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3 years ago

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Be sure to answer all parts. Show the overall reaction for formation of racemic 3−bromohexane from (E)−3−hexene by entering the

structures of the products and reactants in the template provided. Part 1 out of 3 draw structure ... arr draw structure ... + draw structure ... (E)−3−hexene (S)−3−bromohexane (R)−3−bromohexane Racemic product = 50% (S) and 50% (R)

1 answer:

natima [27]3 years ago

3 0

Answer:

(E)-3-hexene treated with bromine to form racemic mixture.

Explanation:

(E)-3-hexene treated with bromine to form (S)−3−bromohexane and (R)−3−bromohexane.

The chemical reaction is as follows.

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3 years ago

How many moles of oxygen are needed to burn 2.7 moles of methyl

QveST [7]

Answer:

3.38 moles of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4CH2OH + 5O2 → 4CO2 + 6H2O

From the balanced equation above,

4 moles of CH2OH required 5 moles of O2 for complete combustion.

Finally, we shall determine the number of mole of O2 needed to react with 2.7 moles of CH2OH. This can be obtained as follow:

From the balanced equation above,

4 moles of CH2OH required 5 moles of O2 for complete combustion.

Therefore, 2.7 moles of CH2OH will require = (2.7 × 5)/4 = 3.38 moles of O2 for complete combustion.

Thus, 3.38 moles of O2 is required.

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3 years ago

Which of the following would be a value measured using a scale in lab?

Neko [114]

Answer:

A. Actual Yield

Since all of the others are calculated not measured outright.

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3 years ago

For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The

Alik [6]

Answer : The rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-2}M^{-2}s^{-1}$

Thus, the value of the rate constant 'k' for this reaction is $7.5\times 10^{-2}M^{-2}s^{-1}$

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1$

$\text{Rate}=4.25\times 10^{-2}Ms^{-1}$

Therefore, the rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

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3 years ago

1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?

Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

<u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L

V (final volume) = 473 mL → 0.473 L

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)

P (final pressure) = ? (in atm)

<u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

$P_0*V_0 = P*V$

$1*1 = P*0.473$

$1 = 0.473\:P$

$0.473\:P = 1$

$P = \dfrac{1}{0.473}$

$\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}$

<u><em>Answer: </em></u>

<u><em>The new pressure of the gas is 2.11 atm </em></u>

___________________________________

$\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

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3 years ago