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Montano1993 [528]
3 years ago
6

When two atoms form a covalent bond, they share electrons from all of their orbitals. All of their orbitals, in turn, combine to

form a single molecular orbital. True or false?
Chemistry
1 answer:
erma4kov [3.2K]3 years ago
8 0

Answer:

It's false.

Explanation:

Molecular orbital theory states that the number of molecular orbitals is equal to the number of atomic orbitals that overlap. The lowest energy molecular orbital is formed when two atomic orbitals that are in phase overlap, forming a bonding molecular orbital. However, another molecular orbital is also formed, called an anti-binding orbital.

So if an "n" quantity of atomic orbitals is combined, an "n" quantity of molecular orbitals is formed.

Have a nice day!

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Box 1
Alex73 [517]

10g

Explanation:

Box 1, Mass of A = 10g

Box 2, Mass of B = 5g

Box 3, = 1A + 1B

Unknown:

Mass of B that would combine with mass of 20g of A

Solution:

   Mass ratio of A to B:

   \frac{mass of A}{mass of B} = mass ratio

           \frac{10}{5} = mass ratio

        The mass ratio of A to B = 2: 1

Now, number of B that will combine with 20g of A;

       

           \frac{mass of A}{mass of B} = mass ratio

               \frac{20}{mass of B} = \frac{2}{1}

                     Mass of B = 10g

10g of B would combine with 20g of A

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7 0
3 years ago
Read 2 more answers
What do dipole-dipole forces do?
g100num [7]
An ion-dipole force is a type of intermolecular force in which forces of attraction or repulsion occur between neighboring ions, molecules or atoms.
3 0
3 years ago
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A container of gas is initially at 0.25 atm and 0 ˚C. What will the pressure be at 125 ˚C?
Pani-rosa [81]

Answer:

0.37atm

Explanation:

Given parameters:

Initial pressure  = 0.25atm

Initial temperature  = 0°C  = 273K

Final temperature  = 125°C  = 125 + 273  = 398K

Unknown:

Final pressure  = ?

Solution:

To solve this problem, we use a derivative of the combined gas law;

           \frac{P1}{T1}  = \frac{P2}{T2}

  P and T are pressure and temperature

  1 and 2 are initial and final values

        \frac{0.25}{273}   = \frac{P2}{398}  

         P2  = 0.37atm

3 0
2 years ago
Consider the reaction: 2A(g)+B(g)→3C(g). When A is changing at a rate of -0.110M⋅s−1, How fast is C increasing?
mr_godi [17]
From the reaction above, the rate is given by the following formula:
r = -(1/2) dA / dt = - dB / dt = (1/3) dC/ dt
Note that A and B charge is negative due to they decrease with time
given dA / dt = -0.110 M/s
hence dB / dt = -0.110 / 2 = -0.055 M/s
dC / dt = (-3/2) (-0.110) = 0.165 M/s 
5 0
3 years ago
1.33 dm3 of water at 70°C are saturated by 2.25
astraxan [27]

Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the

amount amount of solute that will be deposited is 1,927.413 grams.

<h3>How can the amount of solute deposited be found?</h3>

The volume of water 1.33 dm³ of water 70 °C.

The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C  = 2.25 moles

At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles

Therefore;

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;

1.33 dm³ contains 2.25 moles.

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;

1.33 dm³ contains 0.53 moles

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles

The number of moles that precipitate out = The amount of solute deposited

Which gives;

Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles

The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g

The molar mass of Pb(NO₃)₂ = 331 g/mol

The amount of solute deposited = Number of moles × Molar mass

Which gives;

The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>

Learn more about saturated solutions here:

brainly.com/question/2624685

5 0
2 years ago
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