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1 year ago

Draw structures from the following names, and determine which compounds are optically active:(c) 1,2-dibromo-2-methylbutane

Chemistry

1 answer:

dolphi86 [110]1 year ago

8 0

The given compound 1,2-dibromo-2-methylbutane is an optically active compound .

Because this compound does not have plane of symmetry (POS) and center of symmetry (COS) i.e. does not have di-symmetry . And also forms non superimposable mirror image . the compound is optically active .

It has chiral center.

Here , the chiral carbon is attached to the 4 distinct groups such as : methyl , ethyl , bromine , bromomethane .

<h3>What is di-symmetry?</h3>

Di-symmetry is that which have no center of symmetry and plane of symmetry and alternate axis of symmetry .

<h3>Chiral center :</h3>

Have Sp3 hybridized center (4sigma bond ) .

4 distinct group is attached to the chiral atom. form non -superimposable mirror image .

<h3>What is optical isomerism ?</h3>

Same molecular formula and same structural formula . also have same physical and chemical properties .

They differ in their behavior towards plane polarized light (ppl) .

Learn more about chiral center here:

brainly.com/question/9522537

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2 years ago

Increasing the concentration of reactants generally increases the rate of a reaction because particles

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3 years ago

10 To become positively charged, an atom must: A Galn a proton B Lose a proton C С. Galn an electron D Lose an electron }

Studentka2010 [4]

Answer:

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Explanation:

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3 years ago

The equilibrium constant, Kc, for the following reaction is 5.10×10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibri

levacccp [35]

<u>Answer:</u> The equilibrium concentration of HCl is $2.26\times 10^{-3}M$

<u>Explanation:</u>

We are given:

Moles of $NH_4Cl(s)$ = 0.564 moles

Volume of vessel = 1.00 L

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Molarity of $NH_4Cl=\frac{0.564}{1}=0.564M$

The given chemical equation follows:

$NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)$

<u>Initial:</u> 0.564

<u>At eqllm:</u> 0.564-x x x

The expression of $K_c$ for above equation follows:

$K_c=[NH_3][HCl]$

The concentration of pure solid and pure liquid is taken as 1.

We are given:

$K_c=5.10\times 10^{-6}$

Putting values in above equation, we get:

$5.10\times 10^{-6}=x\times x\\\\x=2.26\times 10^{-3}M,-2.26\times 10^{-3}M$

Negative sign is neglected because concentration cannot be negative.

So, $[HCl]=2.26\times 10^{-3}M$

Hence, the equilibrium concentration of HCl is $2.26\times 10^{-3}M$

5 0

3 years ago

Question 11 (10 points)

andrey2020 [161]

Answer:

The correct answer is option a.

Explanation:

When aluminum hydroxide reacts with of nitrous acid it gives of aluminum nitrite and of water.

$Al(OH)_3+3HNO_2\rightarrow Al(NO_2)_3+3H_2O$

According to above reaction ,when 1 mole of aluminum hydroxide reacts with 3 moles of nitrous acid it gives 1 mole of aluminum nitrite and 3 moles of water.

Hence, the correct answer is option a.

4 0

3 years ago