Answer:
1.07 g
Explanation:
Half-life of Pu-234 = 4.98 hours
Initially present = 45 g
mass remains after 27 hours = ?
Solution:
Formula
mass remains = 1/ 2ⁿ (original mass) ……… (1)
Where “n” is the number of half lives
To find "n" for 27 hours
n = time passed / half-life . . . . . . . .(2)
put values in equation 2
n = 27 hr / 4.98 hr
n = 5.4
Mass after 27 hr
Put values in equation 1
mass remains = 1/ 2ⁿ (original mass)
mass remains = 1/ 2^5.4 (45 g)
mass remains = 1/ 42.2 (45 g)
mass remains = 0.0237 x 45 g
mass remains = 1.07 g
Answer:1-methoxy-2,4-dinitrobenzene
Explanation:
The nitro groups are strongly electron withdrawing and promote nucleophilic substitution reactions where one of the original substituents is removed and replaced by a strong nucleophile such as the methoxy group. The mechanism of the reaction is attached below. The electron withdrawing nitrogroup assists the formation of the intermediate in the reaction as shown.
4) would be your correct answer