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3 years ago

Draw 1,2,3,4,5,6-hexachlorocyclohexane with :

Chemistry

1 answer:

nevsk [136]3 years ago

4 0

Answer:

This is required answer.

Explanation:

Given that,

1,2,3,4,5,6-hexachlorocyclohexane

(a). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions.

When we say that all the chloro groups in axial position that means axial bonds are vertical.

(b). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions.

When we say that all the chloro groups in equatorial position that means axial bonds are horizontal.

Hence, This is required answer.

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Explanation:

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3 years ago

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How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction

nekit [7.7K]

Answer : The volume of $NaOH$ required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?$

Putting values in above equation, we get:

$2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL$

Hence, the volume of $NaOH$ required to neutralize is, 340 mL

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