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3 years ago

Write METAL, NONMETAL, DIATOMIC MOLECULE ,METALLOID

1 answer:

3 0

Alkali metal - Group 1 metal
Alkaline earth metal - Group 2 metal
Iron triad - metal
Halogens - most are diatomic and are nonmetals - Cl2, Br2, I2
Noble gases - nonmetals and are not reactive due to full valence shell.
Boron group - metalloid
Oxygen group - oxygen molecule - a diatomic molecule composed of oxygen atoms, which are nonmetals.

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Name an element or elements in the periodic table that you would expect to be chemically similar to potassium. check all that ap

Inga [223]

It would have to be all elements in the same family, so since potassium is an Alkali Metal the similar chemicals would be...Lithium, Sodium, Rubidium, Cesium, Francium.

7 0

3 years ago

An ionic solid is placed in water. which information is described by the solubility product constant?

Illusion [34]

Answer:

$\boxed{\text{C. the equilibrium between the solid and its ions in solution}}$

Explanation:

$\rm MX(s) $\, \rightleftharpoons \,$ M$^{+}$(aq) + $^{-}$(aq); $K_{\text{sp}}$ = [M$^{+}$][X$^{-}$]\\\\\text{$K_{\text{sp}}$ gives us information on}\\\\\boxed{\textbf{ the equilibrium between the solid and its ions in solution}}$

It tells us nothing about the amount of precipitate that will form or the temperature at which the equilibrium occurs.

8 0

3 years ago

Read 2 more answers

What types of elements- metals,

crimeas [40]

Answer:

Noble gases

Explanation:

The noble gases are non-metals that requires the highest amount of energy to remove an electron from their shells.

The reason for this difficult is that their electronic configuration confers a stable configuration them.

The ionization energy is the energy required to remove the most loosely held electrons in an atom.
Due to the special stability of noble gases, it is very difficult to remove electrons from an atom of noble gases.

7 0

3 years ago

What is the bond order of li2−? express the bond order numerically?

jeka94

Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (ABMO). Bond Order is calculated as,

Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2

Bond Order = (4 - 3) ÷ 2

Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2
Or,
Bond Order = 0.5

4 0

3 years ago

Given the partial equation:

Nikolay [14]

Answer : The balanced chemical equation in acidic medium will be,

$IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$IO_3^-(aq)+Sn^{2+}(aq)\rightarrow I^-(aq)+Sn^{4+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-$

First balance the main element in the reaction.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-$

Now balance oxygen atom on both side.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-+6H^+\rightarrow I^-+3H_2O$

Now balance the charge.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}+2e^-$

Reduction : $IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O$

The charges are not balanced on both side of the reaction. Thus, we are multiplying oxidation reaction by 2 and the adding both equation, we get the balanced redox reaction.

Oxidation : $2Sn^{2+}\rightarrow 2Sn^{4+}+4e^-$

Reduction : $IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O$

The balanced chemical equation in acidic medium will be,

$IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)$

5 0

3 years ago