Answer:
Reduction
Explanation:
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Mn⁺⁷ +3e⁻ → Mn⁴⁺
Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.
Examples:
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Proton and neutron, which are both approximately 1 amu
Answer:
1 and 2
Trigonal planar
Tetrahedral
Trigonal Planar
Linear
Bent ( v- shape)
Explanation:
The highlighted atoms has a delocalized lone pair - 1 and 2
All sp2-hybridized carbon atoms have geometry- Trigonal planar
All sp3-hybridized carbon atoms have geometry - Tetrahedral
The nitrogen atom has geometry - Trigonal Planar
The oxygen atom of the C=O bond has geometry - Linear
and the other oxygen atom has geometry - Bent ( v- shape)
P.S - The correct question is -
Every science experiment should follow the basic principles of proper investigation so that the results presented at the end are seen as credible.
Observation and Hypothesis. ...
Prediction and Modeling. ...
Testing and Error Estimation. ...
Result Gathering and Presentation. ...
Conclusions. ...
Law Formation.
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L
2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)
V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)
V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}
V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
