Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
Explanation:
Given: 
= 61 L, 
= 183 K, 
= 0.60 atm
At STP, the value of pressure is 1 atm and temperature is 273.15 K.
Now, formula used to calculate the new volume is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
Answer:
Rate of the reaction is 0.2593 M/s
-0.5186 M/s is the rate of the loss of ozone.
Explanation:
The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.
Rate of formation of oxygen : 
Rate of the reaction(R) =![\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B2%7D%5Cfrac%7Bd%5BO_3%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![R=\frac{1}{3}\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
Rate of formation of oxygen=3 × (R)
Rate of the reaction(R): 
Rate of the reaction is 0.2593 M/s
Rate of disappearance of the ozone:
![R=-\frac{1}{2}\frac{d[O_3]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BO_3%5D%7D%7Bdt%7D)
![\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_3%5D%7D%7Bdt%7D%3D-2%5Ctimes%20R%3D-2%5Ctimes%200.2593%5Ctimes%20M%2Fs%3D-0.5186M%2Fs)
-0.5186 M/s is the rate of the loss of ozone.
Answer:
319.8 m/min
Explanation:
533 cm/s
We can convert 533 cm/s to m/min by doing the following:
First, we shall convert 533 cm/s to m/s. This can be obtained as illustrated below:
Recall:
100 cm/s = 1 m/s
Therefore,
533 cm/s = 533 cm/s /100 cm/s × 1 m/s
533 cm/s = 5.33 m/s
Finally, we shall convert 5.33 m/s to m/min. This can be obtained as follow:
1 m/s = 60 m/min
Therefore,
5.33 m/s = 5.33 m/s / 1 m/s × 60 m/min
5.33 m/s = 319.8 m/min
Therefore, 533 cm/s is equivalent to 319.8 m/min
H2SO4 ---> 2H^+ + SO4^2-
Hence n H+ = 9 mols
Mass of H = nM = (9*1) = 9g
Alternately
mass of H2SO4= nM= 4.5*98= 441
Mass of H= mass h2so4 * molar mass of H/molar mass of h2so4
Mass of H= 441 * 2/98 = 9g
