Answer:
oxygen is produced at the anode and hydrogen gas is produced at the cathode
Explanation:
Answer:
if you know what they have in common please comment what here
Explanation:
Answer: 2.17 g of bromide product would be formed
Explanation:
The reaction of calcium bromide with lithium oxide will be:
![CaBr_2+Li_2O\rightarrow 2LiBr+CaO](https://tex.z-dn.net/?f=CaBr_2%2BLi_2O%5Crightarrow%202LiBr%2BCaO)
To calculate the moles :
![\text{Moles of calcium bromide}=\frac{2.5g}{199.9g/mol}=0.0125moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20calcium%20bromide%7D%3D%5Cfrac%7B2.5g%7D%7B199.9g%2Fmol%7D%3D0.0125moles)
As lithium oxide is in excess, calcium bromide is the limiting reagent.
According to stoichiometry :
1 mole of
produce = 2 moles of ![LiBr](https://tex.z-dn.net/?f=LiBr)
Thus 0.0125 moles of
will require=
of ![NH_3](https://tex.z-dn.net/?f=NH_3)
Mass of ![LiBr=moles\times {\text {Molar mass}}=0.025moles\times 86.8g/mol=2.17g](https://tex.z-dn.net/?f=LiBr%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.025moles%5Ctimes%2086.8g%2Fmol%3D2.17g)
Thus 2.17 g of bromide product would be formed
Answer:The oxidation state of carbon in NaHCO3 is +4. Sodium has a +1 oxidation state.
Explanation:Hydrogen has a +1 oxidation state. Oxygen has a -2 oxidation state