Answer:
The odor of a substance is a physical property. That would be your answer.
Explanation:
Physical Properties
Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:
color (intensive)
density (intensive)
volume (extensive)
mass (extensive)
boiling point (intensive): the temperature at which a substance boils
melting point (intensive): the temperature at which a substance melts
Chemical Properties
Remember, the definition of a chemical property is that measuring that property must lead to a change in the substance’s chemical structure. Here are several examples of chemical properties:
Heat of combustion is the energy released when a compound undergoes complete combustion (burning) with oxygen. The symbol for the heat of combustion is ΔHc.
Chemical stability refers to whether a compound will react with water or air (chemically stable substances will not react). Hydrolysis and oxidation are two such reactions and are both chemical changes.
Flammability refers to whether a compound will burn when exposed to flame. Again, burning is a chemical reaction—commonly a high-temperature reaction in the presence of oxygen.
The preferred oxidation state is the lowest-energy oxidation state that a metal will undergo reactions in order to achieve (if another element is present to accept or donate electrons).
Answer:
NH3
Explanation:
NH3 molecules make strong intermolecular H bondings and also because of the increase in the size of atomic radius.
Answer:
D They have the same properties.
Explanation:
The same no. of isotopes doesnot describes because isotopes have same symbol but didderent atomic mass.
Answer:
26,000 KL/kg will be the density of the Elephant.
Answer:
1.750 M
Explanation:
In case of titration , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = concentration of acid ,
V₁ = volume of acid ,
M₂ = concentration of base,
V₂ = volume of base .
from , the question ,
M₁ = ? M
V₁ = 7.14 mL
M₂ = 0.250 M
V₂ = 50 mL
Using the above formula , the molarity of acid , can be calculated as ,
M₁V₁ = M₂V₂
M₁ * 7.14 mL = 0.250 M * 50mL
M₁ = 1.750 M
Hence, the molarity of the hydrochloric acid = 1.750 M
