Answer:
Here are a few more examples:
Smoke and fog (Smog)
Dirt and water (Mud)
Sand, water and gravel (Cement)
Water and salt (Sea water)
Potassium nitrate, sulfur, and carbon (Gunpowder)
Oxygen and water (Sea foam)
Petroleum, hydrocarbons, and fuel additives (Gasoline)
Heterogeneous mixtures possess different properties and compositions in various parts i.e. the properties are not uniform throughout the mixture.
Examples of Heterogeneous mixtures – air, oil, and water, etc.
Examples of Homogeneous mixtures – alloys, salt, and water, alcohol in water, etc.
Explanation:
Answer:
(NH₄)₂SO₃
Explanation:
Data Given:
formula for "ammonium sulfite”
Solution
Ammonium sulfite consist of following atoms
Nitrogen = N
Hydrogen = H
Sulfur = S
Oxygen = O
Now we will look for the symbols of ammonium and sulfite
Ammonium = NH₄⁺
ammonium consists one nitrogen and 4 hydrogen atoms. it have overall 1 positive charge.
Sulfite = SO₃²⁻
it have one sulfur and 2 oxygen atoms. the overall charge is 2-.
Now we will write Both ammonium and sulfite ions together. the charges will are the combining power and will be written at the base of the alternate position formula unit.
(NH₄)₂SO₃
So 2 ammonium ions combine with one sulfite ion.
So the formula for “ammonium sulfite” is (NH₄)₂SO₃
Answer:
ΔH°rxn = - 162.5 kJ
Explanation:
Hello,
In this case, we use the Hess law to compute the required enthalpy of the reaction for the chlorine with ozone.
1. At first, we invert the original (1) equation in order to place the chlorine at the reactants, so the enthalpy sign is inverted to positive:
Cl ( g ) + 2O2 ( g ) ⟶ClO ( g ) + O3 ( g ) ΔH°rxn = 122.8 kJ
2. Then we do not modify the second reaction:
2O3 ( g ) ⟶ 3O2 ( g ) ΔH°rxn = −285.3 kJ
Next, the add the aforementioned reactions:
Cl ( g ) + 2O2 ( g ) + 2O3 ( g ) ⟶ ClO ( g ) + O3 ( g ) + 3O2 ( g )
In order to obtain (3):
O3 ( g ) + Cl ( g ) ⟶ ClO ( g ) + O2 ( g )
And the enthalpy of reaction results:
ΔH°rxn = 122.8 kJ − 285.3 kJ
ΔH°rxn = - 162.5 kJ
Best regards.
Answer: Family 16 is most likely to accept 2 electrons.
Explanation:
Family 16 has the electronic configuration 
. As to attain stability the p orbital will readily accept 2 electrons.
Therefore, family 16 is most likely to accept 2 electrons. Family 16 is also known as chalcogens.
Whereas, family 2 has the electronic configuration 
. As the s orbital is completely filled so there is no need to accept 2 electrons.
Family 12, and family 18 has the electronic configuration ![[Ar] nd^{10} ns^{2}](https://tex.z-dn.net/?f=%5BAr%5D%20nd%5E%7B10%7D%20ns%5E%7B2%7D)
and ![[He]ns^{2} np^{6}](https://tex.z-dn.net/?f=%5BHe%5Dns%5E%7B2%7D%20np%5E%7B6%7D)
respectively.
Both family 12 and 18 has completely filled orbitals so it will not accept any electrons.
Chlorine goes from Cl₂ to ClO⁻.
Oxidation state of Cl in Cl₂ is zero whereas in ClO⁻ it is +1. Hence change in oxidation state of Chlorine is 1.
Calculation of oxidation number of Cl in ClO⁻ or HClO⁻ :
Let x be oxidation number of Cl in ClO⁻ the. Now since the net charge on ClO⁻ is -1, sum of oxidation number of all must be equal to -1.
Therefore,
x + (-2) = -1 .....[ oxidation number of O is -2]
∴ x = 2-1 = +1
Therefore oxidation number of Cl in ClO⁻ is +1
