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EleoNora [17]
3 years ago
5

You replicate the CEC analysis of a secondary alcohol you performed in the lab using a reverse phase TLC plate producing the TLC

plate below. Based on this TLC plate and the mnemonic in your notes, what can you conclude about the stereochemistry of the alcohol? (1 pts)
Chemistry
1 answer:
Dafna1 [17]3 years ago
4 0

Answer:

Answer for the question:

You replicate the CEC analysis of a secondary alcohol you performed in the lab using a reverse phase TLC plate producing the TLC plate below. Based on this TLC plate and the mnemonic in your notes, what can you conclude about the stereochemistry of the alcohol? (1 pts)

is given below which explains the best option for the answer.

Explanation:

The enantiomer of the alcohol cannot be determined.

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How many liters of hydrogen are required to react completely with 2.4L of oxygen to form water? 2H2 + O2 --> 2H2O
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Answer:

2.4 mole of oxygen will react with 2.4 moles of hydrogen

Explanation:

As we know

1 liter = 1000 grams

2H2 + O2 --> 2H2O

Weight of H2 molecule = 2.016 g/mol

Weight of water = 18.01 gram /l

2 mole of oxygen react with 2 mole of H2

2.4 mole of oxygen will react with 2.4 moles of hydrogen

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Consider four elements from Group 17: fluorine in the second period, chlorine in the third period, bromine in the fourth period,
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Which of the following traits are inherited from parents when the genetic material found in genes is passed to offspring on the
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The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall
miv72 [106K]

Answer:

The value is  C_p  = 42. 8 J/K\cdot mol

Explanation:

From the question we are told that  

     \gamma = \frac{C_p }{C_v}

The  initial volume of the  fluorocarbon gas is  V_1 = V

 The final  volume of the fluorocarbon gas isV_2 = 2V

  The initial  temperature of the fluorocarbon gas is  T_1  =  298.15 K

  The final  temperature of the fluorocarbon gas is T_2  =  248.44 K

   The initial  pressure is P_1  = 202.94\  kPa

    The final   pressure is  P_2  =  81.840\  kPa

Generally the equation for  adiabatically reversible expansion is mathematically represented as

       T_2 =  T_1  * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }

Here R is the ideal gas constant with the value  

        R =  8.314\  J/K \cdot mol

So  

   248.44 =   298.15  * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }

=> C_v  =  31.54 J/K\cdot mol

Generally adiabatic reversible expansion can also be mathematically expressed as

    P_2 V_2^{\gamma} = P_1 V_1^{\gamma}

=>[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}    

=>  2^{\gamma} =  2.56

=>    \gamma =  1.356

So

     \gamma  =  \frac{C_p}{C_v} \equiv  1.356 = \frac{C_p}{31.54}

=>    C_p  = 42. 8 J/K\cdot mol

3 0
3 years ago
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