Based on what you know about the unknown metal sample propose a balanced chemical equation for (4 points) a. the formation of the Group 1 metal oxide b. the formation of the Group 2 metal oxide

Group 1.

2Na(s) + O₂(g) → Na₂O(g) ;  2K(s) + O₂(g) → K₂O(g)

4Li (s)+ O₂(g) →   2Li₂O(g) ;  3 Rb(s) + RbO₂(g) → 2Rb₂O(g)

2Fr(s) + O₂(g) → Fr₂O(g) ;  2Cs+ O₂(g)  → Cs₂O(g)

Group 2.

2Mg(s) + O₂(g) → 2MgO(s) ; 2Be(s) + O₂(g) → 2BeO(s)

2Ca(s) + O₂(g) → 2CaO(s) ; 2Sr(s) + O₂(g) → 2SrO(s)

2Ba(s) + O₂(g) → 2BaO(s) ; 2Ra(s) + O₂(g) → 2RaO(s)

Explanation:

Elements from group 1 are: Li, Na, K, Rb, Cs and Fr

When they react to oxygen, they produce the correspond oxide.

2Na(s) + O₂(g) → Na₂O(g)

2K(s) + O₂(g) → K₂O(g)

4Li (s)+ O₂(g) →   2Li₂O(g)

When Rb reacts with the oxygen, it makes the superoxide

Rb(s)+ O₂(g) → RbO₂(g)

The superoxide reduces to Rb₂O by excess of Rb, according to:

3 Rb(s) + RbO₂(g) → 2Rb₂O(g)

2Fr(s) + O₂(g) → Fr₂O(g)

2Cs+ O₂(g)  → Cs₂O(g)

In the presence of excess oxygen, Na and K forms peroxide

2Na(s) + O₂(g) → Na₂O₂(g)

2K(s) + O₂(g) → K₂O₂(g)

K also forms, the peroxide when it burns into the air

K(s) + O₂(g) → KO₂

Group 2. 2Mg(s) + O₂(g) → 2MgO(s)

2Be(s) + O₂(g) → 2BeO(s) It does not react in water, so we can't produce, the Be(OH)₂ but the hidroxide surely, can be decomposed:

2Be(OH)₂ → 2BeO + 2H₂O

2Ca(s) + O₂(g) → 2CaO(s)

2Sr(s) + O₂(g) → 2SrO(s)

2Ba(s) + O₂(g) → 2BaO(s)

2Ra(s) + O₂(g) → 2RaO(s)

under a microscope, cancer cells may look very different from normal cells. they often have different sizes and some may be larger than normal while others are smaller. cancer cells are often abnormally shaped and the control centre of the cell (the nucleus) may have an abnormal appearance.

i think? ?

a electronegativity decreases beacuse the size of nucleus is bigger.