Which element is likely more reactive, and why

If a solid line represents a covalent bond and a dotted line represents intermolecular attraction, which of these choices shows

blsea [12.9K]

Generally, a hydrogen bond can be characterized as a proton shared by two lone electron pairs. It occurs when a hydrogen (H) atom, covalently bound to a highly electronegative atom such as nitrogen (N), oxygen (O), or fluorine (F), experiences the electrostatic field of another highly electronegative atom nearby.

Among the choices in the bond (-N...H-O) one side of the Hydrogen is bonded to a highly electronegative atom with a lone pair (-N) and the other side is directly bonded with a highly electronegative atom (O-).

So -N...H-O- shows a hydrogen bond.

4 0

3 years ago

This problem has been solved!

Vanyuwa [196]

Answer:

NaCl and water: Ion - Dipolo forces

NaCl and Hexane: Ion-ion force between Na+ and Cl− ions and London dispersion force between two hexane molecules

Explanation:

NaCl and water:

The ion-dipole force is established between an ion and a polar molecule. Polar molecules are dipoles, they have a positive end and a negative end.

H2O has an important charge separation in its atoms (the H has a positive partial charge and the O has a negative partial charge) and this causes permanent electrical dipoles in the water molecules.

Sodium chloride is an ionic compound formed of positive and negative charge ions, Na + and Cl-. Depending on their charge, these ions will be attracted to opposite charges in the water molecules (H attracts chloride ions and O attracts sodium ions), causing the salt to dissolve in water.

NaCl and Hexane:

The dispersion forces of London occur between apolar molecules, and they occur because when two molecules approach a distortion of the electronic clouds of both originates, generating in them, transient induced dipoles, due to the movement of the electrons, so it allows interact with each other.

Hexane is a non-polar molecule, which are those that have no charge separation within the molecules. Then there is London dispersion force between two hexane molecules. 

On the other hand, the ion-ion force is produced between ions of the same or different charge, where ions with charges of opposite sign attract each other and ions with charges of the same sign repel each other. This is the force that occurs between the NaCl ions.

5 0

3 years ago

A main group metal was studied and found to exhibit the following properties:

Alecsey [184]

Answer: we learned this not to long ago i think its a,c

Explanation:

6 0

3 years ago

What volume of 0.194 MNa3PO4 solution is necessary to completely react with 85.5 mL of 0.109 MCuCl2 ?

ira [324]

Answer:

35.9 ml

Explanation:

Start with the balanced equation:

3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)

This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-

∴ 1 mole CuCl2 will react with 2/3 moles Na3PO4

We know that concentration = moles/volume i.e:

c= n/v

∴n=c×v

∴nCuCl2=0.107×91.01000=9.737×10−3

I divided by 1000 to convert ml to L

∴nNa3PO4=9.737×10−3×23=6.491×10−3

v=nc=6.491×10−30.181=35.86×10−3L

∴v=35.86ml

4 0

1 year ago

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

8090 [49]

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is $P_A$ and vapour pressure of component B is $P_B$ .

$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

$P_A >P^0_A\cdot x_A$ , $P_B>P^0_B\cdot x_B$

Therefore, $P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B$

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

6 0

3 years ago