Answer:
Explanation:
An electrophilic addition reaction occurs when an electrophile attacks a substrate, with the end result being the inclusion of one or many comparatively straightforward molecules along with multiple bonds.
In the given question, the hydrogen bromide provides the electrophile while the bromide is the nucleophile. The mechanism proceeds with the attack of the electrophile on the carbon, followed by deprotonation. This process is continued with a formation of carbocation and the bromide(nucleophile) finally bonds to the carbocation to form a stable product.
The first diagram showcases the possible various starting molecules for the synthesis while the second diagram illustrates their mechanism.
Color change bubbles and i think dissolving
Answer:
P.E for the 4kg bowling ball held 2.5 meters above head is 100j, while the other is 60j so the 4kg ball has more potential energy
Answer:
is there a table or an equation for that ??
Answer:

Explanation:
1. Calculate the rate constant
The integrated rate law for first order decay is

where
A₀ and A_t are the amounts at t = 0 and t
k is the rate constant

2. Calculate the half-life
