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2 years ago

What is a common reason for observing an increased chemical shift of a c-h proton

1 answer:

3 0

Deshielding due to an electronegative element close by is the common reason for observing increased chemical shift of a c-h proton

<h3>What is a chemical shift? </h3>

The resonance frequency of a proton in relation to a standard compound is represented by chemical shift. Chemical shift, which is measured in ppm and is represented by the sign (δ), (parts per million).The chemical shift in a proton NMR spectrum provides details about the targeted proton's chemical surroundings. The structure of the investigated substance, especially electronegative components or effects, has a significant impact on the chemical shift value. Electronegative elements' ability to remove electron density from the proton, which raises the chemical shift value, is one explanation for this. The proton is more exposed to the magnetic field that is being applied externally as a result of this process, which is referred to as de-shielding.

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N2(g) + O2(g) → NO(g) ΔHrxn = +90.3 kJ NO(g) + Cl2(g) → NOCl(g) ΔHrxn = –38.6 kJ What is the value of ΔHrxn for the decompositio

Ivanshal [37]

Answer : The value of $\Delta H_{rxn}$ is, -103.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

$2NOCl(g)\rightarrow N_2(g)+O_2(g)+Cl_2(g)$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\rightarrow NO(g)$ $\Delta H_1=+90.3kJ$

(2) $NO(g)+\frac{1}{2}Cl_2(g)\rightarrow NOCl(g)$ $\Delta H_2=-38.6kJ$

Now reversing and multiplying by 2 of reaction 1 and 2 then adding all the equations, we get :

(1) $2NO(g)\rightarrow N_2(g)+O_2(g)$ $\Delta H_1=2\times (-90.3kJ)=-180.6kJ$

(2) $2NOCl(g)\rightarrow 2NO(g)+Cl_2(g)$ $\Delta H_2=2\times (+38.6kJ)=+77.2kJ$

The expression for $\Delta H_{rxn}$ will be,

$\Delta H=\Delta H_1+\Delta H_2$

$\Delta H=(-180.6)+(+77.2)$

$\Delta H=-103.4kJ$

Therefore, the value of $\Delta H_{rxn}$ is, -103.4 kJ

4 0

3 years ago

Read 2 more answers

Put 5.1 x 10^-7 into standard notation.

Luden [163]

Answer:

My answer to the question is 0.00000051

6 0

3 years ago

Read 2 more answers

The table below provides some information about an unidentified element. Based on this information, the unidentified element is

Leona [35]

In the physical properties, it is mentioned that the element has 4 valence electrons. The elements in the periodic table are arranged such that, the number of valence electrons present in the neutral atoms belonging to a particular group is equal to the group number.

Thus, the unidentified element can be best classified as a nonmetal in period 4

Ans C)

5 0

3 years ago

What is the half life of an atom that goes from 16000 g to 125 g in 163.24 days.

Alenkinab [10]

Answer:

$23.32\ \text{days}$

Explanation:

N = Final mass of atom = 125 g

$N_0$ = Initial mass of atom = 16000 g

t = Time taken = 163.24 days

$t_{1/2}$ = Half life

We have the relation

$N=N_0\dfrac{1}{2}^{\dfrac{t}{t_{1/2}}}\\\Rightarrow 125=16000\times \dfrac{1}{2}^{\dfrac{163.24}{t_{1/2}}}\\\Rightarrow \dfrac{125}{16000}=\dfrac{1}{2}^{\dfrac{163.24}{t_{1/2}}}\\\Rightarrow \ln0.0078125=\dfrac{163.24}{t_{1/2}}\ln0.5\\\Rightarrow t_{1/2}=\dfrac{163.24\times\ln0.5}{\ln0.0078125}\\\Rightarrow t_{1/2}=23.32\ \text{days}$

The half life of the atom is $23.32\ \text{days}$ .

3 0

3 years ago

Object a has a mass of 12 G and a volume of 8 cm3 object be has a mass of 20 G and a volume of 8 cm3 which object has a greater

solniwko [45]

Answer:

Option C = object B by 1 gram per cubic cm.

Explanation:

Given data:

Mass of object A = 12 g

Volume of object A = 8 cm³

Mass of object B = 20 g

Volume of object B = 8 cm³

Densities = ?

Solution:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Density of object A:

d = m/v

d = 12 g/ 8 cm³

d = 1.5 g/cm³

Density of object B:

d = m/v

d = 20 g/ 8 cm³

d = 2.5 g/cm³

object b has high density.

5 0

3 years ago