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ddd [48]
1 year ago
13

Consider the molecule BF3. (a) What is the electron config- uration of an isolated B atom? (b) What is the electron con- figurat

ion of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B–F bonds in BF3? (d) What valence orbitals, if any, remain unhybrid- ized on the B atom in BF3?
Chemistry
1 answer:
anzhelika [568]1 year ago
8 0

For the given molecule, we are asked to give-

  1. The electron configuration of an isolated B atom
  2. The electron configuration of an isolated F atom
  3. Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
  4. valence orbitals, if any, remain unhybridized on the B atom.
  • The electron configuration of an isolated B atom:

as atomic number of B is 5

electronic configuration will be [He] 2s² 2p¹

  • The electron configuration of an isolated F atom:

as atomic number of F is 9

electronic configuration will be  [He] 2s² 2p5

  • Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.

as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.

  • valence orbitals, if any, remain unhybridized on the B atom will be 1

To know more about hybrisisation:

brainly.com/question/23038117

#SPJ4

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B. if 6.73 g of na2co3 is dissolved in enough water to make 250. ml of solution, what is the molar concentration of sodium carbo
BARSIC [14]

Answer:- [Na_2CO_3]=0.254M , [Na^+]=0.508 M , [CO_3^2^-]=0.254M

Solution:- We are asked to calculate the molarity of sodium carbonate solution as well as the sodium and carbonate ions.

Molarity is moles of solute per liter of solution. We have been given with 6.73 grams of sodium carbonate and the volume of solution is 250.mL.  Grams are converted to moles and mL are converted to L and finally the moles are divided by liters to get the molarity of sodium carbonate.

Molar mass of sodium carbonate is 105.99 gram per mol. The calculations for the molarity of sodium carbonate are shown below:

\frac{6.73gNa_2CO_3}{250.mL}(\frac{1mol}{105.99g})(\frac{1000mL}{1L})

= 0.254MNa_2CO_3

So, molarity of sodium carbonate solution is 0.254 M.

sodium carbonate dissociate to give the ions as:

Na_2CO_3(aq)\rightarrow 2Na^+(aq)+CO_3^2^-

There is 1:2 mol ratio between sodium carbonate and sodium ion. So, the molarity of sodium ion will be two times of sodium carbonate molarity.

[Na^+]=2(0.254M) = 0.508 M

There is 1:1 mol ratio between sodium carbonate and carbonate ion. So, the molarity of carbonate ion will be equal to the molarity of sodium carbonate.

[CO_3^2^-]=0.254M


5 0
3 years ago
How does an enzyme’s active site relate to its substrate?
Fofino [41]

Answer:

The enzyme's active site must be the mold shape of the substrate.

Explanation:

An enzyme and a substrate bind together in order to work. If the shapes of the 2 are different, they will not be able to bond together. If the shapes of the 2 fit glove-in-hand, then they will be able to bond together.

7 0
3 years ago
Read 2 more answers
AWNSER FAST PLEASE
Ann [662]

Answer: The coefficients are 2, 2 and 1.

Explanation: According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants.

The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical equation for the given reaction is:

2H2o➡️2h2+o2

3 0
2 years ago
Identify the major attractive force in HF molecules.A) London Dispersion Forces (LDFs)
Llana [10]

Answer:

C) hydrogen bonding

Explanation:

All atoms and molecules have London Dispersion Forces between them, but they are usually overshadowed but the much stronger forces. In this scenario the major attractive force in HF molecules are hydrogen bonds. Hydrogen bonds are electrostatic forces of attraction found when Hydrogen is bonded to a more electronegative atom such as Oxygen, Chlorine and Fluorine.

6 0
3 years ago
Dinitrogen pentoxide is used in the preparation of explosives. If 7.93 mol of
Tpy6a [65]

The volume of O₂ produced: 84.6 L

<h3>Further explanation</h3>

Given

7.93 mol of  dinitrogen pentoxide

T = 48 + 273 = 321 K

P = 125 kPa = 1,23365 atm

Required

Volume of O₂

Solution

Decomposition reaction of dinitrogen pentoxide

2N₂O₅(g)→4NO₂(g)+O₂ (g)

From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :

= 0.5 x mol N₂O₅

= 0.5 x 7.93

= 3.965 moles

The volume of O₂ :

\tt V=\dfrac{nRT}{P}\\\\V=\dfrac{3.965\times 0.082\times 321}{1.23365}\\\\V=84.6~L

5 0
3 years ago
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