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AnnZ [28]
1 year ago
9

Give a possible explanation for the relative amounts of the isometric methyl nitrobenzoates formed in the nitration reaction. Co

nsider mechanistic effects of the substituents.
Chemistry
1 answer:
vovikov84 [41]1 year ago
4 0

Answer:

The electrophilic aromatic substitution reaction nitration is used to nitrate methyl benzoate and acetanilide with a nitronium ion. Crystallization was used to purify the product. The melting point was used to determine its purity and the regiochemistry of the products.

Explanation:

Methyl m-Nitrobenzoate is formed in this

reaction rather that ortho/para isomers

because of the ester group of your starting

product of methylbenzoate. The functional

group of ester is a electron withdrawing group

causing nitrobenzene (N02) to become in the

meta position. Thus N02 is a deactivating

group causing itself to be a meta director.

Basically you must look at the substituents

that are attached to your starting benzene ring

in order to figure out whether your reaction

with be ortho/para directors or meta

directors. If the substituents are electron

withdrawing groups then you will be left with

meta as your product but if your substituents

are electron donating groups then your

product will be ortho/para.

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How many moles of ions are in 285 ml of 0.0150 m mgcl2?
Solnce55 [7]
MgCl₂)= Mg²⁺ + 2Cl⁻
V(MgCl₂)=285cm³=0,285dm³
c(MgCl₂)=0,015 mol/dm³
n(MgCl₂)=c·V= 0,015 mol/dm³ · 0,285dm³ = 0,0042 mol
n(Mg²⁺)=n(MgCl₂)=0,0042 mol
n(Cl⁻)=2n(MgCl₂)=0,0084 mol
7 0
2 years ago
Read 2 more answers
It is 762 miles from here to Chicago. An obese physics teacher jogs at a rate of 5.0 miles every 20.0 minutes. How long would it
Lina20 [59]

Answer:

3,048 minutes

Explanation:

762 divided by 5

that number times 20

5 0
1 year ago
A galvanic cell employs the reaction Mg2+(aq) + Cu(s) → Mg(s) + Cu2+(aq) and NaNO3 is the salt used in the salt bridge. During t
sveta [45]

Answer:

b. Na+ leaves the salt bridge and enters enters the cathode

Explanation:

A galvanic cell or electrochemical cell depicts an oxidation -reduction half reactions (redox) reaction. it consists of two half cells ; one for the reduction reaction which involves the gain of electrons and the other for the oxidation reaction which involves the loss of electrons.  One half cell contains the anode and oxidation occurs at the anode while the other half cell contains the cathode and reduction occurs at the cathode. The anode is usually connected to the cathode, a salt bridge is added to complete the circuit and allow current to flow. The salt bridge serves as a counter ions, they do not interfere with the electrochemical reaction but provides a passage for the migration of ions thereby preventing the cells from reaching equilibrium too quickly and thus the electrons in the salt are able to move along with any electrons.

In this galvanic cell, Cu at the anode losses two electrons to become Cu2+, and the electrons moves from the anode to the cathode where Mg2+ gain these two electrons to become negatively charged. Positively charged ions in the salt brigde Na+ will move to the cathode to pick negatively charged ions from the cathode solution. this helps to remove the strong negative charge from the cathode and allows the electrons to continue to move to the cathode.

8 0
2 years ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
2 years ago
Consider the balanced equation of K I KI reacting with P b ( N O 3 ) 2 Pb(NOX3)X2 to form a precipitate. 2 K I ( a q ) + P b ( N
professor190 [17]

Answer: 50.7 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of KI

\text{Number of moles}=molarity\times {\text {Volume in L]}=0.417M\times 0.528L=0.220moles

The balanced chemical equation is:

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)

KI is the limiting reagent as it limits the formation of product and Pb(NO_3)_2 is in excess.

According to stoichiometry :

2 moles of KI give =  1 mole of PbI_2

Thus 0.220 moles of KI give=\frac{1}{2}\times 0.220=0.110moles  of PbI_2

Mass of PbI_2=moles\times {\text {Molar mass}}=0.110moles\times 461.01g/mol=50.7g

Thus 50.7 g of PbI_2 will be formed.

6 0
2 years ago
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