Answer : The excess reactant in the combustion of methane in opem atmosphere is 
molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,
First we have to calculate the moles of methane.
= 
= 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give 
moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is molecule.
Answer:
<h2>10.5 gram(cm)^-3 is the right answer.</h2>
D is the correct answer, you do not want to keep the fumes from numerous frogs in a room. There is no logical reason to keep the fumes inside.
Answer:
shorter
longer
Explanation:
The carbon-carbon bond length in ethylene is <u>shorter</u> than the carbon-carbon bond length in ethane, and the HCH bond angle in ethylene is <u>longer</u> the HCH bond angle in ethane.
The objective of this question is to let us understand the concept of Bond Length and Bond angle among the unsaturated aliphatic hydrocarbons (i.e alkanes, alkenes and alkynes).
The variation in bond angles of unsaturated aliphatic hydrocarbons can be explained by two concepts; The valence shell electron pair repulsion (VSEPR) model and hybridization.
The VSEPR model determines the total number of electron pairs surrounding the central atom of a species. The total number of electron pairs consist of the bond pairs and lone pairs. All the electron pairs( lie charge ) will then orient themselves in such a way to minimize the electrostatic repulsion between them.
As the number of the lone pairs increases from zero to 2 ; the bond angles diminish progressively.
However;
Hybridization is the mixing or blending of two or more pure atomic orbitals (s,p and d) to form two or more hybrid atomic orbitals that are identical in shape and energy . e.g sp, sp² , sp³ hybrid orbitals etc .
The shape of the geometry of this compound hence determines their bond angle.
The shape of the geometry of ethane is tetrahedral which is 109.5° in bond angle while that of ethylene is trigonal planar which is 120°.
This is why the HCH bond angle in ethylene is longer the HCH bond angle in ethane .
