Answer:
H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)
Explanation:
H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)
A careful observation of the equation above, shows that the equation is already balanced.
To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:
H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)
Fireworks owe their colors to reactions of combustion of the metals present. When Mg and Al burn, they emit a white bright light, whereas iron emits a gold light. Besides metals, oxygen is necesary for the combustion. The decomposition reactions of barium nitrate and potassium chlorate provide this element. At the same time, barium can burn emitting a green light.
(a) Barium nitrate is a <em>salt</em> formed by the <em>cation</em> barium Ba²⁺ and the <em>anion</em> nitrate NO₃⁻. Its formula is Ba(NO₃)₂. Potassium chlorate is a <em>salt</em> formed by the <em>cation</em> potassium K⁺ and the <em>anion</em> chlorate ClO₃⁻. Its formula is KClO₃.
(b) The balanced equation for the decomposition of potassium chloride is:
2KClO₃(s) ⇄ 2KCl(s) + 3O₂(g)
(c) The balanced equation for the decomposition of barium nitrate is:
Ba(NO₃)₂(s) ⇄ BaO(s) + N₂(g) + 3O₂(g)
(d) The balanced equations of metals with oxygen to form metal oxides are:
- 2 Mg(s) + O₂(g) ⇄ 2 MgO(s)
- 4 Al(s) + 3 O₂(g) ⇄ 2 Al₂O₃(s)
- 4 Fe(s) + 3 O₂(g) ⇄ 2 Fe₂O₃(s)
The options
Select one:
a. a 3- ion forms.
b. the noble gas configuration of argon is achieved.
c. the result is a configuration of 1s2 2s2 2p6.
d. the atom gains five electrons.
Answer:
c. the result is a configuration of 1s2 2s2 2p6.
Explanation:
Aluminium atom has atomic number of 13 , hence the number of electron is 13 for a neutral atom of aluminium. When aluminium atom reacts with other elements it usually gives out three electron to attain the octet configuration.
The cation representation of aluminium is Al3+ because it has loss three electron to attain the octet rule. Aluminium will be left with 10 electrons after losing 3 of it electrons. The electronic configuration will be represented as follows after losing three electrons;
1S² 2S² 2P∧6 .
At this stage the octet rule has been achieved as it will be represented as
2 8. The first energy shell now contains two electron and the second energy shell contains 8 electrons.
The configuration of Neon has been formed in the process.
Could I hear the options..
(P1V1/T1)=(P2V2/T2)
I just took the test and this should be correct
