I believe it’s answer B because even in space gravity plays roles.
Answer:
(a) Rate at which 
is formed is 0.050 M/s
(b) Rate at which 
is consumed is 0.0250 M/s.
Explanation:
The given reaction is:-
The expression for rate can be written as:-
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
Given that:- ![\frac{d[NO]}{dt}=-0.050\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-0.050%5C%20M%2Fs)
(Negative sign shows consumption)
![-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-(-0.050\ M/s)=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%28-0.050%5C%20M%2Fs%29%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\frac{d[NO_2]}{dt}=0.050\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D0.050%5C%20M%2Fs)
(a) Rate at which is formed is 0.050 M/s
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-0.050%5C%20M%2Fs%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![\frac{d[O_2]}{dt}=0.0250\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D0.0250%5C%20M%2Fs)
(b) Rate at which is consumed is 0.0250 M/s.
Answer:
You have a chemical reaction that is occurring
The answer is 19.9 grams cadmium.
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
heat lost by cadmium = heat gained by the water
-qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
-qcadmium = qwater
-(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
mcadmium = 19.9 grams
Answer:
Likely the reason behind carrying them in liquid for is to condense them. liquid takes up far less space than gases do and will last far longer. they can be transformed back into gases through heating them back up. To keep them solid would require frigid temperatures but liquid will transfer more easily through space.
