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ser-zykov [4K]
3 years ago
6

Consider a disubstituted aromatic compound. The parent name is benzene and there is a chloro and bromo substituent. Disubstitute

d benzenes can be described using the terms ortho, meta and para, depending on their relative distance from each other. The terms are often just abbreviated as o, m and p. In addition, the IUPAC name can use locant numbers instead of the descriptor. Br
Select the correct names for the structure.
1. 1-bromo-3-chlorobenzene
2. 3-bromo-1-chlorobenzene
3. meta-bromochlorobenzene
4. o-bromochlorobenzene
5. ortho-bromochlorobenzene
6. m-bromochlorobenzene
Chemistry
1 answer:
Marina86 [1]3 years ago
8 0

Answer:The correct names for the structure are:

--> 1. 1-bromo-3-chlorobenzene.

--> 3. meta-bromochlorobenzene.

--> 6. m-bromochlorobenzene.

Explanation:

Benzene is the simplest member of the aromatic hydrocarbons. It has a ring structure consisting of six carbon and six hydrogen atoms. This equally means that a benzene can have up to six substituents. One of the chemical properties is that benzene and other members of its series undergo substitution reaction whereby one or more of its six hydrogen atoms is replaced by monoatomic reagents.

Disubstituted benzene consists of two substituents which are described based on either numerical locants or specific words for the three possible forms.

The numerical locant method are used the same naming substitutes of other hydrocarbons. From the question, the numerical locant method was derived through using the following steps:

--> the functional group is benzene

--> there are two substituents which includes bromine( written as bromo) and chlorine ( written as chloro)

--> while placing the number, it's done alphabetically ('1-bromo' comes before '3-chloro') in a clockwise manner. This is to give chorine the lowest locant number.

The second naming method for a disubstituted benzene is the the ortho-, meta-, para- (or their singel letter equivalent) nomenclature method. This is only used for benzene structures.

--> ortho or O : this is used when the substituents are close to each other in the benzene ring.

--> meta or (m) : This is used when the substituents are separated by one carbon in the benzene ring.

--> para or (p): This is used when the substituents are across each other in the benzene ring

From the question, the bromine substituent is separated from the chlorine by one carbon atom, therefore it's meta-bromochlorobenzene or m-bromochlorobenzene.

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What is the molarity of a nitric acid solution if 43.13 mL 0.1000 M KOH solution is needed to neutralize 30.00 mL of the acid so
Leno4ka [110]

Answer:

0.144M

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

HNO3 + KOH —> KNO3 + H20

From the equation,

nA = 1

nB = 1

From the question given, we obtained the following:

Ma =?

Va = 30.00mL

Mb = 0.1000M

Vb = 43.13 mL

MaVa / MbVb = nA/nB

Ma x 30 / 0.1 x 43.13 = 1

Cross multiply to express in linear form

Ma x 30 = 0.1 x 43.13

Divide both side by 30

Ma = (0.1 x 43.13) /30 = 0.144M

The molarity of the nitric acid is 0.144M

8 0
3 years ago
Write balanced equations for each of the following by inserting the correct coefficients in the blanks:
Aleksandr-060686 [28]

Balanced equation :

Cu(NO₃)₂(aq) + 2KOH(aq) → Cu(OH)₂(s)  + 2KNO₃(aq)

Balancing a chemical equation :

A chemical equation shows us the substances involved in a chemical reaction - the substances that react (reactants) and the substances that are produced (products). In general, a chemical equation looks like this:

                                 Reactant →Product

According to the law of conservation of mass, when a chemical reaction occurs, the mass of the products should be equal to the mass of the reactants. Therefore, the amount of the atoms in each element does not change in the chemical reaction. As a result, the chemical equation that shows the chemical reaction needs to be balanced. A balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.

Learn more about balanced equation :

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3 0
2 years ago
What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
dsp73

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

6 0
3 years ago
Will ya'll help me, please
JulsSmile [24]

Answer:

Either A or D

But i think D

hope this helps =3

sorry if im wrong

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3 years ago
1.1 L of nitrogen dioxide were produced in the reaction seen below. How
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22.4 since it's in STp
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