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Olegator [25]
3 years ago
10

The following reaction has been reported in the chemical literature and gives a single organic product in high yield. Write the

structure of the product for the reaction of styrene with bromine in aqueous solution (ignore stereochemistry).
(a) 2-phenylpropene + HCl
(b) Styrene + Beomin in aqueous solution

Chemistry
1 answer:
Alla [95]3 years ago
8 0

Answer:

Explanation:

The principle applied is the Markovnikoff's rule which states that when hydrogen chloride adds to a double bond, the hydrogen atoms join to the carbon that already has the most hydrogen atoms bonded to it. The rule wa postulated by a russian chemist known as Vladimir Markovnikoff.

In the markovnikoff's rule, there are sveral conditions that must be met, one of them is that no free radicals must be involved.

The reaction and the structure of the product is as shown in the attachment.

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What is the chemical name of the compound Na2CO3? Use the list of polyatomic ions and the periodic table to help you answer.
m_a_m_a [10]

Answer:

I believe the answer is B. Sodium carbonate

5 0
2 years ago
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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
8 0
3 years ago
Read 2 more answers
Photosynthesis is a cell process used by various organisms. Which of the following best
Dmitry [639]

Answer:

Carbon dioxide and water are turned into sugar and oxygen

3 0
2 years ago
In an acid-base titration,
Lemur [1.5K]

Answer:

b

Explanation:

An acid-base titration is an experimental procedure used to determined the unknown concentration of an acid or base by precisely neutralizing it with an acid or base of known concentration. ... It is filled with a solution of strong acid (or base) of known concentration.

7 0
3 years ago
Joan’s initial nickel (II) chloride sample was green and weighed 4.3872 g. After the dehydration reaction and removal of excess
ANEK [815]

Answer:

a) yes, it was an hydrate

b) the number of waters of hydration, x = 6

Explanation:

a) yes it was an hydrate because the mass decreased after the process of dehydration which means removal of water thus some water molecules were present in the sample.

b) NiCl2. xH2O

mass if dehydrated NiCl2 = 2.3921 grams

mass of water in the hydrated sample = mass of hydrated - mass of dehydrated = 4.3872 - 2.3921 = 1.9951 g which represent the mass of water that was present in the hydrated sample.

NiCl2.xH2O

mole of dehydrated NiCl2 = m/Mm = 2.3921/129.5994 = 0.01846 mole

mole of water = m/Mm = 1.9951/18.02 = 0.11072 mole

Divide both by the smallest number of mole (which is for NiCl2) to find the coefficient of each

for NiCl2 = 0.01846/0.01846 = 1

for H2O = 0.11072/0.01846 = 5.9976 = 6

thus the hydrated sample was NiCl2. 6H2O

4 0
3 years ago
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