Answer:
0.35 g Li
Explanation:
H2SO4 + 2Li -> Li2SO4 + H2
7 g Li -> 2 g H2
x -> 0.10 g H2
x= (0.10 g H2 * 7 g Li)/ 2 g H2 x= 0.35 g Li
Answer:
Ca^2+(aq) + SO4^2-(aq) → CaSO4(s)
Explanation:
Step 1: The unbalanced equation
CaCl2(aq) + Na2SO4 (aq) → ...
Step 2: Balancing the equation
CaCl2(aq) + Na2SO4 (aq) → CaSO4 + NaCl
On the left side we have 2 times Cl (in CaCl2), On the right side we have 1 time Cl (in NaCl)
On the left side we have 2 times Na (in Na2SO4), on the right side we have 1 time Na (in NaCl)
To balanced the amount of Na and Cl we have to multiply NaCl, on the right side, by 2. Now the equation is balanced.
CaCl2(aq) + Na2SO4 (aq) → CaSO4(s) + 2NaCl(aq)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will look like this
Ca^2+(aq) + SO4^2-(aq) → CaSO4(s)
Answer:
The answer to your question is below
Explanation:
1.- It is a metal attach to an oxygen, then the name would be name of the metal + oxide.
Zinc oxide
2.- It is a metal attach to a non metal, then the name would be name of the metal + name of the non metal ending in "ide".
Manganese phosphide
3.- it is also a metal attach to a non metal, then the name would be name of the metal + name of the non metal with ending "ide".
Calcium phosphide
4.- K₂S
5.- PbBr₄
6.- Ba₃N₂