I think the answer is 4.20 hours and minutes planning the last road trip .
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.
Answer
Find out the how much prize money did she win.
To proof
Let us assume that the total amount prize money did she win be u.
As given
Mrs.Gill won certain prize money in a cooking competition.
She spent half of the prize money on the clothes
one third on grocery
(L.C.M of (2,3) =6 )
gave away the remaining Rs.2000 to an orphanage
u =Rs 12000
Rs 12000 prize money did she win.
Hence proved
Check the picture below.
so, let's notice, is really just a 2x20 rectangle with a quarter of a semicircle with a radius of 11.
![\bf \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \implies A=\pi 11^2\implies A=121\pi \implies \stackrel{\textit{one quarter of that}}{\boxed{A=\cfrac{121\pi }{4}}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\underline{\textit{area of the figure}}}{\stackrel{\textit{rectangle's area}}{(2\cdot 20)}+\stackrel{\textit{circle's quart's area}}{\cfrac{121\pi }{4}}\qquad \approx \qquad 135.03\implies \stackrel{\textit{rounded up}}{135}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cimplies%20A%3D%5Cpi%2011%5E2%5Cimplies%20A%3D121%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bone%20quarter%20of%20that%7D%7D%7B%5Cboxed%7BA%3D%5Ccfrac%7B121%5Cpi%20%7D%7B4%7D%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Cunderline%7B%5Ctextit%7Barea%20of%20the%20figure%7D%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Brectangle%27s%20area%7D%7D%7B%282%5Ccdot%2020%29%7D%2B%5Cstackrel%7B%5Ctextit%7Bcircle%27s%20quart%27s%20area%7D%7D%7B%5Ccfrac%7B121%5Cpi%20%7D%7B4%7D%7D%5Cqquad%20%5Capprox%20%5Cqquad%20135.03%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B135%7D%7D)
D would not correctly solve the problem it is adding more kittens instead of eliminating the ones already known
