Answer:
Experiments to determine mechanisms involve looking at indirect evidence to help support or disprove a proposed mechanism.
Most intermediates are not typically isolated to determine reaction mechanisms.
Carbocations are very reactive and are typically not isolated for analysis.
Scientists can prove that a specific mechanism exists.
Evidence of intermediates sometimes can be seen using techniques such as nuclear magnetic resonance spectroscopy
Explanation:
The study of reaction mechanism and chemical kinetics often form the main thrust of study in organic, inorganic and physical chemistry.
We often want to know the actual processes involved in the conversion of one specie to another. Unfortunately, this information may have to be obtained indirectly by certain chemical reactions or by the use of new instrumental methods such as nuclear magnetic resonance spectroscopy.
Many organic reactions have carbocation intermediates. These carbocations are relatively short-lived and are transient intermediates which are rarely isolated unless they are isolated in a molecular cage using a macromolecule or in superacids.
By intensive study, scientists can proof or disprove the authenticity of any proposed mechanism.
We must know that a transition state has partial bonds. It is often an extremely short-lived specie which cannot be isolated.
Answer: It's frequency also increases.
Explanation:
The sound is perceived as louder if the amplitude increases, and softer if the amplitude decreases. ... The amplitude of a wave is related to the amount of energy it carries. A high amplitude wave carries a large amount of energy; a low amplitude wave carries a small amount of energy.
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<h3><u>Answer;</u></h3>
0.5 M HBr, pOH = 13.5 ; Has the lowest pH
<h3><u>Explanation;</u></h3>
From the question;
pH = -Log [OH]
or pH = 14 - pOH
Therefore;
For 0.5 M HBr
[H+] = 0.5 M
pH = - Log [0.5]
= 0.30
For; pOH = 13.5
pH = 14 - pOH
= 14 -13.5
= 0.5
For; 0.05 M HCl
pH = - log [H+]
[H+] = 0.05
pH = - Log [0.05]
= 1.30
For; pOH = 12.7
pH = 14 -pOH
= 14 -12.7
= 1.30
For; 0.005 M KOH,
pOH = - log [OH]
[OH-] = 0.005
pOH = - Log 0.005
= 2.30
pH = 14 - 2.30
= 11.7
For; pOH = 2.3
pH = 14 -pOH
= 14- 2.3
= 11.7
Answer:
Freezing T° of solution = - 8.98°C
Explanation:
We apply Freezing point depression to solve this problem, the colligative property that has this formula:
Freezing T° of pure solvent - Freezing T° of solution = Kf . m
Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water
m = molality (moles of solute / 1kg of solvent)
We convert the mass of solvent from g to kg
1.5 g . 1kg/1000g = 0.0015 kg
We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g
0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles
Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m
- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent
-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C
Freezing T° of solution = - 8.98°C
