<u>ANSWER:
</u>
Rate per annum at which CI will amount from RS 2000 to RS 2315.35 in 3 years is 5%
<u>SOLUTION:
</u>
Given,
P = RS 2000
C.I = RS 2315.35
T = 3 years
We need to find the rate per annum. i.e. R = ?
We know that,
When interest is compound Annually:

Where p = principal amount
r = rate of interest
n = number of years



![$1+\frac{R}{100}=\sqrt[3]{1.157}$](https://tex.z-dn.net/?f=%241%2B%5Cfrac%7BR%7D%7B100%7D%3D%5Csqrt%5B3%5D%7B1.157%7D%24)



R = 5%
Hence, rate per annum is 5 percent.
Answer:
129/2
Step-by-step explanation:
1 129
64 --- = -------
2 2
To find the improper fraction, you multiply the denominator by the whole number and then add the numerator. In the end, you put that answer over the starting denominator.
64 X 2 = 128 128 + 1 = 129 129 / 2 is the answer
True because tan(20)=2.1 and cot(20)=is o.44
The reaction is missing and it's ;
N2H4 + H2 ---> 2NH3
It asks for the total pressure too.
Answer:
A) Rate of change for NH3 = 152 torr/h
B) The total pressure in the vessel will remain the same.
Step-by-step explanation:
N2H4 + H2 ---> 2NH3
1 mole of N2H4 yields 2 moles of NH3.
From the question, the rate given for N2H4 is 76 torr/h.
Thus, The rate of change for NH3 will be = 2 x 76 = 152 torr/h
Now, on the reaction side, 1 mole of N2H4 reacts with 1 mole of H2. So we have 2 moles on the left hand side.
While on the product side, 2 moles of NH3 are produced.
So the total pressure will remain the same because for every 2 moles on the reaction side, 2 moles are gotten on the product side.