The reaction of benzaldehyde with acetone and sodium hydroxide produces DibenzalacetoneThis is an example of an aldol condensation reaction.
Chemical reactions often involve color changes, temperature changes, gas evolution, or precipitate formation. Simple examples of everyday reactions are digestion, combustion, and cooking. As the name suggests, simple reactants produce or synthesize more complex products. The basic form of a synthetic reaction is A + B → AB. A simple example of a synthetic reaction is the formation of water from its elements hydrogen and oxygen: 2 H2(g) + O2(g) → 2 H2O(g).
A physical reaction is a reaction in which a change in the physical properties of matter or substances occurs. Physical properties include density, mass, and volume. The definition of a physical reaction is a reaction in which molecules undergo molecular rearrangements but do not change chemically.
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Answer:
The oxidation state of the reducing agent has changed from 0 to +2.
Explanation:
reducing agent is anything that loses electron or gains oxygen
in this case, zinc
Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
Answer:
I hope it helps..... you can stop at the full-stop
Answer:
2.28 × 10^-3 mol/L
Explanation:
The equation for the equilibrium is
CN^- + H2O ⇌ HCN + OH^-
Ka = 4.9 × 10^-10
KaKb = Kw
4.9 × 10^-10 Kb = 1.00 × 10^-14
Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5
Now, we can set up an ICE table
CN^- + H2O ⇌ HCN + OH^-
I/(mol/L) 0.255 0 0
C/(mol/L) -x +x +x
E/(mol/L) 0.255 - x x x
Ka = x^2/(0.255 - x) = 2.05 × 10^-5
Check for negligibility
0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255
x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6
x = sqrt(5.20 × 10^-6) = 2.28 × 10^-3
[OH^-] = x mol/L = 2.28 × 10^-3 mol/L
