Answer:
They are all alkali earth metals.
Explanation:
Their valence shell each has 2 electrons. Also, they are all shiny, silvery-white, somewhat reactive metals at standard temperature and pressure. They form alkaline solutions, hydroxides, when reacting with water and their oxides are found in the earth’s crust.
You are exposed to nuclear radiation every day is true!
Answer:
H₂S(aq)+ 2 LiOH(aq) → Li₂S(aq) + 2 H₂O(l)
6 HI(aq) + 2 Al(s) → 2 AlI₃(aq) + 3 H₂(g)
2 H₂SO₄(aq) + TiO₂(s) → Ti(SO₄)₂(aq) + 2 H₂O(l)
H₂CO₃(aq) + 2 LiOH(aq) → Li₂CO₃(aq) + 2 H₂O(l)
Explanation:
H₂S(aq)+ 2 LiOH(aq) → Li₂S(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
6 HI(aq) + 2 Al(s) → 2 AlI₃(aq) + 3 H₂(g)
This is a single displacement reaction.
2 H₂SO₄(aq) + TiO₂(s) → Ti(SO₄)₂(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
H₂CO₃(aq) + 2 LiOH(aq) → Li₂CO₃(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
Given the following:
Concentration of residual (NH4)+ = 0.12 - 0.03 = 0.09 mol / L
Concentration of created NH3 = 0.03 mol / L
The (NH4) + reacts with all (OH)- to form NH3 and H2O ( Water )
The resultant solution is NH4+ / NH3 buffer system in which (NH4)+ is acid and the NH3 is base or as called the salt.
For a buffer system :
pH = pKa + Log (acid) / (salt)
= 9.25 + Log [ (NH4)+ ] / [ NH3 ]
= 9.25 + Log [ 0.09 ] / [ 0.03 ]
= 9.25 + Log 3
= 9.25 + 0.4771
pH = 9.7271 or 9.73 which is rounded up to 2 decimal places
