Answer: i iii i i i ii i i ir r
q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q
What is research? Depending on who you ask, you will likely get very different answers to this seemingly innocuous question. Some people will say that they routinely research different online websites to find the best place to buy goods or services they want. Television news channels supposedly conduct research in the form of viewer polls on topics of public interest such as forthcoming elections or government-funded projects. Undergraduate students research the Internet to find the information they need to complete assigned projects or term papers. Graduate students working on research projects for a professor may see research as collecting or analyzing data related to their project. Businesses and consultants research different potential solutions to remedy organizational problems such as a supply chain bottleneck or to identify customer purchase patterns. However, none of the above can be considered “scientific research” unless: (1) it contributes to a body of science, and (2) it follows the scientific method. This chapter will examine what these terms mean
A helium balloon displaces the amount of air. The weight of the helium in the balloon + the balloon fabric = lighter than the air that it displaces. Helium is lighter than air, so therefore, if it's in a balloon, the balloon will float.
Molecular weight of air = 29 g/mole, moles of air entering = 5/29 kgmoles/hr= 0.1724 kg moles/hr
Moles of N2= 0.21*0.1724=0.036 kg moles/hr, moles of N2= 0.79*0.1724= 0.1362 kgmoles/hr
Moles of CO2 can be calculated from gas law equation, n= PV/RT
V= 25 m3/hr= 25* 1000 L/hr, P= 2 bar = 2*0.9869 atm,=1.9738 atm R= 0.0821 L.atm/mole.K T= 100 deg.c =100+?273.15= 373.15K
n= number of moles of CO2= 1.9738*25*1000/ (0.0821*373.15) =1611 moles/hr= 1.611 kg moles/hr
total moles of mixture = 0.1724 +1.611 =1.7834 kg moles/hr
Moles % CO2 in the mixed stram = 100*1.611/1.7834 = 90.33%
Answer is 90.33%
Answer : The rate of formation of 
is, 
Explanation : Given,
Rate of disappearance of 
= 
The given rate of reaction is,
The expression for rate of reaction :
![\text{Rate of disappearance}=-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[Cl_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D)
![\text{Rate of formation}=\frac{1}{2}\frac{d[NOCl]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D)
From this we conclude that,
![\frac{1}{2}\frac{d[NOCl]}{dt}=-\frac{d[Cl_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D)
![\frac{d[NOCl]}{dt}=2\times \frac{d[Cl_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D2%5Ctimes%20%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D)
Now put the value of rate of disappearance of , we get:
![\frac{d[NOCl]}{dt}=2\times (4.24\times 10^{-2}M/s)=8.48\times 10^{-2}M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D2%5Ctimes%20%284.24%5Ctimes%2010%5E%7B-2%7DM%2Fs%29%3D8.48%5Ctimes%2010%5E%7B-2%7DM%2Fs)
Therefore, the rate of formation of is,
