Answer:
group I: alkali metals
group II (beryllium to radium): alkaline earth metals
group II (scandium to zinc , yttrium to cadmium, lanthanoid series to mercury, actinoid series to copernicium): transition metals
group VII (fluorine to astatine): halogens
group VIII: (helium to radon): noble gases
Explanation:
Answer:
Molarity of 40 ml of NaoH is 0.3125 mL
Explanation:
As we know
Molarity of acid * volume of acid = molarity of base * volume of base
Substituting the given values, we get

Molarity of 40 ml of NaoH is 0.3125 mL
Answer:
The answer is 0.01 M
Explanation:
The problem is solved by applying the expression for ionic product of water as follows:
Kw = [H₃O⁺] [OH⁻]
Where Kw is the ionic product of water and it is 1.10⁻¹⁴ M at⁴ 25ºC. As [OH⁻]= 1.10⁻¹² M, [H₃O⁺] will be:
[H₃O⁺]= Kw/ [OH⁻]= 1.10⁻¹⁴M/1.10⁻¹²M= 0.01 M
Answer:
sp²
Explanation:
You need to look at how many electron orbitals around the atom. Looking at the structure below, you can see that there are three electron orbitals. This gives you an sp² hybridization.