Volume = 3 cm × 3cm × 3cm
= 27 cm ³
Mass = 213 g
Density =
= <span>
</span>
<span> = 7.89 g / cm³
Thus answer is D.</span>
The subscript 3 outside of the parentheses means to multiply everything in the parentheses by 3, so it becomes a matter of addition:
3*(CH3) * (1 carbon atom)/(CH3) = 3 carbon atoms in (CH3)3
CH has one carbon atom and one hydrogen atom, so:
1 carbon atom in CH
Add both of them together, and you get 4 carbon atoms in isobutane
Answer:
Stoichiometry measures these quantitative relationships, and is used to determine the amount of products and reactants that are produced or needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry.
Explanation:
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
