In an Aop if Sio 35 and Sq ² 28 find aio S = 35 Son9
Answers
Step-by-step explanation:
The balanced chemical reaction is A+2B⟶C.
5 moles of A will react with 15×2=10 moles of B.
However, only 8 moles of B are present.
Thus, B is the limiting reagent.
Hence, 8 moles of B will give 28×1=4 moles of C
Answer:
T_{10}=24T
10
=24
Step-by-step explanation:
It is given that S_{10}=150S
10
=150 ⇒\frac{10}{2}(2a+9d)=150
2
10
(2a+9d)=150 (using \frac{n}{2}(2a+(n-1)d)=S_{n}
2
n
(2a+(n−1)d)=S
n
)
⇒2a+9d=302a+9d=30 (1)
and S_{9}=\frac{9}{2}(2a+8d)=126S
9
=
2
9
(2a+8d)=126
⇒2a+8d=282a+8d=28 (2)
From equations (1) and (2), we get
2a+9d-2a-8d=30-282a+9d−2a−8d=30−28
⇒d=2d=2
Putting the value of =2 in equation (1), we get
2a+18=302a+18=30
a=6a=6
Now, T_{10}=a+(n-1)dT
10
=a+(n−1)d
=6+(10-1)26+(10−1)2
=6+186+18
=2424
Thus, T_{10}=24T
10
=24