The molecular weight of LiOH would be 23.95 g/mol, so the amount of LiOH in mol would be: 1.64g/(<span>23.95 g/mol)= 0.069 mol
The reaction of LiOH with HCl would be:
</span><span>HCl + LiOH = H2O + LiCl
The coefficient of LiOH:HCL is 1:1 so you need the same amount of HCl to neutralize LiOH.
HCl= LiOH
volume* 0.15M= </span>0.069 mol
volume= 0.069 mol/ (0.15 mol/ 1000ml)
volume= 459.29 ml
<u>Answer:</u> The concentration of reactant after the given time is 0.0205 M
<u>Explanation:</u>
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 11.0 min = 660 s (Conversion factor: 1 min = 60 s)
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = 0.400 M
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![4.50\times 10^{-3}s^{-1}=\frac{2.303}{660s}\log\frac{0.400}{[A]}](https://tex.z-dn.net/?f=4.50%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B660s%7D%5Clog%5Cfrac%7B0.400%7D%7B%5BA%5D%7D)
![[A]=0.0205M](https://tex.z-dn.net/?f=%5BA%5D%3D0.0205M)
Hence, the concentration of reactant after the given time is 0.0205 M
For example Existing rock is weathered into sediment, then compacted, cemented and finally Lithified (Lithification)
Answer:
54.0 g.
Explanation:
- From the given data and the balanced reaction:
<em>2H₂ + O₂ → 2H₂O.</em>
<em></em>
When 3.00 moles of hydrogen molecules and 1.50 moles of oxygen molecules react, they form 3.00 moles of water according to the balanced reaction.
<em>∴ The no. of grams in 3.0 moles of water = no. of moles x molar mass </em>= (3.0 mol)(18.0 g/mol) = <em>54.0 g.</em>
Answer: heck the chemistry app itll help you i dont know this answer but the app will tell u!
