Iodate has a charge of -1, this means its chemical symbol
must be IO3(-1) while Lanthanum is La(+3). Therefore making the compound
La(IO3)3.
However sulfate has a chemical symbol of SO4(-2), it has
charge of -2, therefore the formula for lanthanum sulfate is:
<span>La2(SO4)<span>3</span></span>
Explanation:
1)  + 7 H_2(g)](https://tex.z-dn.net/?f=%202%20Al%28s%29%20%2B%202%20NaOH%28aq%29%20%2B%206%20H_2O%28l%29%20%5Clongleftrightarrow%202%20Na%5BAl%28OH%29_4%5D%28aq%29%20%2B%207%20H_2%28g%29)
![Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNa%5BAl%28OH%29_4%5D%5D%5E2%2A%5BH_2%5D%5E7%7D%7B%5BNaOH%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:
2) 
![Kc=\frac{[H_2SO_4]}{[SO_3]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2SO_4%5D%7D%7B%5BSO_3%5D%5E2%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:
3) 
![Kc=\frac{1}{[O_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B1%7D%7B%5BO_2%5D%5E3%7D)
The Kc for the reverse reaction is the inverse of the Kc of the reaction:
Answer:
specialized if you add one more I cause I'm pretty sure there is supposed to be one more i.
Explanation:
9/4 Be +2 (the 9 and 4 are stacked next to Be). Atomic #: 4
Mass #: 9, # protons: 4, # neutrons: 5, #electrons: 2.
31/15 P (31 is stacked over 15 next to the P). Atomic #: 15,
Mass #: 31, # protons: 15, # neutrons: 16, # electrons: 15.
