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spayn [35]
2 years ago
10

Methylene chloride (CH2Cl2) has fewer chlorine atoms than chloroform (CHCl3). Nevertheless, methylene chloride has a larger mole

cular dipole moment than chloroform. Explain.
Chemistry
1 answer:
icang [17]2 years ago
3 0

Answer:

Explanation has been given below.

Explanation:

  • Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
  • Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
  • First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
  • Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
  • Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
  • Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
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Be sure to answer all parts. ΔH o f of hydrogen chloride [HCl(g)] is −92.3 kJ/mol. Given the following data, determine the ident
Akimi4 [234]

Answer:

NH₄Cl(s) → NH₃(g) + HCl(g)

ΔH°rxn 74,89 kJ/mol

Explanation:

The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):

<em>(1) </em>¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol

It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).

For the reactions:

<em>(2) </em>N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol

<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

The sum of -(2) + (3) gives:

<em>-(2) </em>2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol

<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)

ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol

Now, the sum of -(2) + (3) + 2×(1)

<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol

<em>2×(1)  </em>H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol

<em>-(2) + (3) + 2×(1) </em>2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)

ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol

The reaction of:

<em>NH₄Cl(s) → NH₃(g) + HCl(g)</em>

<em>Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol</em>

I hope it helps!

8 0
2 years ago
A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi
viktelen [127]

Actual question from source:-

A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette.  A blank had an absorbance of 0.029.  The absorbance of an unknown solution of compound A was 0.375.  Find the concentration of A in the unknown.

Answer:

Molar absorptivity of compound A = 1502.53\ {Ms}^{-1}

Explanation:

According to the Lambert's Beer law:-

A=\epsilon l c

Where, A is the absorbance

 l is the path length  

\epsilon is the molar absorptivity

c is the concentration.  

Given that:-

c = 3.96\times 10^{-4}\ M

Path length = 1.000 cm

Absorbance observed = 0.624

Absorbance blank = 0.029

A = 0.624 - 0.029 = 0.595

So, applying the values in the Lambert Beer's law as shown below:-

0.595=\epsilon\times 1.000\ cm\times 3.96\times 10^{-4}\ M

\epsilon=\frac{0.595}{3.96\times 10^{-4}}\ {Ms}^{-1}=1502.53\ {Ms}^{-1}

<u>Molar absorptivity of compound A = 1502.53\ {Ms}^{-1}</u>

4 0
2 years ago
Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d
PilotLPTM [1.2K]

Answer:

k = 23045 N/m

Explanation:

To find the spring constant, you take into account the maximum elastic potential energy that the spring can support. The kinetic energy of the car must be, at least, equal to elastic potential energy of the spring when it is compressed to its limit. Then, you have:

K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2    (1)

M: mass of the car = 1050 kg

k: spring constant = ?

v: velocity of the car = 8 km/h

x: maximum compression of the spring = 1.5 cm = 0.015m

You solve the equation (1) for k. But first you convert the velocity v to m/s:

v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}

k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}

The spring constant is 23045 N/m

3 0
2 years ago
The ph of a solution that contains 0.818 m acetic acid (ka = 1.76 x 10-5) and 0.172 m sodium acetate is __________.
crimeas [40]
PH is calculated using <span>Handerson- Hasselbalch equation,

                                      pH  =  pKa  +  log [conjugate base] / [acid]

Conjugate Base  =  Acetate (CH</span>₃COO⁻)
Acid                    =  Acetic acid (CH₃COOH)
So,
                                      pH  =  pKa  +  log [acetate] / [acetic acid]

We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,

                                     pKa  =  -log Ka
Putting value of Ka,
                                     pKa  =  -log 1.76 × 10⁻⁵

                                     pKa  =  4.75
Now,
Putting all values in eq. 1,
                     
                                     pH  =  4.75 + log [0.172] / [0.818]

                                     pH  =  4.072
4 0
2 years ago
What can properties of substances tell us about the electrical forces within those<br> substances?
nydimaria [60]

Answer:

HS-PS1-3

Explanation:

Plan and carry out an investigation to gather evidence to compare the structure of substances at the bulk scale to infer the strength of electrical forces between particles. Examples of bulk properties of substances could include the melting point and boiling point, vapor pressure, and surface tension.

5 0
2 years ago
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