Upper M g (s) plus 2 upper H right arrow upper M g (a q) plus upper H subscript 2 (g).
Mg + 2H⁺ → Mg²⁺ + H₂
Explanation:
The net ionic equation is the sum of the oxidation half reactions.
The half-reactions are:
Mg → Mg²⁺ + 2e⁻ Oxidation half reaction
2H⁺ + 2e⁻ → H₂ reduction half reaction
Combining the reactions gives:
Mg + 2H⁺ + 2e⁻ → H₂ + Mg²⁺ + 2e⁻
Since the electrons are balanced, they cancel out of the equation,
This leaves behind;
Mg + 2H⁺ → Mg²⁺ + H₂
Solution:
Ionic equation brainly.com/question/2947744
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Answer:
Fluorine, chlorine, bromine and iodine are grouped together because they have the same number of valence electrons.
Explanation:
Fluorine, chlorine, bromine and iodine are all found in group 7, which is the second-last column from the right. Group 7 elements are also called the "Halogens" family.
The group number also tells you the number of valence electrons that the elements have in that group. Valence electrons mean the outermost electrons (See picture).
For example, fluorine has two shells (the circles with dots on them). The outermost electrons, or valence electrons, are the dots on the biggest circle. There are 7 dots, so there are 7 valence electrons, which corresponds with Group "7".
A full shell (except the for first shell) is when there are 8 dots. Since 7 is so close to 8, Halogens are very reactive.
Fluorine, chlorine, bromine and iodine all have 7 valence electrons and are in the Halogens family, which are very reactive.
Reaction of Cu(NO₃)₂ with each salt is as follow,
1) with KNO₃;
Cu(NO₃)₂ + KNO₃ → Cu(NO₃)₂ + KNO₃
Both salt products are water soluble.
2) With CuSO₄;
Cu(NO₃)₂ + CuSO₄ → CuSO₄ + Cu(NO₃)₂
Again both Salt products are water soluble.
3) With K₂SO₄;
Cu(NO₃)₂ + K₂SO₄ → CuSO₄ + 2 KNO₃
Again both salt products are water soluble.
4) With K₂S;
Cu(NO₃)₂ + K₂S → CuS + 2 KNO₃
In this case CuS is water insoluble, hence precipitates out.
Result:
Option-4 is the correct answer.
A conductor transmits heat and electrical good.
Answer:
A. Methanol
B. 2-chloropropan-1-ol
C. 2,2-dichloroethanol
D. 2,2-difluoropropan-1-ol
Explanation:
Primary alcohols are stronger acids than secondary alcohols which are stronger than tertiary alcohols.
This trend is so because of the stability of the alkoxide ion formed(stabilising the base, increases the acidity). A more stabilised alkoxide ion is a weaker conjugate base (dissociation of an acid in water).
By electronic factors, When there are alkyl groups donating electrons, the density of electrons on th O- will increase a d thereby make it less stable.
By stearic factors, More alkyl group bonded to the -OH would mean the bulkier the alkoxide ion which would be harder to stabilise.
Down the group of the periodic table, basicity (metallic character) decreases as we go from F– to Cl– to Br– to I– because that negative charge is being spread out over a larger volume that is electronegativity decreases down the group.
Electronegative atoms give rise to inductive effect and a decrease in indutive effects leads to a decrease in acidity. Therefore an Increasing distance from the -OH group lsads to a decrease in acidity.
From above,
A. Methanol
B. 2-chloropropan-1-ol
C. 2,2-dichloroethanol
D. 2,2-difluoropropan-1-ol
