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Liula [17]
3 years ago
8

Compare the strength of intermolecular forces in no2 to the strength of intermolecular forces in n2

Chemistry
2 answers:
matrenka [14]3 years ago
7 0
N2  is non-polar molecules,
NO2 is polar molecules that have slightly positive charge on N and slightly negative on O, so <span>intermolecular forces are stronger  in NO2</span>
Crazy boy [7]3 years ago
4 0

Answer: The intermolecular forces "Dipole-dipole forces" in NO2 is more stronger than "London-dispersion forces" in N2.

Explanation:

  • Dipole-dipole forces is considered as an attractive force (5.0 kJ to 20.0 kJ) of the positive end of one polar molecule with the negative end of another polar molecule and it is found in NO2.
  • The London dispersion force is the weakest intermolecular forces. It results when the electrons in two neighboring atoms positions that make the atoms form temporary dipoles and it is found in N2. This force is sometimes called an induced dipole attraction.
  • So, The dipole-dipole forces are more stronger than that of London-dispersion forces.
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Solve for ‘x’: 12(x-7)+3(2x+2)=50x-2
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Answer:

Explanation:

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12(x-7) = 12x -84

3(2x+2) = 6x+6

Total 18x-78

18x-78 = 50x-2

-78= 32x-2

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19/8 =x

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5 0
4 years ago
How much energy is required when a 142.1 gram sample of aluminum goes from 25.5°C to 46°C? (the specific heat of aluminum is 0.9
vfiekz [6]

Answer:

2621.75 j heat is required to increase the temperature 25.5°C to 46°C.

Explanation:

Given data:

Mass of sample = 142.1 g

Initial temperature = 25.5°C

Final temperature = 46°C

Specific heat capacity of Al = 0.90 J/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 46°C - 25.5°C

ΔT = 20.5°C

Q = 142.1 × 0.90 J/g.°C × 20.5°C

Q = 2621.75 j

Thus,  2621.75 j heat is required to increase the temperature 25.5°C to 46°C.

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A solution is made containing 4.6 g of sodium chloride per 250g of water.
lozanna [386]

Answer:

\large \boxed{1.81 \, \%}

Explanation:

\text{Mass percent} = \dfrac{\text{mass of solute} }{\text{mass of solution}} \times 100 \, \%

Data:

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Mass of water = 250    g

Calculations:

Mass of solution = mass of NaCl + mass of water = 4.6 g + 250 g = 254.6 g.

\text{\% m/m} = \dfrac{\text{4.6 g} }{\text{254.6 g}} \times 100 \, \% = \mathbf{1.81 \, \%}\\\\\text{The percent by mass of NaCl is $\large \boxed{\mathbf{1.81 \, \%}}$}

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Answer:

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