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Lesechka [4]
2 years ago
10

What type of intermolecular forces are expected between 4-methylcyclohexanone molecules?

Chemistry
1 answer:
Keith_Richards [23]2 years ago
7 0

The answer is dipole-dipole and dipole-induced dipole forces.

The dipole-induced dipole is a kind of interaction induced by a polar molecule by disturbing the arrangement of electrons.

  • In methyl cyclohexanone molecules, there is a permanent dipole moment due to dipole moment vectors not canceling.
  • There is induction of dipole by disturbing the electronic arrangement.
  • A permanent dipole moment is created in this interaction.
  • Dipole-dipole interactions are defined as the forces that is formed from the close linkage of permanent or induced dipoles.
  • These forces are called Van der Waal forces.
  • Proteins contain a large number of these interactions, which vary considerably in strength.

To learn more about dipole-dipole interactions visit:

brainly.com/question/14173758

#SPJ4

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A student is participating in the school science fair. She wants to investigate household cleaners.
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Answer:

A

Explanation:

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3 years ago
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Calculate the number of moles in 2.67 g of carbon trihydride
Scorpion4ik [409]

Answer:0.178 moles

Explanation: carbon trihydride seems to be an unusual name for the methyl group CH3–

ionic wt 15

moles = 2.67/15 = 0.178

8 0
2 years ago
The balanced equation for the reaction of ammonia and oxygen is the following. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) The stand
ale4655 [162]

Answer:

ΔS° = 180.5 J/mol.K

Explanation:

Let's consider the following reaction.

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.

ΔS° = ∑np × S°p - ∑nr × S°r

where,

ni are the moles of reactants and products

S°i are the standard molar entropies of reactants and products

ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))

ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol

ΔS° = 180.5 J/K

This is the change in the entropy per mole of reaction.

7 0
3 years ago
How many atoms of lodine(I) are in 0.156 grams of I?
CaHeK987 [17]

Answer: iodine 131 iodine 132 isotopes

Explanation:

4 0
3 years ago
What is the longest wavelength in the Balmer series? (Hint: the Rydberg constant for Hydrogen is 1.096776×107 1/m, and the Balme
boyakko [2]

<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 1.096776\times 10^7m^{-1}

n_f = Higher energy level = n_i+1=(2+1)=3

n_i= Lower energy level = 2    (Balmer series)

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m

Converting this into nanometers, we use the conversion factor:

1m=10^9nm

So, 6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm

Hence, the longest wavelength of light is 656.5 nm

4 0
3 years ago
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