How was a first person how made earth

g In the elimination reaction of t-butanol and t-butyl bromide. A. They have a common intermediate and require a strong base B.

11111nata11111 [884]

Answer:

Both reactions share a common intermediate and differ only in the leaving group

Explanation:

The elimination reaction of tertiary alkyl halides usually occur by E1 mechanism. In E1 mechanism, the substrate undergoes ionization leading to the loss of a leaving group and formation of a carbocation.

Loss of a proton from the carbocation completes the reaction mechanism yielding the desired alkene.

In the cases of t-butanol and t-butyl bromide, the mechanism is the same. The both reactions proceed by E1 mechanism. The leaving groups in each case are water and chloride ion respectively.

6 0

3 years ago

What is the volume of 3.14 M aqueous acid solution that contains 9.25 g of HCLO4

Amanda [17]

The answer is basic

7 0

3 years ago

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent

elixir [45]

Answer : The initial rate for a reaction will be $3.8\times 10^{-4}Ms^{-1}$

Explanation :

Rate law : It 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.

The chemical equation will be:

$A+B+C\rightarrow P$

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:

$6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0$

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:

$6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1$

$k=7.6\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

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

$\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1$

$\text{Rate}=3.8\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.8\times 10^{-3}Ms^{-1}$

6 0

3 years ago

By increasing the length of heat exchanger, the effectiveness and cold stream output temperature of the heat exchanger

Dominik [7]

Answer:

Effectiveness and cold stream output temperature of the heat exchange Increases. So, Answer is b) Increases.

Explanation:

We have a heat exchanger, and it is required to compare the effectiveness and cold stream output if the length is increased.

Heat exchangers are engineering devices used to transfer energy. Thermal energy is transferred from Fluid 1 - Hot fluid (HF) to a Fluid 2 - Cold Fluid (CF). Both fluids 1 and 2 can flow with different values of mass flow rate and different specific heat. When the streams go inside the heat exchanger Temperature of Fluid 1 (HF) will decrease, at the same time Temperature of the Fluid 2 (CF) will increase.

In this case, we need to analyze the behavior taking into account different lengths of heat exchangers. If the length of the heat exchanger increases, it means the transfer area will increases. Heat transfer will increase if the transfer area increases. In this sense, the increasing length is the same than increase heat transfer.

If the heat transfer increases, it means Fluid 1 (HF) will reduce its temperature, and at the same time Fluid 2 (CF) will increase its temperature.

Finally, Answer is b) Effectiveness and cold stream output temperature increases when the length of the heat exchanger is increased.

5 0

3 years ago

1s22s22p63s23p3<br> what element does this represent

s344n2d4d5 [400]

Answer:

Its phosphorus (P)

Explanation:

In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.

3 0

2 years ago