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ehidna [41]
3 years ago
12

Lesson 04.01: Chemical vs. Physical Changes and Properties Define and identify chemical and physical properties and changes. Giv

e examples of physical and chemical properties and changes. Compare and contrast chemical and physical properties and changes.
Chemistry
1 answer:
Sonbull [250]3 years ago
8 0

A chemical property and a physical property are related to chemical and physical changes of matter.

Answer: A physical property is an aspect of the matter that can be observed or measured without changing it. Examples of physical properties include color, molecular weight, and volume.

A chemical property may only be observed by changing the chemical identity of a substance.

This property measures the potential for undergoing a chemical change. Examples of chemical properties include reactivity, flammability and oxidation states.


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NikAS [45]

Answer:

Structures are given below.

Explanation:

  • Treatment of 2-bromo-2-methylbutane with KOH in ethanol will give elimination of HBr through E2 mechanism.
  • H atoms adjacent to Br will be eliminated.
  • 2-bromo-2-methylbutane has two possible adjacent H atoms that can be eliminated giving mixture of products.
  • Product of this elimination reaction is alkene. Here saytzeff fule is followed during elimination. So most substituted alkene will be major product.
  • Structure of alkenes are given below.

5 0
3 years ago
For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the
ladessa [460]

<span>Answer is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁<span>).
k</span>₁<span> = 0,000643 1/s.
k</span>₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

1/T₁<span> = 1/622 K = 0,0016 1/K.
1/T</span>₂<span> = 1/666 K = 0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>

4 0
3 years ago
Extended networks of pi bonds can cause a delocalization of electrons. Explain how this occurs.
algol [13]
The answer is: Electrons are shared in each pi bond. If the pi bonds flip back and forth between the adjacent p-orbitals on the two sides of an atom, the shared electrons in the p-orbitals can become delocalized.



Hope this help
7 0
3 years ago
Please help me someone ASAP!!!!
vagabundo [1.1K]

Answer:

Explanation:

g

6 0
2 years ago
The analysis of compound only magnesium, phosphorus and oxygen showed 36.23% MgO and 63.77 % P2O5. set up the simplest formula
Rzqust [24]

Answer:

dfgs

Explanation:

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2 years ago
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