Answer:
The answer is 375.54 g of AgBr
Explanation:
Mass (g) = Concentration (mol/L) x volume (L) x Molecular Weight of AgBr (g/mol)
Mass = 2M x 1L x 187.77 g/mol
Mass = 375.54g
Answer:
a.
![Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BHCO_3%5E-%5D%5BOH%5E-%5D%7D%7B%5BCO_3%5E%7B2-%7D%5D%7D)
b.
![Keq=[O_2]^3](https://tex.z-dn.net/?f=Keq%3D%5BO_2%5D%5E3)
c.
![Keq=\frac{[H_3O^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
d.
![Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
Explanation:
Hello there!
In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:
a.
![Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BHCO_3%5E-%5D%5BOH%5E-%5D%7D%7B%5BCO_3%5E%7B2-%7D%5D%7D)
b.
![Keq=[O_2]^3](https://tex.z-dn.net/?f=Keq%3D%5BO_2%5D%5E3)
c.
![Keq=\frac{[H_3O^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
d.
![Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
Regards!
44g per mole of CO2(1 mole=44g)
Answer:
Diffusion plays an essential function in cell as it's molecules move from higher to lower concentration and by which cell exchange toxic gases for life sustaining gases.
Explanation: