3 years ago

Why are alkali metals more reactive than alkaline earth metals?

1 answer:

5 0

A: It takes more energy to remove two valence electrons from an atom than one valence electron. This makes alkaline Earth metals with their two valence electrons less reactive than alkali metals with their one valence electron.

You might be interested in

Can rocks include organic materials, but minerals cannot true or false

Alexus [3.1K]

Answer:

false

Explanation:

Because rocks cannot include organic materials

6 0

3 years ago

Brass and bronze are both (space) of copper because copper is the (space)

Vanyuwa [196]

I believe the first space is 1. alloys, and the second space is also 1. main. I don't know for sure but I hope this helps.

8 0

3 years ago

Read 2 more answers

Use your periodic table of elements. Take any element from the second column, except for beryllium and calcium, and answer the f

Alja [10]

Answer:

Radium (Ra)

Metal

Alkaline Earth Metal

Protons & Electrons: 88

Neutrons: 138

Explanation:

6 0

3 years ago

Read 2 more answers

How many protons, electrons, and neutrons does the following isotope contain? 79Br+

stellarik [79]

The number of protons = atomic number in Periodic table, so 35

The number of neutrons = mass number - number of protons = 79-35=44

The number of electrons = atomic number - 1 = 35-1 =34

6 0

3 years ago

Read 2 more answers

Considering the following precipitation reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq),

patriot [66]

Answer: The ions which are not spectator ions are $Pb^{2+}$ and $I^-$

Explanation:

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The given chemical equation is:

$Pb^{2+}(aq)+NO_3^-(aq)+2K^+(aq)+2I^-(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^-(aq)$

The ions which are present on both the sides of the equation are $K^+$ and $NO_3^-$ and are spectator ions.

Thus the net ionic equation is:

$Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)$

Hence, the correct answer is $Pb^{2+}$ and $I^-$

5 0

3 years ago

Read 2 more answers