physical Classification:
Matter can exist in three physical states:
- Solids - have definite volume and definite shape
- Liquids - have definite volume but not definite shape.
- Gases - have neither definite volume nor definite shape.
(B) Chemical Classification:
Based upon the composition, matter can be divided into two main types:
- Pure Substances are defined as a single substance (or matter) which cannot be separated by simple physical methods. Pure substances can be further classified as (i) Elements (ii) Compounds
- Mixtures are the combination of two or more elements or compounds which are not chemically combined together and may also be present in any proportion.