Answer:
18.65004 grams H2O
Explanation:
First, we need to write down the balanced chemical equation for the decomposition reaction:
2LiOH -> H2O + Li2O
Since we have grams of LiOH and we need to know the grams of water, we need to convert to moles since we can only compare moles to moles.
The amu of LiOH is 23.947.
The given grams of LiOH is 63.. To convert to moles, we will divide 63 by 23.947..
This gives us 2.6310 moles LiOH..
To convert to moles of H2O (and later grams of H2O), we will use the mole fractions from the balanced equation...
When we look at the balanced equation we can see that 2 moles of LIOH can produce 1 mol of Water, so:
2.6310 moles
= 1.3155 moles H2O
Now we will convert from moles to grams (we must multiply by the amu)
1.3155 moles H2O = 18.65 grams H2O
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.
<span>Movement of water from an area of lower solute concentration to an area of higher solute concentration</span>
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