Plz help ASAP Dhdhdg

Give the structure that corresponds to the following molecular formula and 1H NMR spectrum: C4H10O2: δ 1.36 (3H, d, J = 5.5 Hz);

kondor19780726 [428]

Answer:

For the determination of a structure through its NMR it is necessary to know its molecular formula as well as the delta values, its coupling and the shift of each signal.

The separation produced is called coupling constant J and is measured in Hz. If the split is produced by two equal protons (equal J) a triple signal known as triplet is produced and if produced by three equal protons, the signal is quadruple and is known as quadruplet. The magnitude of the coupling is varied, depending on the relative disposition of the coupled protons (elevations that separate them, arrangement, spatial arrangement)

OH CH2 CH2 CH2 CH2 OH

(A) (B) (C)

1,4-butanediol

In the case of the molecule to study the signal at 1.36 shows a doublet, which corresponds to the hydrogen (C), is split in two for each different proton on the same carbon or on neighboring carbons.

At 3.32 ocurrs a singlet, wich belong to hidrogen (B). The last signal is a quintet, at 4.63 belonging to the hydrogen (C)

Explanation

Nuclear magnetic resonance NMR is a physical phenomenon based on the mechanical-quantum properties of atomic nuclei. NMR also refers to the family of scientific methods that explore this phenomenon to study molecules, macromolecules, as well as tissues and whole organisms.

NMR takes advantage of the fact that atomic nuclei resonate at a frequency directly proportional to the force of a magnetic field exerted, in accordance with the Larmor precession frequency equation, to subsequently disturb this alignment with the use of an alternating magnetic field, of orthogonal orientation.

The behavior of the nuclei in the magnetic field can be influenced in multiple ways, to give different types of information, but the basic information obtained is:

Frequency at which each particular nuclei comes out, displacement.
Number of nucleis of each type, integral.
Number and arrangement of nearby nuclei, multiplicity.

6 0

2 years ago

True or false: atoms are electrically neutral; that is they do not have a charge.

AfilCa [17]

Its true, By definition, an atom is electrically neutral it has the same number of protons as it does electrons, plus some number of neutrons depending on the isotope

3 0

2 years ago

2. When you transfer energy into a substance, the molecules' kinetic energy

GenaCL600 [577]

I don’t! Know bit I guess it’s C

8 0

2 years ago

How many molecules are in 1 mole of H2O

Olegator [25]

Answer:

A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance. The mole is the SI unit for the amount of a substance. There are, therefore, 6.02 × 10 23 water molecules in a mole of water molecules. Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen.

8 0

1 year ago

The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

Na: 23
H: 1
C: 12
O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So the limiting reagent is sodium bicarbonate.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

So the grams of excess reactant that do not participate in the reaction are 0333 g.

3 0

2 years ago