It’s is 130 that is the answer!!!!
Answer:
Correct option is
B
5 liters of CH
4
(g)NO
2
at STP
No. of molecules=
22.4
5
mol=
22.4
5
×N
A
molecules
A) 5ℊ of H
2
(g)
No. of moles=
2
5
mol=
2
5
×N
A
molecules
B) 5l of CH
4
(g)
No. of moles of CH
4
=
22.4
5
mol=
22.4
5
N
A
molecules
C) 5 mol of O
2
=5N
A
O
2
molecules
D) 5×10
23
molecules of CO
2
(g)
Molecules of 5l NO
2
(g) at STP=5l of CH
4
(g) molecules at STP
Therefore, option B is correct.
In this case, we are going to assume that there are 100 atoms to make things easier.
Let R% be the abundance of n-15. With this in mind, we calculate the abundance of n-14 to be 100%-R%
14.0031*(100-R)% + 15.001 * R%= 14.00674
In this case, we can delete or ignore the % sign since we do not want to carry it around, however, we need to keep in mind that the final answer is in %
14.0031*(100-R) + 15.001 * R= 14.00674
1400.31-14.0031R+15.001R=1400.674
0.9979R=0.364
R=0.3648
Then, the abundance of n-15 is 0.3648%