Answer:
![m_{LiNO_3}=314gLiNO_3](https://tex.z-dn.net/?f=m_%7BLiNO_3%7D%3D314gLiNO_3)
Explanation:
Hello!
In this case, for the given chemical reaction:
![Pb(SO_4 )_2, + 4 LiNO_3 \rightarrow Pb(NO_3)_4, + 2 Li_2 SO_4](https://tex.z-dn.net/?f=Pb%28SO_4%20%29_2%2C%20%2B%204%20LiNO_3%20%5Crightarrow%20Pb%28NO_3%29_4%2C%20%2B%202%20Li_2%20SO_4)
We can see a 4:2 mole ratio between lithium nitrate (68.946 g/mol) and lithium sulfate (109.94 g/mol); in such a way, via stoichiometry, the required mass of lithium nitrate to make 250 g of lithium sulfate turns out:
![m_{LiNO_3}=250gLi_2SO_4*\frac{1molLi_2SO_4}{109.94gLi_2SO_4} *\frac{4molLiNO_3}{2molLi_2SO_4} *\frac{68.946gLiNO_3}{1molLiNO_3}\\\\m_{LiNO_3}=314gLiNO_3](https://tex.z-dn.net/?f=m_%7BLiNO_3%7D%3D250gLi_2SO_4%2A%5Cfrac%7B1molLi_2SO_4%7D%7B109.94gLi_2SO_4%7D%20%2A%5Cfrac%7B4molLiNO_3%7D%7B2molLi_2SO_4%7D%20%2A%5Cfrac%7B68.946gLiNO_3%7D%7B1molLiNO_3%7D%5C%5C%5C%5Cm_%7BLiNO_3%7D%3D314gLiNO_3)
Best regards!
The key principle here is to find the mole ratio of each element to know the empirical formula. First we need to convert the mass to moles by using the molar mass of each element. The molar mass of S is 32 and O is 16 grams/mole. The moles of S and O given the mass is 0.03125 and 0.09375 moles, respectively. The compound would need 3 moles of O per S. the compound is SO3
Answer:
Explanation:
Part two of Dalton's theory had to be modified after mass spectrometry experiments demonstrated that atoms of the same element can have different masses because the number of neutrons can vary for different isotopes of the same element. ... Scientists have even developed the technology to see the world on an atomic level!
hoped i helped you :)
Explanation:
The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.
This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated
The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.