1,38×10²² = 0,138×10²³
0,138×10²³ ----- 1,5g
6,02×10²³ ------ X
X = (1.5×6,02×10²³)/0,138×10²³
X = 65,435 g/mol
It's ZINC (Zn)
:•)
I believe this question has the following five choices to
choose from:
>an SN2 reaction has occurred with inversion of
configuration
>racemization followed by an S N 2 attack
>an SN1 reaction has taken over resulting in inversion
of configuration
>an SN1 reaction has occurred due to carbocation
formation
>an SN1 reaction followed by an S N 2 “backside”
attack
The correct answer is:
an SN1 reaction has occurred due to carbocation formation
8. 4 grams of water will be produced.
Answer:
The percent isotopic abundance of Ir-193 is 60.85 %
The percent isotopic abundance of Ir-191 is 39.15 %
Explanation:
we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193
First of all we will set the fraction for both isotopes
X for the isotopes having mass 193
1-x for isotopes having mass 191
The average atomic mass is 192.217
we will use the following equation,
193x + 191(1-x) = 192.217
193x + 191 - 191x = 192.217
193x- 191x = 192.217 - 191
2x = 1.217
x= 1.217/2
x= 0.6085
0.6085 × 100 = 60.85 %
60.85% is abundance of Ir-193 because we solve the fraction x.
now we will calculate the abundance of Ir-191.
(1-x)
1-0.6085 =0.3915
0.3915× 100= 39.15 %
60 I think bcz if there is 1&2 they differ 10 times
