type the question please. The picture is to hard to see
Answer:
+2
Explanation:
Alkaline earth metals are present in group 2 of periodic table. There are six elements in second group. Beryllium, magnesium, calcium, strontium, barium and radium.
All have two valance electrons.
Electronic configuration of Beryllium:
Be = [He] 2s²
Electronic configuration of magnesium.
Mg = [Ne] 3s²
Electronic configuration of calcium.
Ca = [Ar] 4s²
Electronic configuration of strontium.
Sr = [Kr] 5s²
Electronic configuration of barium.
Ba = [Xe] 6s²
Electronic configuration of radium.
Ra = [ Rn] 7s²
They are present in group two and have same number of valance electrons (two valance electrons) and show oxidation state +2 by loosing two valance electrons. They also show similar reactivity.
They react with oxygen and form oxide.
2Ba + O₂ → 2BaO
2Mg + O₂ → 2MgO
2Ca + O₂ → 2CaO
this oxide form hydroxide when react with water,
BaO + H₂O → Ba(OH)₂
MgO + H₂O → Mg(OH)₂
CaO + H₂O → Ca(OH)₂
With sulfur,
Mg + S → MgS
Ca + S → CaS
Ba + S → BaS
Dissolution means to make the compound apart, So when we have ionic compounds like NaI which has metal and non-metal ions, It separates into parts of positive ions and negative ions. After we separate this compound apart we will put the charge of each on above its symbol and then start to balance the equation of the dissolution.
So the dissolution equation of NaI is:
NaI(s) → Na^+(s) + I^-(Aqu)
Answer:
ano po yan? hind ko po maintindihan
Answer:
The substance that remained on the filter paper is Al(OH).
Explanation:
- Filter paper is the substance that is used in laboratory to separate the solid objects. It doesn't filter aqueous and gaseous products.
- So in our experiment the product formed is aluminum hydroxide Al(OH) and sodium chloride (NaCl).
- Between two products, sodium chloride is in aqueous form as indicated in the question. So it wont remain in the filter paper.
- Hence aluminum hydroxide being only solid product remains on the filter paper.